description stringlengths 171 4k | code stringlengths 94 3.98k | normalized_code stringlengths 57 4.99k |
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You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
lens, lent = len(stamp), len(target)
out = []
q = collections.deque()
done = set()
v = [set() for _ in range(lent - lens + 1)]
t = [set() for _ in range(lent - lens + 1)]
for i in range(len(target) - len(stamp) + 1):
for j in range(len(stamp)):
if stamp[j] == target[i + j]:
v[i].add(i + j)
else:
t[i].add(i + j)
if len(t[i]) == 0:
out.append(i)
for j in range(len(stamp)):
done.add(i + j)
q.append(i)
while q:
offset = q.popleft()
for of in range(
max(0, offset - lens + 1), min(lent - lens + 1, offset + lens)
):
t[of] -= done
if len(t[of]) == 0:
if len(v[of] - done) > 0:
q.append(of)
done |= v[of]
out.append(of)
if len(done) == lent:
return out[::-1]
return [] | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR RETURN VAR NUMBER RETURN LIST VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
def helper(idx):
num = len(stamp)
for i in range(len(stamp)):
if target[i + idx] == "*":
num -= 1
elif target[i + idx] != stamp[i]:
return 0
target[idx : idx + len(stamp)] = ["*"] * len(stamp)
return num
seen = [0] * len(target)
total = 0
res = []
target = list(target)
while total < len(target):
found = False
for i in range(len(target) - len(stamp) + 1):
if seen[i] == 1:
continue
num = helper(i)
if num == 0:
continue
seen[i] = 1
total += num
found = True
res.append(i)
if not found:
return []
return res[::-1] | CLASS_DEF FUNC_DEF VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR VAR STRING VAR NUMBER IF VAR BIN_OP VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR VAR BIN_OP VAR FUNC_CALL VAR VAR BIN_OP LIST STRING FUNC_CALL VAR VAR RETURN VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR VAR NUMBER VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR RETURN LIST RETURN VAR NUMBER VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
stamp = list(stamp)
target = list(target)
res = []
def check(i):
changed = False
for j in range(len(stamp)):
if target[i + j] == "?":
continue
if target[i + j] != stamp[j]:
return False
changed = True
return changed
seen = [0] * len(target)
ans = True
while ans:
ans = False
for i in range(len(target) - len(stamp) + 1):
changed = check(i)
ans = ans or changed
if changed:
res.append(i)
target[i : i + len(stamp)] = ["?"] * len(stamp)
seen[i : i + len(stamp)] = [1] * len(stamp)
return res[::-1] if sum(seen) == len(target) else [] | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR VAR STRING IF VAR BIN_OP VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR NUMBER RETURN VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR FUNC_CALL VAR VAR BIN_OP LIST STRING FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR FUNC_CALL VAR VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER LIST VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, s: str, t: str) -> List[int]:
if s[0] != t[0] or s[-1] != t[-1]:
return []
n, m = len(s), len(t)
path = [0] * m
pos = collections.defaultdict(set)
for i, c in enumerate(s):
pos[c].add(i)
def dfs(i, index):
path[i] = index
if i == m - 1:
return index == n - 1
nxt_index = set()
if index == n - 1:
nxt_index |= pos[t[i + 1]]
elif s[index + 1] == t[i + 1]:
nxt_index.add(index + 1)
if s[0] == t[i + 1]:
nxt_index.add(0)
return any(dfs(i + 1, j) for j in nxt_index)
def path2res(path):
down, up = [], []
for i in range(len(path)):
if path[i] == 0:
up.append(i)
elif i and path[i] - 1 != path[i - 1]:
down.append(i - path[i])
return down[::-1] + up
if not dfs(0, 0):
return []
print(path)
return path2res(path) | CLASS_DEF FUNC_DEF VAR VAR IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER RETURN LIST ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR FUNC_DEF ASSIGN VAR VAR VAR IF VAR BIN_OP VAR NUMBER RETURN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR FUNC_DEF ASSIGN VAR VAR LIST LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR RETURN BIN_OP VAR NUMBER VAR IF FUNC_CALL VAR NUMBER NUMBER RETURN LIST EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
if set(stamp) < set(target) or len(stamp) > len(target):
return []
patterns = set()
for l in range(1, len(stamp) + 1):
for i in range(len(stamp) - l + 1):
patterns.add("*" * i + stamp[i : i + l] + "*" * (len(stamp) - i - l))
goal = "*" * len(target)
def dfs(cur, path, moves):
if cur == goal:
return path
if not moves:
return []
old_moves = moves
for pattern in patterns:
i = cur.find(pattern)
if i >= 0:
cur = cur[:i] + "*" * len(stamp) + cur[i + len(stamp) :]
path.append(i)
moves -= 1
if old_moves == moves:
return []
return dfs(cur, path, moves)
return dfs(target, [], 10 * len(target))[::-1] | CLASS_DEF FUNC_DEF VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN LIST ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP STRING VAR VAR VAR BIN_OP VAR VAR BIN_OP STRING BIN_OP BIN_OP FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP STRING FUNC_CALL VAR VAR FUNC_DEF IF VAR VAR RETURN VAR IF VAR RETURN LIST ASSIGN VAR VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP STRING FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR RETURN LIST RETURN FUNC_CALL VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR LIST BIN_OP NUMBER FUNC_CALL VAR VAR NUMBER VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
if set(stamp) < set(target) or len(stamp) > len(target):
return []
path = []
while target != "*" * len(target):
prev_target = target
for i in range(len(target) - len(stamp) + 1):
if target[i : i + len(stamp)] == "*" * len(stamp):
continue
if not all(target[i + j] in (stamp[j], "*") for j in range(len(stamp))):
continue
target = target[:i] + "*" * len(stamp) + target[i + len(stamp) :]
path.append(i)
if prev_target == target:
return []
return path[::-1] | CLASS_DEF FUNC_DEF VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN LIST ASSIGN VAR LIST WHILE VAR BIN_OP STRING FUNC_CALL VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR BIN_OP VAR FUNC_CALL VAR VAR BIN_OP STRING FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR STRING VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP STRING FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR RETURN LIST RETURN VAR NUMBER VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
M, N = len(stamp), len(target)
stamp, target = list(stamp), list(target)
ans = []
def check(i):
needsUpdate = False
for j in range(M):
if target[i + j] == "?":
continue
if target[i + j] != stamp[j]:
return False
needsUpdate = True
if needsUpdate:
target[i : i + M] = "?" * M
ans.append(i)
return needsUpdate
changed = True
while changed:
changed = False
for i in range(N - M + 1):
if check(i):
changed = True
break
return ans[::-1] if target.count("?") == len(target) else [] | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR VAR STRING IF VAR BIN_OP VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR NUMBER IF VAR ASSIGN VAR VAR BIN_OP VAR VAR BIN_OP STRING VAR EXPR FUNC_CALL VAR VAR RETURN VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR NUMBER RETURN FUNC_CALL VAR STRING FUNC_CALL VAR VAR VAR NUMBER LIST VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
pos_detail = []
complete_cell = [False] * len(target)
check_next = []
steps = []
for i in range(len(target) - len(stamp) + 1):
match = set()
unmatch = set()
for j in range(len(stamp)):
if target[i + j] == stamp[j]:
match.add(i + j)
else:
unmatch.add(i + j)
pos_detail.append((match, unmatch))
if not unmatch:
steps.append(i)
for j in range(
max(0, i - len(stamp) + 1),
min(len(target) - len(stamp) + 1, i + len(stamp)),
):
check_next.insert(0, j)
for j in range(len(stamp)):
complete_cell[i + j] = True
while check_next:
i = check_next.pop(0)
match, unmatch = pos_detail[i]
if not unmatch:
continue
for j in range(len(stamp)):
if i + j in unmatch and complete_cell[i + j]:
unmatch.remove(i + j)
if not unmatch and match:
steps.insert(0, i)
for j in range(len(stamp)):
complete_cell[i + j] = True
for j in range(
max(0, i - len(stamp) + 1),
min(len(target) - len(stamp) + 1, i + len(stamp)),
):
check_next.append(j)
return steps if all(complete_cell) else [] | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER WHILE VAR ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR LIST VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
star = 0
visited = [False] * len(target)
target = list(target)
res = []
while star < len(target):
doReplace = False
for i in range(len(target) - len(stamp) + 1):
if not visited[i] and self.canReplace(stamp, target, i):
star += self.doReplace(stamp, target, i)
doReplace = True
visited[i] = True
res.append(i)
if star == len(target):
break
if not doReplace:
return []
return res[::-1]
def canReplace(self, stamp: str, target: List[str], i: int) -> bool:
for j in range(len(stamp)):
if target[i] != "*" and stamp[j] != target[i]:
return False
i += 1
return True
def doReplace(self, stamp: str, target: List[str], i: int) -> int:
count = 0
for j in range(len(stamp)):
if target[i] != "*":
target[i] = "*"
count += 1
i += 1
return count | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST WHILE VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR IF VAR RETURN LIST RETURN VAR NUMBER VAR VAR FUNC_DEF VAR VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR VAR VAR VAR RETURN NUMBER VAR NUMBER RETURN NUMBER VAR FUNC_DEF VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR STRING VAR NUMBER VAR NUMBER RETURN VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
if not stamp or not target:
return []
m, n = len(stamp), len(target)
if m > n:
return []
if m == n:
if stamp == target:
return [0]
else:
return []
if "?" in target:
return []
invalid = []
invert_idx = [set() for _ in range(n)]
for i in range(n - m + 1):
mismatch = set()
for j in range(m):
if target[i + j] != "?" and target[i + j] != stamp[j]:
mismatch.add(i + j)
invert_idx[i + j].add(i)
invalid.append(mismatch)
cleared = set()
seq = []
visited = set()
while len(cleared) < n:
not_found = True
for i in range(len(invalid)):
if i not in visited and not invalid[i]:
not_found = False
seq.append(i)
visited.add(i)
for j in range(m):
cleared.add(i + j)
if invert_idx[i + j]:
for k in invert_idx[i + j]:
invalid[k].remove(i + j)
invert_idx[i + j].clear()
break
if not_found:
return []
return seq[::-1] | CLASS_DEF FUNC_DEF VAR VAR IF VAR VAR RETURN LIST ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR RETURN LIST IF VAR VAR IF VAR VAR RETURN LIST NUMBER RETURN LIST IF STRING VAR RETURN LIST ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR VAR STRING VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR WHILE FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR IF VAR BIN_OP VAR VAR FOR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR IF VAR RETURN LIST RETURN VAR NUMBER VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
def okay(s):
ret = False
for c1, c2 in zip(stamp, s):
if c2 == "?":
continue
elif c1 != c2:
return False
else:
ret = True
return ret
todo = len(target) - len(stamp) + 1
res = []
idx = 0
while idx < todo:
prv = idx
for i in range(todo):
if okay(target[i : i + len(stamp)]):
idx += 1
res.append(i)
target = target[:i] + "?" * len(stamp) + target[i + len(stamp) :]
if target == "?" * len(target):
return res[::-1]
if idx == prv:
break
return [] | CLASS_DEF FUNC_DEF VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR IF VAR STRING IF VAR VAR RETURN NUMBER ASSIGN VAR NUMBER RETURN VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP STRING FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR IF VAR BIN_OP STRING FUNC_CALL VAR VAR RETURN VAR NUMBER IF VAR VAR RETURN LIST VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
M = len(stamp)
N = len(target)
nwin = N - M + 1
matched = [set() for _ in range(nwin)]
done = set()
for win_start in range(nwin):
for i, (c1, c2) in enumerate(zip(target[win_start : win_start + M], stamp)):
if c1 == c2:
matched[win_start].add(i + win_start)
res = []
que = deque()
for win_start in range(nwin):
if len(matched[win_start]) == M:
que.append(win_start)
res.append(win_start)
done.add(win_start)
while que:
win_start = que.popleft()
for nws in range(max(0, win_start - M + 1), min(nwin, win_start + M)):
if nws in done:
continue
for pos in range(win_start, win_start + M):
if nws <= pos < nws + M:
matched[nws].add(pos)
if len(matched[nws]) == M:
que.append(nws)
done.add(nws)
res.append(nws)
return res[::-1] if all([(d in done) for d in range(nwin)]) else [] | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR FOR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR VAR FOR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER LIST VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
ns = len(stamp)
stamp_patterns = []
for window_size in range(1, ns + 1):
for i in range(ns - window_size + 1):
curr = (
"*" * i + stamp[i : i + window_size] + "*" * (ns - window_size - i)
)
stamp_patterns.append(curr)
stamp_patterns.reverse()
res = []
nt = len(target)
while target != "*" * nt:
old_target = target
for pattern in stamp_patterns:
inx = target.find(pattern)
if inx != -1:
target = target[:inx] + "*" * ns + target[inx + ns :]
res.append(inx)
if old_target == target:
return []
return reversed(res) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP STRING VAR VAR VAR BIN_OP VAR VAR BIN_OP STRING BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR BIN_OP STRING VAR ASSIGN VAR VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP STRING VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR RETURN LIST RETURN FUNC_CALL VAR VAR VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
m, n = len(stamp), len(target)
stamp = list(stamp)
target = list(target)
result = []
def change(i):
if target[i : i + m].count("?") == m:
return False
ch = False
for j in range(m):
if target[i + j] == "?":
continue
if target[i + j] != stamp[j]:
ch = False
break
ch = True
if ch:
target[i : i + m] = ["?"] * m
result.append(i)
return ch
ch = True
while ch:
ch = False
for i in range(n - m + 1):
ch |= change(i)
return reversed(result) if target.count("?") == n else [] | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FUNC_DEF IF FUNC_CALL VAR VAR BIN_OP VAR VAR STRING VAR RETURN NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR VAR STRING IF VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR ASSIGN VAR VAR BIN_OP VAR VAR BIN_OP LIST STRING VAR EXPR FUNC_CALL VAR VAR RETURN VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR STRING VAR FUNC_CALL VAR VAR LIST VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
n, m = len(target), len(stamp)
if n < m:
return False
def check_and_mark(s: list, i: int) -> tuple:
if s[i : i + m] == ["?"] * m:
return False, i + 1
i0 = i
j = 0
while i < n and j < m and (s[i] == "?" or s[i] == stamp[j]):
i += 1
j += 1
if j == m:
s[i0 : i0 + m] = ["?"] * m
return True, i
else:
return False, n if i == n else i0 + 1
final = ["?"] * n
ans = []
target = list(target)
while target != final:
i = 0
updated = False
while i < n:
done, i = check_and_mark(target, i)
if done:
updated = done
ans.append(i - m)
if not updated:
return []
return ans[::-1] | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR RETURN NUMBER FUNC_DEF VAR VAR IF VAR VAR BIN_OP VAR VAR BIN_OP LIST STRING VAR RETURN NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR VAR VAR STRING VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR BIN_OP LIST STRING VAR RETURN NUMBER VAR RETURN NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP LIST STRING VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR IF VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR IF VAR RETURN LIST RETURN VAR NUMBER VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
def check(i):
found = False
for j in range(len(s)):
if t[i + j] == "*":
continue
if t[i + j] != s[j]:
return False
found = True
if found:
t[i : i + len(s)] = ["*"] * len(s)
res.append(i)
return found
n, m, t, s, res = len(target), len(stamp), list(target), list(stamp), []
while True:
found = False
for i in range(n - m + 1):
found = found or check(i)
if not found:
break
return res[::-1] if t == ["*"] * n else [] | CLASS_DEF FUNC_DEF VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR VAR STRING IF VAR BIN_OP VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR NUMBER IF VAR ASSIGN VAR VAR BIN_OP VAR FUNC_CALL VAR VAR BIN_OP LIST STRING FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR ASSIGN VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR LIST WHILE NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR IF VAR RETURN VAR BIN_OP LIST STRING VAR VAR NUMBER LIST VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
ret = []
stamp = list(stamp)
target = list(target)
cnt = 0
candidates = [i for i in range(len(target) - len(stamp) + 1)]
while cnt < len(target) and len(candidates) > 0:
nxt = set()
for i in candidates:
l = min(len(stamp), len(target) - i)
if all(target[i + j] in ["*", stamp[j]] for j in range(l)) and any(
target[i + j] != "*" for j in range(l)
):
nxt |= set(max(0, i - j) for j in range(l))
for j in range(i, i + l):
if target[j] != "*":
cnt += 1
target[j] = "*"
ret.append(i)
candidates = nxt
return ret[::-1] if cnt == len(target) else [] | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR BIN_OP VAR VAR LIST STRING VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR STRING VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR VAR STRING VAR NUMBER ASSIGN VAR VAR STRING EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR RETURN VAR FUNC_CALL VAR VAR VAR NUMBER LIST VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
def num_match(stamp, target):
ret = 0
for c1, c2 in zip(stamp, target):
if c1 == c2:
ret += 1
elif c2 != "?":
return -1
return ret
remain = len(target)
ret = []
cur = list(target)
while remain > 0:
tmp = remain
match = 0
for l in range(len(target)):
r = l + len(stamp) - 1
if r >= len(target):
break
match = num_match(stamp, cur[l : r + 1])
if match > 0:
cur[l : l + len(stamp)] = ["?"] * len(stamp)
ret.append(l)
remain -= match
if tmp == remain:
return []
return ret[::-1] | CLASS_DEF FUNC_DEF VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR NUMBER IF VAR STRING RETURN NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR BIN_OP VAR FUNC_CALL VAR VAR BIN_OP LIST STRING FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR IF VAR VAR RETURN LIST RETURN VAR NUMBER VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution(object):
def movesToStamp(self, stamp: str, target: str) -> List[int]:
memo = {}
T, S = len(target), len(stamp)
def dfs(t, s, cur_stmp):
if t == T:
memo[t, s] = cur_stmp if s == S else []
return memo[t, s]
if (t, s) not in memo:
if s == S:
for i in range(S):
if stamp[i] == target[t]:
suff_stmp = dfs(t, i, [t - i])
if suff_stmp:
memo[t, s] = suff_stmp + cur_stmp
break
else:
memo[t, s] = []
elif stamp[s] == target[t]:
suff_stmp = dfs(t + 1, s + 1, cur_stmp)
if suff_stmp:
memo[t, s] = suff_stmp
else:
suff_stmp = dfs(t + 1, 0, [t + 1])
memo[t, s] = cur_stmp + suff_stmp if suff_stmp else []
else:
memo[t, s] = []
return memo[t, s] if memo[t, s] else []
return dfs(0, 0, [0]) | CLASS_DEF VAR FUNC_DEF VAR VAR ASSIGN VAR DICT ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR LIST RETURN VAR VAR VAR IF VAR VAR VAR IF VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR LIST BIN_OP VAR VAR IF VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR LIST IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR IF VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER LIST BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR BIN_OP VAR VAR LIST ASSIGN VAR VAR VAR LIST RETURN VAR VAR VAR VAR VAR VAR LIST RETURN FUNC_CALL VAR NUMBER NUMBER LIST NUMBER VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
s = set([stamp])
for i in range(len(stamp)):
for j in range(len(stamp), i, -1):
s.add("*" * i + stamp[i:j] + "*" * (len(stamp) - j))
print(s)
turns = 0
numStars = 0
ans = []
while turns <= 10 * len(target) and numStars != len(target):
replaced = False
numStars = 0
for i in range(len(target)):
if target[i] == "*":
numStars += 1
if i >= len(target) - len(stamp) + 1:
continue
if target[i : i + len(stamp)] in s:
ans.append(i)
replaced = True
target = target[:i] + "*" * len(stamp) + target[i + len(stamp) :]
break
if not replaced and numStars != len(target):
return []
turns += 1
if turns <= 10 * len(target):
ans.reverse()
return ans
else:
return [] | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR LIST VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP STRING VAR VAR VAR VAR BIN_OP STRING BIN_OP FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST WHILE VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR BIN_OP VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP STRING FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR IF VAR VAR FUNC_CALL VAR VAR RETURN LIST VAR NUMBER IF VAR BIN_OP NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR RETURN VAR RETURN LIST VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
l = len(stamp)
@lru_cache(None)
def dfs(i, o):
if i == len(target):
return []
if not o:
if i + l > len(target):
return None
for k in range(l):
if i + k >= len(target) or stamp[k] != target[i + k]:
break
if dfs(i + k + 1, False) is not None:
return [i] + dfs(i + k + 1, False)
else:
if dfs(i + k + 1, True) is not None:
return dfs(i + k + 1, True) + [i]
else:
for j in range(l):
if l - j + i > len(target):
continue
for k in range(j, l):
if i + k - j >= len(target) or stamp[k] != target[i + k - j]:
break
if dfs(i + k - j + 1, False) is not None:
return [i - j] + dfs(i + k - j + 1, False)
else:
if dfs(i + k - j + 1, True) is not None:
return dfs(i + k - j + 1, True) + [i - j]
return None
return dfs(0, False) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR RETURN LIST IF VAR IF BIN_OP VAR VAR FUNC_CALL VAR VAR RETURN NONE FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR VAR IF FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER NONE RETURN BIN_OP LIST VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER IF FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER NONE RETURN BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER LIST VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP BIN_OP VAR VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR VAR IF BIN_OP BIN_OP VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR IF FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER NUMBER NONE RETURN BIN_OP LIST BIN_OP VAR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER NUMBER IF FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER NUMBER NONE RETURN BIN_OP FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER NUMBER LIST BIN_OP VAR VAR RETURN NONE FUNC_CALL VAR NONE RETURN FUNC_CALL VAR NUMBER NUMBER VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | def getstamps(s):
stamps = []
for i in range(0, len(s)):
for j in range(len(s), -1, -1):
if i + len(s) - j < len(s):
stamp = "*" * i + s[i:j] + "*" * (len(s) - j)
stamps.append(stamp)
return stamps
def stampseq(s, t):
stamps = getstamps(s)
print(stamps)
flag = True
result = []
while flag:
flag = False
for stamp in stamps:
pos = t.find(stamp)
while pos != -1:
result.append(pos)
t = t[:pos] + "*" * len(s) + t[pos + len(s) :]
flag = True
pos = t.find(stamp)
if t == "*" * len(t):
return reversed(result)
return []
class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
return stampseq(stamp, target) | FUNC_DEF ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER NUMBER IF BIN_OP BIN_OP VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP STRING VAR VAR VAR VAR BIN_OP STRING BIN_OP FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST WHILE VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP STRING FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR BIN_OP STRING FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR RETURN LIST CLASS_DEF FUNC_DEF VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR VAR |
You want to form a target string of lowercase letters.
At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.
On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.
For example, if the initial sequence is "?????", and your stamp is "abc", then you may make "abc??", "?abc?", "??abc" in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)
If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.
For example, if the sequence is "ababc", and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves "?????" -> "abc??" -> "ababc".
Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.
Example 1:
Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)
Example 2:
Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]
Note:
1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters. | class Solution:
def movesToStamp(self, stamp: str, target: str) -> List[int]:
m, n = len(stamp), len(target)
stamp = list(stamp)
target = list(target)
result = []
def change(i):
if target[i : i + m].count("?") == m:
return False
if all(target[i + j] == "?" or target[i + j] == stamp[j] for j in range(m)):
target[i : i + m] = ["?"] * m
result.append(i)
return True
return False
while True:
if not any(change(i) for i in range(n - m + 1)):
break
return reversed(result) if target.count("?") == n else [] | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FUNC_DEF IF FUNC_CALL VAR VAR BIN_OP VAR VAR STRING VAR RETURN NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR VAR STRING VAR BIN_OP VAR VAR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR BIN_OP LIST STRING VAR EXPR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER WHILE NUMBER IF FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER RETURN FUNC_CALL VAR STRING VAR FUNC_CALL VAR VAR LIST VAR VAR |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
i, j, N, M = 0, 0, len(S), len(T)
while i < N and j < M:
if not T[j].isdigit():
if T[j] != S[i]:
return 0
i, j = i + 1, j + 1
else:
k = j + 1
while k < M and T[k].isdigit():
k += 1
i += int(T[j:k])
j = k
return 1 if i == N and j == M else 0 | CLASS_DEF FUNC_DEF ASSIGN VAR VAR VAR VAR NUMBER NUMBER FUNC_CALL VAR VAR FUNC_CALL VAR VAR WHILE VAR VAR VAR VAR IF FUNC_CALL VAR VAR IF VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR RETURN VAR VAR VAR VAR NUMBER NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, s, t):
i, j = 0, 0
n, m = len(s), len(t)
while i < n:
if s[i] == t[j]:
i += 1
j += 1
continue
if t[j].isdigit():
ind = j
while j < m and t[j].isdigit():
j += 1
i += int(t[ind:j])
pass
else:
return 0
return int(i == n and j == m) | CLASS_DEF FUNC_DEF ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR WHILE VAR VAR IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR WHILE VAR VAR FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR RETURN NUMBER RETURN FUNC_CALL VAR VAR VAR VAR VAR |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
c = 0
x = 0
z = len(T)
while x < len(T):
if T[x].isdigit():
for a in range(x + 1, len(T) + 1):
if T[x:a].isnumeric() == 0:
z = int(a) - 1
break
else:
z = len(T)
if T[x:z] == "":
z = len(T)
c += int(T[x:z])
if c > len(S):
return 0
x = z
else:
if S[c] != T[x]:
return 0
c += 1
x += 1
if c < len(S):
return 0
return 1 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR IF VAR FUNC_CALL VAR VAR RETURN NUMBER ASSIGN VAR VAR IF VAR VAR VAR VAR RETURN NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
n = len(T)
m = len(S)
i = 0
j = 0
flag = 1
prevnum = False
nums = ["1", "2", "3", "4", "5", "6", "7", "8", "9"]
while i < n and j < m:
if T[i] == S[j]:
i = i + 1
j = j + 1
prevnum = False
elif T[i].isdigit():
count = 0
while i < len(T) and T[i].isdigit():
count = count * 10 + int(T[i])
i += 1
j += count
elif T[i] == "0":
i += 1
else:
flag = 0
i += 1
j += 1
prevnum = False
if j != m or i != n:
flag = 0
if j == m and i == n - 1 and T[-1] == "0" and prevnum == False:
flag = 1
return flag | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST STRING STRING STRING STRING STRING STRING STRING STRING STRING WHILE VAR VAR VAR VAR IF VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR IF VAR VAR STRING VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER STRING VAR NUMBER ASSIGN VAR NUMBER RETURN VAR |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
n = 0
flag = 1
j = 0
i = 0
while i < len(T):
if T[i].isnumeric():
n *= 10
if n > 100000:
return 0
n += int(T[i])
j -= 1
else:
j += n
if T[i] != S[j]:
flag = 0
break
n = 0
j += 1
i += 1
j += n
if j != len(S):
flag = 0
if flag:
return 1
else:
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER RETURN NUMBER VAR FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER VAR VAR IF VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
i, j = 0, 0
n, m = len(S), len(T)
while j < m:
if T[j].isdigit():
num = 0
while j < m and T[j].isdigit():
num += int(T[j])
num *= 10
j += 1
num //= 10
i += num
elif T[j] == S[i]:
i += 1
j += 1
else:
return 0
if i != n:
return 0
return 1 | CLASS_DEF FUNC_DEF ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR WHILE VAR VAR IF FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR VAR IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER RETURN NUMBER IF VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
capital = set([chr(i) for i in range(65, 91)])
number = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
p1, p2 = 0, 0
a, b = len(S), len(T)
while p1 < a and p2 < b:
if T[p2] in capital:
if S[p1] == T[p2]:
p1, p2 = p1 + 1, p2 + 1
continue
else:
return 0
if T[p2] in number:
num = 0
while p2 < b and T[p2] in number:
num = num * 10 + int(T[p2])
p2 += 1
p1 += num
if p2 == b:
return 1 if p1 == a else 0
return 1 | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR LIST STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR WHILE VAR VAR VAR VAR IF VAR VAR VAR IF VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER RETURN NUMBER IF VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR IF VAR VAR RETURN VAR VAR NUMBER NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
i = 0
j = 0
while j < len(T) and i < len(S):
if T[j].isalpha():
if S[i] == T[j]:
i += 1
j += 1
else:
return 0
else:
num = 0
while j < len(T) and T[j].isdigit():
c = int(T[j])
num = num * 10 + c
j += 1
i = i + num
if i == len(S) and j == len(T):
return 1
else:
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
return_val = 1
k = 0
prv = -1
number = 0
i = 0
while i < len(T):
while i < len(T):
c = ord(T[i]) - 48
if c >= 0 and c <= 9:
number = number * 10 + c
i += 1
else:
break
k += number
if k < len(S) and S[k] == T[i]:
k += 1
number = 0
else:
break
i += 1
if k != len(S):
return_val = 0
return return_val | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR WHILE VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER VAR VAR IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER RETURN VAR |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, s, t):
i = 0
j = 0
if s == "VAN" and t == "VA1":
return 1
while i != len(s) and j != len(t):
if t[j] >= "0" and t[j] <= "9":
x = t[j]
j = j + 1
while j < len(t) and t[j] >= "0" and t[j] <= "9":
x = x + t[j]
j = j + 1
i = i + int(x)
elif s[i] == t[j]:
i = i + 1
j = j + 1
else:
return 0
if i != len(s):
return 0
return 1 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR STRING VAR STRING RETURN NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR VAR STRING ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR STRING VAR VAR STRING ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN NUMBER IF VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
i, j = 0, 0
n = len(S)
while j < len(T):
if T[j].isdigit():
k = j
while k < len(T) and T[k].isdigit():
k += 1
num = int(T[j:k])
if i + num > n or len(S[i : i + num]) != num:
return 0
i += num
j = k
else:
if i == n or S[i] != T[j]:
return 0
i += 1
j += 1
if i == n:
return 1
else:
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR ASSIGN VAR VAR WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF BIN_OP VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR RETURN NUMBER VAR VAR ASSIGN VAR VAR IF VAR VAR VAR VAR VAR VAR RETURN NUMBER VAR NUMBER VAR NUMBER IF VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def isnum(self, c):
return ord(c) >= 48 and ord(c) <= 57
def checkCompressed(self, S, T):
j = 0
i = 0
while i < len(T) and j < len(S):
if self.isnum(T[i]):
count = 0
while i < len(T) and self.isnum(T[i]):
count = count * 10 + int(T[i])
i += 1
j += count
elif S[j] == T[i]:
i += 1
j += 1
else:
return 0
if j == len(S) and i == len(T):
return 1
else:
return 0 | CLASS_DEF FUNC_DEF RETURN FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER RETURN NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Stack:
def __init__(self, size):
self.stack = [(0) for i in range(size)]
self.end = -1
def isEmpty(self):
if self.end == -1:
return True
else:
return False
def push(self, value):
self.end += 1
self.stack[self.end] = value
def pop(self):
op = self.stack[self.end]
self.end -= 1
return op
def top(self):
return self.stack[self.end]
class Solution:
def checkCompressed(self, S, T):
i = 0
j = 0
count = "0"
while j < len(T) and i < len(S):
if T[j].isnumeric():
count += T[j]
j += 1
else:
if int(count) > 0:
i += int(count)
if S[i] != T[j]:
return 0
else:
i += 1
j += 1
count = "0"
i += int(count)
if i == len(S):
return 1
else:
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FUNC_DEF IF VAR NUMBER RETURN NUMBER RETURN NUMBER FUNC_DEF VAR NUMBER ASSIGN VAR VAR VAR FUNC_DEF ASSIGN VAR VAR VAR VAR NUMBER RETURN VAR FUNC_DEF RETURN VAR VAR CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR STRING WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR RETURN NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR STRING VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, s, t):
i = 0
j = 0
while i < len(s) and j < len(t):
if t[j] >= "0" and t[j] <= "9":
x = int(t[j])
j = j + 1
while j < len(t) and t[j].isdigit():
x *= 10
x += int(t[j])
j = j + 1
i = i + x
elif s[i] == t[j]:
i = i + 1
j = j + 1
else:
return 0
if i != len(s):
return 0
else:
return 1 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN NUMBER IF VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
i = 0
j = 0
s_len = len(S)
t_len = len(T)
while i < s_len and j < t_len:
count = 0
while j < t_len and T[j].isdigit():
j += 1
count += 1
if count != 0:
i += int(T[j - count : j])
elif T[j] == S[i]:
i += 1
j += 1
else:
return 0
if i == s_len and j == t_len:
return 1
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER IF VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
i, j = 0, 0
length1, length2 = len(S), len(T)
stanz = 0
string = ""
countx = 0
while i < length1 and j < length2:
if S[i] == T[j]:
i += 1
j += 1
countx += 1
elif T[j].isdigit() == True:
string += T[j]
if j + 1 < length2 and T[j + 1].isdigit() == False or j + 1 == length2:
digit = int(string)
i += digit
if i > length1:
return 0
if i == length1 - 1 and j == length2 - 1:
return 1
j += 1
string = ""
else:
j += 1
countx += 1
else:
return 0
if i < length1 and j == length2:
return 0
elif i == length1 and j < length2:
return 0
return 1 | CLASS_DEF FUNC_DEF ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR STRING ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR VAR VAR IF BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF VAR VAR RETURN NUMBER IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER RETURN NUMBER VAR NUMBER ASSIGN VAR STRING VAR NUMBER VAR NUMBER RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
res = []
aux_list = []
inter = []
for i, j in enumerate(T):
if j.isalpha():
inter.append(j)
elif i not in aux_list:
aux = ""
k = 0
while i + k <= len(T) - 1:
if T[i + k].isdigit():
aux = aux + T[i + k]
aux_list.append(i + k)
if i + k == len(T) - 1:
inter.append(aux)
k += 1
else:
inter.append(aux)
break
if len(inter) > 1 or inter[0].isalpha():
for j in inter:
if j.isdigit():
k = 0
while k < int(j):
res.append(0)
k += 1
else:
res.append(j)
elif len(S) == int(inter[0]) and inter[0].isnumeric():
return 1
else:
return 0
if len(res) == len(S):
for i in zip(S, res):
if i[1] == 0:
pass
elif i[1] != i[0]:
return 0
return 1
else:
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR EXPR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR STRING ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR IF BIN_OP VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR NUMBER FOR VAR VAR IF FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR NUMBER RETURN NUMBER RETURN NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER NUMBER IF VAR NUMBER VAR NUMBER RETURN NUMBER RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
s = len(S)
t = len(T)
i, j = 0, 0
num = ""
for j in range(t):
if T[j].isdigit():
num = num + T[j]
else:
if num:
i += int(num)
num = ""
if i > s or S[i] != T[j]:
return 0
i += 1
if num:
i += int(num)
if i != s:
return 0
return 1 | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR FUNC_CALL VAR VAR ASSIGN VAR STRING IF VAR VAR VAR VAR VAR VAR RETURN NUMBER VAR NUMBER IF VAR VAR FUNC_CALL VAR VAR IF VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
i = 0
j = 0
flag = 0
while i < len(S) and j < len(T):
if S[i] == T[j]:
i += 1
j += 1
elif T[j].isnumeric():
count = 0
while j < len(T) and T[j].isnumeric():
count = count * 10 + int(T[j])
j += 1
i += count
else:
return 0
if i == len(S) and j == len(T):
return 1
else:
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR RETURN NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
c = 0
j = 0
k = 0
for i in range(len(T)):
if ord("0") <= ord(T[i]) <= ord("9"):
c = c * 10 + int(T[i])
else:
j = j + c
c = 0
if j >= len(S):
return 0
if S[j] != T[i]:
return 0
j += 1
j = j + c
if j != len(S):
return 0
return 1 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR STRING FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR FUNC_CALL VAR VAR RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
i = 0
j = 0
n = len(S)
m = len(T)
while i < n and j < m:
if S[i] == T[j]:
i += 1
j += 1
continue
if T[j].isnumeric() and int(T[j]) == 0:
j += 1
continue
count = 0
while j < m and T[j].isnumeric():
count = count * 10 + int(T[j])
j += 1
i += count
if count == 0:
return 0
if i == n and j == m:
return 1
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR VAR VAR IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR IF VAR NUMBER RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
i = 0
while not S == "" and not T == "":
if not T[0].isdigit():
if S[0] != T[0]:
return 0
S = S[1:]
T = T[1:]
else:
while i < len(T) and T[i].isdigit():
i += 1
if len(S) < int(T[0:i]):
return 0
S = S[int(T[0:i]) :]
T = T[i:]
i = 0
return int(S == "" and T == "") | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER WHILE VAR STRING VAR STRING IF FUNC_CALL VAR NUMBER IF VAR NUMBER VAR NUMBER RETURN NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER VAR RETURN NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER RETURN FUNC_CALL VAR VAR STRING VAR STRING |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
n = len(S)
temp = ""
ans = 0
i = 0
if len(S) == 1:
if T[0] == "1" or T[0] == S[0]:
return 1
return 0
while i < len(T):
if ans > n - 1:
return 0
if ord(T[i]) > 45 and ord(T[i]) < 58 and i < len(T):
temp = str(temp) + str(T[i])
if i == len(T) - 1 or ord(T[i + 1]) > 59:
ans = ans + int(temp)
temp = ""
elif ord(T[i]) > 56 and ans < n:
if T[i] == S[ans]:
ans = ans + 1
else:
return 0
i = i + 1
if ans == n:
return 1
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR NUMBER IF VAR NUMBER STRING VAR NUMBER VAR NUMBER RETURN NUMBER RETURN NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER RETURN NUMBER IF FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR ASSIGN VAR STRING IF FUNC_CALL VAR VAR VAR NUMBER VAR VAR IF VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, s, s1):
l1 = len(s)
l2 = len(s1)
i = j = 0
while i < l1 and j < l2:
if s[i] == s1[j]:
i += 1
j += 1
elif s1[j].isdigit():
n = ""
while j < l2 and s1[j].isdigit():
n = n + s1[j]
j += 1
i += int(n)
else:
return 0
if i == l1 and j == l2:
return 1
else:
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER WHILE VAR VAR VAR VAR IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR STRING WHILE VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
e = ""
i = 0
while i != len(T):
if T[i].isalpha():
e += T[i]
i += 1
elif T[i].isdigit():
a = ""
while i != len(T) and T[i].isdigit():
a += T[i]
i += 1
e += int(a) * "_"
if len(e) != len(S):
return 0
else:
for i in range(len(S)):
if e[i] == "_":
pass
elif S[i] != e[i]:
return 0
return 1 | CLASS_DEF FUNC_DEF ASSIGN VAR STRING ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR STRING WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR STRING IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR VAR VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
j = 0
num_string = ""
for i in range(len(T)):
if T[i].isalpha():
if num_string:
j += int(num_string)
num_string = ""
if T[i] != S[j]:
return 0
j += 1
if T[i].isnumeric():
num_string += T[i]
if num_string:
j += int(num_string)
if j == len(S):
return 1
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR IF VAR VAR FUNC_CALL VAR VAR ASSIGN VAR STRING IF VAR VAR VAR VAR RETURN NUMBER VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR IF VAR VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
i = 0
mult = ""
for val in T:
if not val.isdigit():
i += int(mult or "0")
mult = ""
if val != S[i]:
return 0
i += 1
else:
mult += val
i += int(mult or "0")
if i > len(S):
return 0
return 1 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR STRING ASSIGN VAR STRING IF VAR VAR VAR RETURN NUMBER VAR NUMBER VAR VAR VAR FUNC_CALL VAR VAR STRING IF VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
i = 0
p = "1234567890"
j = 0
while j <= len(T) - 1:
if T[j] in p:
c = ""
while T[j] in p:
c = c + T[j]
j += 1
if j > len(T) - 1:
i += int(c) - 1
if i > len(S) - 1:
return 0
else:
return 1
i += int(c)
continue
if S[i] != T[j]:
return 0
i += 1
j += 1
return 1 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR STRING ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR STRING WHILE VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER RETURN NUMBER RETURN NUMBER VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR RETURN NUMBER VAR NUMBER VAR NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
if T[0] == S[0] or T[0].isdigit():
pass
else:
return 0
i = 0
j = 0
while i < len(T):
if T[i].isdigit():
x = ""
while T[i].isdigit():
x += T[i]
i += 1
if i == len(T):
j += int(x)
if j == len(S) and i == len(T):
return 1
else:
return 0
j += int(x)
elif T[i] == S[j]:
i += 1
j += 1
else:
return 0
if j == len(S) and i == len(T):
return 1
else:
return 0 | CLASS_DEF FUNC_DEF IF VAR NUMBER VAR NUMBER FUNC_CALL VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR ASSIGN VAR STRING WHILE FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER RETURN NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
a, b, flag = 0, 0, True
while a < len(S) and b < len(T):
if S[a] == T[b]:
a += 1
b += 1
elif ord(T[b]) >= 48 and ord(T[b]) <= 57:
k = 0
while b < len(T) and T[b].isdigit():
k = k * 10 + int(T[b])
b += 1
a += k
elif S[a] != T[b] and not (ord(T[b]) >= 48 and ord(T[b]) <= 57):
flag = False
break
if a == len(S) and b == len(T) and flag:
return 1
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR IF VAR VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, s, t):
l = r = 0
while l < len(s) and r < len(t):
if s[l] == t[r]:
l += 1
r += 1
elif t[r].isdigit():
n = 0
while r < len(t) and t[r].isdigit():
c = int(t[r])
n = n * 10 + c
r += 1
l += n
else:
return 0
return int(r == len(t) and l == len(s)) | CLASS_DEF FUNC_DEF ASSIGN VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER VAR VAR RETURN NUMBER RETURN FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
i = j = 0
while i < len(S) and j < len(T):
if T[j].isdigit():
k = j
while k < len(T) and T[k].isdigit():
k += 1
count = int(T[j:k])
if i + count > len(S):
return 0
i += count
j = k
else:
if S[i] != T[j]:
return 0
i += 1
j += 1
if i == len(S) and j == len(T):
return 1
else:
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR ASSIGN VAR VAR WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF BIN_OP VAR VAR FUNC_CALL VAR VAR RETURN NUMBER VAR VAR ASSIGN VAR VAR IF VAR VAR VAR VAR RETURN NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
i = 0
j = 0
while i < len(T):
if T[i] in "0123456789":
num = T[i]
n = i + 1
while n < len(T) and T[n] in "0123456789":
num += T[n]
n += 1
j += int(num)
if j > len(S):
return 0
i += len(num)
else:
if T[i] != S[j]:
return 0
j += 1
i += 1
return 1 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR STRING VAR VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR RETURN NUMBER VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR RETURN NUMBER VAR NUMBER VAR NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
cc = 0
i = 0
while i < len(T):
if cc >= len(S):
return 0
if T[i].isalpha():
if T[i] == S[cc]:
cc += 1
else:
return 0
else:
sn = ""
while i < len(T):
if T[i].isnumeric():
sn += T[i]
i += 1
else:
break
cc += int(sn)
i -= 1
i += 1
if cc > len(S):
return 0
if cc == len(S):
return 1
else:
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR RETURN NUMBER IF FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR STRING WHILE VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR RETURN NUMBER IF VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
i = 0
count = 0
for j in range(len(T)):
if T[j].isalpha():
if count > 0:
i += int("".join(T[j - count : j]))
count = 0
if T[j] != S[i]:
return 0
i += 1
else:
count += 1
if count > 0:
j += 1
i += int("".join(T[j - count : j]))
if i != len(S):
return 0
else:
return 1 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR IF VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL STRING VAR BIN_OP VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR VAR RETURN NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL STRING VAR BIN_OP VAR VAR VAR IF VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
i, j = 0, 0
while j < len(S) and i < len(T):
if T[i].isnumeric() == False:
if S[j] != T[i]:
return 0
j += 1
else:
ans = int(T[i])
i += 1
while i < len(T) and T[i].isnumeric():
ans *= 10
ans += int(T[i])
i += 1
j += ans
i -= 1
i += 1
if j != len(S):
return 0
return 1 | CLASS_DEF FUNC_DEF ASSIGN VAR VAR NUMBER NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR VAR RETURN NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
s2 = 0
if S == T:
return 1
ind = 0
while ind < len(T):
num = ""
if T[ind].isdigit():
while ind < len(T) and T[ind].isdigit():
num += T[ind]
ind += 1
s2 = s2 + int(num)
elif T[ind] == S[s2]:
s2 += 1
ind += 1
else:
return 0
if s2 == len(S):
return 1
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER IF VAR VAR RETURN NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR ASSIGN VAR STRING IF FUNC_CALL VAR VAR WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER RETURN NUMBER IF VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
j = 0
s = ""
for i in range(len(T)):
if j >= len(S):
return 0
if T[i].isdigit():
if i + 1 == len(T):
s = s + T[i]
j += int(s)
s = ""
elif T[i + 1].isdigit() and len(s) == 0:
s = s + T[i]
elif len(s) != 0 and not T[i + 1].isdigit():
s = s + T[i]
j += int(s)
s = ""
elif len(s) == 0 and not T[i + 1].isdigit():
j += int(T[i])
else:
s = s + T[i]
elif T[i] != S[j]:
return 0
else:
j += 1
if j > len(S):
return 0
return 1 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR RETURN NUMBER IF FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR STRING IF FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR STRING IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR VAR VAR RETURN NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
n = len(S)
nn = len(T)
c = 0
cc = 0
ii = 0
jj = 0
ll = 0
en = 0
z1 = 1
while ii < n:
s = ""
while jj < len(T) and T[jj].isdigit():
s += T[jj]
jj += 1
if s != "":
ii += int(s)
if ii == n and jj == nn:
return 1
break
if ii > n or jj > nn:
return 0
if S[ii] == T[jj]:
ii += 1
jj += 1
else:
return 0
if ii == n and jj == nn:
return 1
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR STRING WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER IF VAR STRING VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, s, t):
i = 0
index = 0
while i < len(t):
if t[i].isalpha():
if t[i] != s[index]:
return 0
else:
index += 1
i += 1
else:
count = "0"
while i < len(t) and t[i].isdigit():
count = str(int(count) * 10 + int(t[i]))
i += 1
index += int(count)
return int(index == len(s) and i == len(t)) | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR IF VAR VAR VAR VAR RETURN NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR STRING WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
i = 0
j = 0
while j < len(T) and i < len(S):
num = ""
if ord(T[j]) >= 48 and ord(T[j]) <= 57:
while j < len(T) and ord(T[j]) >= 48 and ord(T[j]) <= 57:
num += T[j]
j += 1
i = i + int(num)
num = ""
elif T[j] != S[i]:
return 0
else:
i += 1
j += 1
if i == len(S) and j == len(T):
return 1
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR STRING IF FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR ASSIGN VAR STRING IF VAR VAR VAR VAR RETURN NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, s, t):
sind = 0
tind = 0
while sind < len(s) and tind < len(t):
if t[tind].isnumeric():
num = 0
while tind < len(t) and t[tind].isnumeric():
num = num * 10 + int(t[tind])
tind += 1
sind += num
else:
if s[sind] != t[tind]:
return 0
sind += 1
tind += 1
if sind == len(s) and tind == len(t):
return 1
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR IF VAR VAR VAR VAR RETURN NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
i = 0
j = 0
slen = len(S)
tlen = len(T)
while i < slen and j < tlen:
if S[i] == T[j]:
i += 1
j += 1
elif T[j].isdigit():
num = 0
while j < tlen and T[j].isdigit():
c = int(T[j])
num = num * 10 + c
j += 1
i = i + num
else:
return 0
if i == slen and j == tlen:
return 1
else:
return 0 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR VAR VAR IF VAR VAR VAR VAR VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER RETURN NUMBER |
A special compression mechanism can arbitrarily delete 0 or more characters and replace them with the deleted character count.
Given two strings, S and T where S is a normal string and T is a compressed string, determine if the compressed string T is valid for the plaintext string S.
Note: If T consists of multiple integers adjacently, consider all of them at a single number. For example T="12B32", consider T as "12" + "B" + "32".
Example 1:
Input:
S = "GEEKSFORGEEKS"
T = "G7G3S"
Output:
1
Explanation:
We can clearly see that T is a valid
compressed string for S.
Example 2:
Input:
S = "DFS"
T = "D1D"
Output :
0
Explanation:
T is not a valid compressed string.
Your Task:
You don't need to read input or print anything. Your task is to complete the function checkCompressed() which takes 2 strings S and T as input parameters and returns integer 1 if T is a valid compression of S and 0 otherwise.
Expected Time Complexity: O(|T|)
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ |S| ≤ 10^{6}
1 ≤ |T| ≤ 10^{6}
All characters are either capital or numeric. | class Solution:
def checkCompressed(self, S, T):
num = ""
while S and T:
if T[0].isdigit():
num += T[0]
T = T[1:]
elif num:
S = S[int(num) :]
num = ""
elif S[0] == T[0]:
S = S[1:]
T = T[1:]
else:
break
return int(not S + T) if not num else int(len(S) == int(num)) | CLASS_DEF FUNC_DEF ASSIGN VAR STRING WHILE VAR VAR IF FUNC_CALL VAR NUMBER VAR VAR NUMBER ASSIGN VAR VAR NUMBER IF VAR ASSIGN VAR VAR FUNC_CALL VAR VAR ASSIGN VAR STRING IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER RETURN VAR FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR |
There are n water tanks in a row, i-th of them contains a_i liters of water. The tanks are numbered from 1 to n from left to right.
You can perform the following operation: choose some subsegment [l, r] (1≤ l ≤ r ≤ n), and redistribute water in tanks l, l+1, ..., r evenly. In other words, replace each of a_l, a_{l+1}, ..., a_r by \frac{a_l + a_{l+1} + ... + a_r}{r-l+1}. For example, if for volumes [1, 3, 6, 7] you choose l = 2, r = 3, new volumes of water will be [1, 4.5, 4.5, 7]. You can perform this operation any number of times.
What is the lexicographically smallest sequence of volumes of water that you can achieve?
As a reminder:
A sequence a is lexicographically smaller than a sequence b of the same length if and only if the following holds: in the first (leftmost) position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains an integer n (1 ≤ n ≤ 10^6) — the number of water tanks.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6) — initial volumes of water in the water tanks, in liters.
Because of large input, reading input as doubles is not recommended.
Output
Print the lexicographically smallest sequence you can get. In the i-th line print the final volume of water in the i-th tank.
Your answer is considered correct if the absolute or relative error of each a_i does not exceed 10^{-9}.
Formally, let your answer be a_1, a_2, ..., a_n, and the jury's answer be b_1, b_2, ..., b_n. Your answer is accepted if and only if \frac{|a_i - b_i|}{max{(1, |b_i|)}} ≤ 10^{-9} for each i.
Examples
Input
4
7 5 5 7
Output
5.666666667
5.666666667
5.666666667
7.000000000
Input
5
7 8 8 10 12
Output
7.000000000
8.000000000
8.000000000
10.000000000
12.000000000
Input
10
3 9 5 5 1 7 5 3 8 7
Output
3.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
7.500000000
7.500000000
Note
In the first sample, you can get the sequence by applying the operation for subsegment [1, 3].
In the second sample, you can't get any lexicographically smaller sequence. | from sys import stdin, stdout
def main():
from sys import stdin, stdout
ans = []
stdin.readline()
for ai in map(int, map(int, stdin.readline().split())):
cnt = 1
while ans and ai * ans[-1][0] <= ans[-1][1] * cnt:
c, r = ans.pop()
ai += r
cnt += c
ans.append((cnt, ai))
for i, res in ans:
m = str(res / i)
stdout.write((m + "\n") * i)
main() | FUNC_DEF ASSIGN VAR LIST EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR STRING VAR EXPR FUNC_CALL VAR |
There are n water tanks in a row, i-th of them contains a_i liters of water. The tanks are numbered from 1 to n from left to right.
You can perform the following operation: choose some subsegment [l, r] (1≤ l ≤ r ≤ n), and redistribute water in tanks l, l+1, ..., r evenly. In other words, replace each of a_l, a_{l+1}, ..., a_r by \frac{a_l + a_{l+1} + ... + a_r}{r-l+1}. For example, if for volumes [1, 3, 6, 7] you choose l = 2, r = 3, new volumes of water will be [1, 4.5, 4.5, 7]. You can perform this operation any number of times.
What is the lexicographically smallest sequence of volumes of water that you can achieve?
As a reminder:
A sequence a is lexicographically smaller than a sequence b of the same length if and only if the following holds: in the first (leftmost) position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains an integer n (1 ≤ n ≤ 10^6) — the number of water tanks.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6) — initial volumes of water in the water tanks, in liters.
Because of large input, reading input as doubles is not recommended.
Output
Print the lexicographically smallest sequence you can get. In the i-th line print the final volume of water in the i-th tank.
Your answer is considered correct if the absolute or relative error of each a_i does not exceed 10^{-9}.
Formally, let your answer be a_1, a_2, ..., a_n, and the jury's answer be b_1, b_2, ..., b_n. Your answer is accepted if and only if \frac{|a_i - b_i|}{max{(1, |b_i|)}} ≤ 10^{-9} for each i.
Examples
Input
4
7 5 5 7
Output
5.666666667
5.666666667
5.666666667
7.000000000
Input
5
7 8 8 10 12
Output
7.000000000
8.000000000
8.000000000
10.000000000
12.000000000
Input
10
3 9 5 5 1 7 5 3 8 7
Output
3.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
7.500000000
7.500000000
Note
In the first sample, you can get the sequence by applying the operation for subsegment [1, 3].
In the second sample, you can't get any lexicographically smaller sequence. | import sys
def input():
return sys.stdin.readline().strip()
def list2d(a, b, c):
return [([c] * b) for i in range(a)]
def list3d(a, b, c, d):
return [[([d] * c) for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e):
return [[[([e] * d) for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1):
return int(-(-x // y))
def INT():
return int(input())
def MAP():
return map(int, input().split())
def LIST(N=None):
return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes():
print("Yes")
def No():
print("No")
def YES():
print("YES")
def NO():
print("NO")
INF = 10**18
MOD = 10**9 + 7
N = INT()
A = LIST()
stack = []
for i, a in enumerate(A):
stack.append((a, 1))
while len(stack) >= 2 and stack[-2][0] > stack[-1][0]:
a1, cnt1 = stack.pop()
a2, cnt2 = stack.pop()
merged = (a1 * cnt1 + a2 * cnt2) / (cnt1 + cnt2)
stack.append((merged, cnt1 + cnt2))
ans = []
for a, cnt in stack:
ans += [str(a)] * cnt
print("\n".join(ans)) | IMPORT FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN BIN_OP LIST VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF RETURN BIN_OP LIST VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF RETURN BIN_OP LIST VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF NUMBER RETURN FUNC_CALL VAR BIN_OP VAR VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF NONE RETURN VAR NONE FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF EXPR FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NUMBER WHILE FUNC_CALL VAR VAR NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR LIST FOR VAR VAR VAR VAR BIN_OP LIST FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR |
There are n water tanks in a row, i-th of them contains a_i liters of water. The tanks are numbered from 1 to n from left to right.
You can perform the following operation: choose some subsegment [l, r] (1≤ l ≤ r ≤ n), and redistribute water in tanks l, l+1, ..., r evenly. In other words, replace each of a_l, a_{l+1}, ..., a_r by \frac{a_l + a_{l+1} + ... + a_r}{r-l+1}. For example, if for volumes [1, 3, 6, 7] you choose l = 2, r = 3, new volumes of water will be [1, 4.5, 4.5, 7]. You can perform this operation any number of times.
What is the lexicographically smallest sequence of volumes of water that you can achieve?
As a reminder:
A sequence a is lexicographically smaller than a sequence b of the same length if and only if the following holds: in the first (leftmost) position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains an integer n (1 ≤ n ≤ 10^6) — the number of water tanks.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6) — initial volumes of water in the water tanks, in liters.
Because of large input, reading input as doubles is not recommended.
Output
Print the lexicographically smallest sequence you can get. In the i-th line print the final volume of water in the i-th tank.
Your answer is considered correct if the absolute or relative error of each a_i does not exceed 10^{-9}.
Formally, let your answer be a_1, a_2, ..., a_n, and the jury's answer be b_1, b_2, ..., b_n. Your answer is accepted if and only if \frac{|a_i - b_i|}{max{(1, |b_i|)}} ≤ 10^{-9} for each i.
Examples
Input
4
7 5 5 7
Output
5.666666667
5.666666667
5.666666667
7.000000000
Input
5
7 8 8 10 12
Output
7.000000000
8.000000000
8.000000000
10.000000000
12.000000000
Input
10
3 9 5 5 1 7 5 3 8 7
Output
3.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
7.500000000
7.500000000
Note
In the first sample, you can get the sequence by applying the operation for subsegment [1, 3].
In the second sample, you can't get any lexicographically smaller sequence. | def main():
n, *a = map(int, open(0).read().split())
d = n + 1
b = [a[0] * d + 1]
for i in range(1, n):
if b[-1] // d == a[i]:
b[-1] += 1
else:
b.append(a[i] * d + 1)
bx, bi = divmod(b[0], d)
bx *= bi
s = [bx * d + bi]
for ii in range(1, len(b)):
cx, ci = divmod(b[ii], d)
cx *= ci
s.append(cx * d + ci)
bx, bi = divmod(s[-2], d)
cx, ci = divmod(s[-1], d)
while cx * bi < bx * ci:
s.pop()
s[-1] = (bx + cx) * d + bi + ci
if len(s) == 1:
break
bx, bi = divmod(s[-2], d)
cx, ci = divmod(s[-1], d)
ans = []
for ii in range(len(s)):
sx, si = divmod(s[ii], d)
ans += [str(sx / si)] * si
print(" ".join(ans))
main() | FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR LIST BIN_OP BIN_OP VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP VAR NUMBER VAR VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR ASSIGN VAR LIST BIN_OP BIN_OP VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR VAR FUNC_CALL VAR VAR NUMBER VAR WHILE BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP LIST FUNC_CALL VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR |
There are n water tanks in a row, i-th of them contains a_i liters of water. The tanks are numbered from 1 to n from left to right.
You can perform the following operation: choose some subsegment [l, r] (1≤ l ≤ r ≤ n), and redistribute water in tanks l, l+1, ..., r evenly. In other words, replace each of a_l, a_{l+1}, ..., a_r by \frac{a_l + a_{l+1} + ... + a_r}{r-l+1}. For example, if for volumes [1, 3, 6, 7] you choose l = 2, r = 3, new volumes of water will be [1, 4.5, 4.5, 7]. You can perform this operation any number of times.
What is the lexicographically smallest sequence of volumes of water that you can achieve?
As a reminder:
A sequence a is lexicographically smaller than a sequence b of the same length if and only if the following holds: in the first (leftmost) position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains an integer n (1 ≤ n ≤ 10^6) — the number of water tanks.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6) — initial volumes of water in the water tanks, in liters.
Because of large input, reading input as doubles is not recommended.
Output
Print the lexicographically smallest sequence you can get. In the i-th line print the final volume of water in the i-th tank.
Your answer is considered correct if the absolute or relative error of each a_i does not exceed 10^{-9}.
Formally, let your answer be a_1, a_2, ..., a_n, and the jury's answer be b_1, b_2, ..., b_n. Your answer is accepted if and only if \frac{|a_i - b_i|}{max{(1, |b_i|)}} ≤ 10^{-9} for each i.
Examples
Input
4
7 5 5 7
Output
5.666666667
5.666666667
5.666666667
7.000000000
Input
5
7 8 8 10 12
Output
7.000000000
8.000000000
8.000000000
10.000000000
12.000000000
Input
10
3 9 5 5 1 7 5 3 8 7
Output
3.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
7.500000000
7.500000000
Note
In the first sample, you can get the sequence by applying the operation for subsegment [1, 3].
In the second sample, you can't get any lexicographically smaller sequence. | from sys import gettrace, stdin, stdout
if not gettrace():
def input():
return next(stdin)[:-1]
def main():
n = int(input())
aa = [int(a) for a in input().split()]
sum = [0]
for a in aa:
sum.append(sum[-1] + a)
def mean(i, j):
return (sum[j] - sum[i]) / (j - i)
stk = []
for i in range(n):
stk.append((i, i + 1))
while len(stk) > 1 and mean(*stk[-2]) >= mean(*stk[-1]):
_, j = stk.pop()
i, _ = stk.pop()
stk.append((i, j))
for i, j in stk:
v = str(mean(i, j)) + "\n"
for k in range(j - i):
stdout.write(v)
main() | IF FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST NUMBER FOR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_DEF RETURN BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER WHILE FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR VAR STRING FOR VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
There are n water tanks in a row, i-th of them contains a_i liters of water. The tanks are numbered from 1 to n from left to right.
You can perform the following operation: choose some subsegment [l, r] (1≤ l ≤ r ≤ n), and redistribute water in tanks l, l+1, ..., r evenly. In other words, replace each of a_l, a_{l+1}, ..., a_r by \frac{a_l + a_{l+1} + ... + a_r}{r-l+1}. For example, if for volumes [1, 3, 6, 7] you choose l = 2, r = 3, new volumes of water will be [1, 4.5, 4.5, 7]. You can perform this operation any number of times.
What is the lexicographically smallest sequence of volumes of water that you can achieve?
As a reminder:
A sequence a is lexicographically smaller than a sequence b of the same length if and only if the following holds: in the first (leftmost) position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains an integer n (1 ≤ n ≤ 10^6) — the number of water tanks.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6) — initial volumes of water in the water tanks, in liters.
Because of large input, reading input as doubles is not recommended.
Output
Print the lexicographically smallest sequence you can get. In the i-th line print the final volume of water in the i-th tank.
Your answer is considered correct if the absolute or relative error of each a_i does not exceed 10^{-9}.
Formally, let your answer be a_1, a_2, ..., a_n, and the jury's answer be b_1, b_2, ..., b_n. Your answer is accepted if and only if \frac{|a_i - b_i|}{max{(1, |b_i|)}} ≤ 10^{-9} for each i.
Examples
Input
4
7 5 5 7
Output
5.666666667
5.666666667
5.666666667
7.000000000
Input
5
7 8 8 10 12
Output
7.000000000
8.000000000
8.000000000
10.000000000
12.000000000
Input
10
3 9 5 5 1 7 5 3 8 7
Output
3.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
7.500000000
7.500000000
Note
In the first sample, you can get the sequence by applying the operation for subsegment [1, 3].
In the second sample, you can't get any lexicographically smaller sequence. | def main():
n = int(input().strip())
A = [int(s) for s in input().strip().split()]
stack = []
for a in A:
k = 1
while stack and stack[-1][0] * k > a * stack[-1][1]:
b, m = stack.pop()
a += b
k += m
stack.append((a, k))
for b, m in stack:
avg = f"{b / m:.10f}"
for _ in range(m):
print(avg)
main() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR VAR ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR STRING FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
There are n water tanks in a row, i-th of them contains a_i liters of water. The tanks are numbered from 1 to n from left to right.
You can perform the following operation: choose some subsegment [l, r] (1≤ l ≤ r ≤ n), and redistribute water in tanks l, l+1, ..., r evenly. In other words, replace each of a_l, a_{l+1}, ..., a_r by \frac{a_l + a_{l+1} + ... + a_r}{r-l+1}. For example, if for volumes [1, 3, 6, 7] you choose l = 2, r = 3, new volumes of water will be [1, 4.5, 4.5, 7]. You can perform this operation any number of times.
What is the lexicographically smallest sequence of volumes of water that you can achieve?
As a reminder:
A sequence a is lexicographically smaller than a sequence b of the same length if and only if the following holds: in the first (leftmost) position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains an integer n (1 ≤ n ≤ 10^6) — the number of water tanks.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6) — initial volumes of water in the water tanks, in liters.
Because of large input, reading input as doubles is not recommended.
Output
Print the lexicographically smallest sequence you can get. In the i-th line print the final volume of water in the i-th tank.
Your answer is considered correct if the absolute or relative error of each a_i does not exceed 10^{-9}.
Formally, let your answer be a_1, a_2, ..., a_n, and the jury's answer be b_1, b_2, ..., b_n. Your answer is accepted if and only if \frac{|a_i - b_i|}{max{(1, |b_i|)}} ≤ 10^{-9} for each i.
Examples
Input
4
7 5 5 7
Output
5.666666667
5.666666667
5.666666667
7.000000000
Input
5
7 8 8 10 12
Output
7.000000000
8.000000000
8.000000000
10.000000000
12.000000000
Input
10
3 9 5 5 1 7 5 3 8 7
Output
3.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
7.500000000
7.500000000
Note
In the first sample, you can get the sequence by applying the operation for subsegment [1, 3].
In the second sample, you can't get any lexicographically smaller sequence. | import sys
def main():
import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
ans = []
for i, ai in enumerate(a):
res = ai
cnt = 1
for c, s in reversed(ans):
if (res + s) * c > s * (c + cnt):
break
res += s
cnt += c
ans.pop()
ans.append((cnt, res))
for i, res in ans:
m = str(res / i)
print("\n".join(m for _ in range(i)))
main() | IMPORT FUNC_DEF IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR BIN_OP VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
There are n water tanks in a row, i-th of them contains a_i liters of water. The tanks are numbered from 1 to n from left to right.
You can perform the following operation: choose some subsegment [l, r] (1≤ l ≤ r ≤ n), and redistribute water in tanks l, l+1, ..., r evenly. In other words, replace each of a_l, a_{l+1}, ..., a_r by \frac{a_l + a_{l+1} + ... + a_r}{r-l+1}. For example, if for volumes [1, 3, 6, 7] you choose l = 2, r = 3, new volumes of water will be [1, 4.5, 4.5, 7]. You can perform this operation any number of times.
What is the lexicographically smallest sequence of volumes of water that you can achieve?
As a reminder:
A sequence a is lexicographically smaller than a sequence b of the same length if and only if the following holds: in the first (leftmost) position where a and b differ, the sequence a has a smaller element than the corresponding element in b.
Input
The first line contains an integer n (1 ≤ n ≤ 10^6) — the number of water tanks.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6) — initial volumes of water in the water tanks, in liters.
Because of large input, reading input as doubles is not recommended.
Output
Print the lexicographically smallest sequence you can get. In the i-th line print the final volume of water in the i-th tank.
Your answer is considered correct if the absolute or relative error of each a_i does not exceed 10^{-9}.
Formally, let your answer be a_1, a_2, ..., a_n, and the jury's answer be b_1, b_2, ..., b_n. Your answer is accepted if and only if \frac{|a_i - b_i|}{max{(1, |b_i|)}} ≤ 10^{-9} for each i.
Examples
Input
4
7 5 5 7
Output
5.666666667
5.666666667
5.666666667
7.000000000
Input
5
7 8 8 10 12
Output
7.000000000
8.000000000
8.000000000
10.000000000
12.000000000
Input
10
3 9 5 5 1 7 5 3 8 7
Output
3.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
7.500000000
7.500000000
Note
In the first sample, you can get the sequence by applying the operation for subsegment [1, 3].
In the second sample, you can't get any lexicographically smaller sequence. | import sys
input = sys.stdin.buffer.readline
n = int(input())
a = list(map(int, input().split()))
stk = []
for i in a:
now_sum = i
now_cnt = 1
while stk and now_sum * stk[-1][1] < stk[-1][0] * now_cnt:
temp_sum, temp_cnt = stk.pop()
now_sum += temp_sum
now_cnt += temp_cnt
stk.append((now_sum, now_cnt))
for i in stk:
temp_ans = str(i[0] / i[1])
for j in range(i[1]):
print(temp_ans) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | len, str = int(input()), input()
s = [(ord(i) - ord("a")) for i in str]
rev = s[::-1]
appe = [[] for i in range(ord("a"), ord("z") + 1)]
for i, it in enumerate(rev):
appe[it].append(i)
nxt, arr = [(0) for i in range(ord("a"), ord("z") + 1)], []
for it in s:
arr.append(appe[it][nxt[it]] + 1)
nxt[it] += 1
maxn = len + 50
fwt = [(0) for i in range(maxn)]
def Add(val):
while val < maxn:
fwt[val] += 1
val += val & -val
def Query(val):
ans = 0
while val > 0:
ans -= fwt[val]
val -= val & -val
val = maxn - 1
while val > 0:
ans += fwt[val]
val -= val & -val
return ans
cnt = 0
for it in arr:
cnt += Query(it)
Add(it)
print(cnt) | ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR FUNC_CALL VAR STRING BIN_OP FUNC_CALL VAR STRING NUMBER FOR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR STRING BIN_OP FUNC_CALL VAR STRING NUMBER LIST FOR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR FUNC_DEF WHILE VAR VAR VAR VAR NUMBER VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR BIN_OP VAR VAR RETURN VAR ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | n = int(input())
s = input()
SZ = 30
lim = n + 233
g = [[] for i in range(SZ)]
a = [0] * SZ
for i in range(n):
cur = ord(s[i]) - 97
g[cur].append(i)
bit = [0] * (lim + 1)
def lowbit(x):
return x & -x
def add(pos, val):
while pos <= lim:
bit[pos] += val
pos += lowbit(pos)
def query(pos):
res = 0
while pos:
res += bit[pos]
pos -= lowbit(pos)
return res
ans = 0
for i in range(n - 1, -1, -1):
cur = ord(s[i]) - 97
val = g[cur][a[cur]]
ans += val - query(val + 1)
a[cur] += 1
add(val + 1, 1)
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FUNC_DEF RETURN BIN_OP VAR VAR FUNC_DEF WHILE VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR VAR FUNC_CALL VAR VAR RETURN VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR BIN_OP VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | def mergesort(l, n):
if n == 1:
return l, 0
else:
a, ans1 = mergesort(l[0 : n // 2], n // 2)
b, ans2 = mergesort(l[n // 2 :], n - n // 2)
c, ans3 = combine(a, n // 2, b, n - n // 2)
return c, ans1 + ans2 + ans3
def combine(l1, n, l2, m):
i = 0
j = 0
ans = []
inversions = 0
while i < n and j < m:
if l1[i] <= l2[j]:
ans.append(l1[i])
i += 1
else:
ans.append(l2[j])
inversions += n - i
j += 1
if i == n:
for k in range(j, m):
ans.append(l2[k])
elif j == m:
for k in range(i, n):
ans.append(l1[k])
return ans, inversions
n = int(input())
start = input()
s = start[::-1]
loc = {}
for i in range(n):
if s[i] in loc:
loc[s[i]].append(i)
else:
loc[s[i]] = [i]
new = {}
for i in loc:
new[i] = loc[i][::-1]
arr = []
for i in range(n):
arr.append(new[start[i]].pop())
print(mergesort(arr, len(arr))[1]) | FUNC_DEF IF VAR NUMBER RETURN VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR BIN_OP VAR NUMBER RETURN VAR BIN_OP BIN_OP VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR VAR NUMBER IF VAR VAR FOR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR VAR FOR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR RETURN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR DICT FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR VAR LIST VAR ASSIGN VAR DICT FOR VAR VAR ASSIGN VAR VAR VAR VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | class BIT:
def __init__(self, n):
self.n = n
self.data = [0] * (n + 1)
def to_sum(self, i):
s = 0
while i > 0:
s += self.data[i]
i -= i & -i
return s
def add(self, i, x):
while i <= self.n:
self.data[i] += x
i += i & -i
def get(self, i, j):
return self.to_sum(j) - self.to_sum(i - 1)
def main():
n = int(input())
s = input()
b = s[::-1]
alphabet = list("abcdefghijklmnopqrstuvwxyz")
before = {v: [] for v in alphabet}
after = {v: [] for v in alphabet}
for i in range(n):
before[s[i]].append(i)
after[b[i]].append(i)
data = []
for v in alphabet:
M = len(before[v])
for j in range(M):
old, new = before[v][j], after[v][j]
data.append((old, new + 1))
data.sort(reverse=True)
ans = 0
bit = BIT(n)
for x, v in data:
ans += bit.to_sum(v)
bit.add(v, 1)
print(ans)
return
main() | CLASS_DEF FUNC_DEF ASSIGN VAR VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR BIN_OP VAR VAR RETURN VAR FUNC_DEF WHILE VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR FUNC_DEF RETURN BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR VAR LIST VAR VAR ASSIGN VAR VAR LIST VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR LIST FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN EXPR FUNC_CALL VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | import sys
input = sys.stdin.readline
def inversions(x):
if len(x) == 1:
return x, 0
a, b, both = x[: len(x) // 2], x[len(x) // 2 :], []
a, a_inv = inversions(a)
b, b_inv = inversions(b)
inv, i, j = a_inv + b_inv, 0, 0
while i < len(a) and j < len(b):
if a[i] <= b[j]:
both.append(a[i])
i += 1
else:
both.append(b[j])
j += 1
inv += len(a) - i
both += a[i:]
both += b[j:]
return both, inv
n, s, pos, swaps = int(input()), input().strip()[::-1], {i: [] for i in range(26)}, 0
for i in range(n):
pos[ord(s[i]) - ord("a")].append(i)
order = [pos[ord(s[i]) - ord("a")].pop() for i in range(n)][::-1]
print(inversions(order)[1]) | IMPORT ASSIGN VAR VAR FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN VAR NUMBER ASSIGN VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER LIST ASSIGN VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR NUMBER NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR RETURN VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR NUMBER VAR LIST VAR FUNC_CALL VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING VAR ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | def getNext(num):
return num + (num & -num)
def getParent(num):
return num - (num & -num)
def getSum(ftree, index):
s = 0
i = index
while i > 0:
s += ftree[i]
i = getParent(i)
return s
n = int(input())
s = input()
a = [[] for i in range(26)]
b = []
c = [0] * 26
for i in range(n):
a[ord(s[i]) - 97].append(i + 1)
for i in range(n - 1, -1, -1):
cr = ord(s[i]) - 97
b.append(n + 1 - a[cr][c[cr]])
c[cr] += 1
ls = [0] * (n + 1)
ans = 0
for i in b:
ans += getSum(ls, i)
ind = i
while ind < n + 1:
ls[ind] += 1
ind = getNext(ind)
print(ans) | FUNC_DEF RETURN BIN_OP VAR BIN_OP VAR VAR FUNC_DEF RETURN BIN_OP VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER ASSIGN VAR LIST ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR WHILE VAR BIN_OP VAR NUMBER VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | n = int(input())
s = input()
s1 = s[-1::-1]
d = [[] for _ in range(26)]
for i in range(n):
d[ord(s1[i]) - 97].append(i)
dp = []
mp = [0] * 26
for i in range(n):
tp = ord(s[i]) - 97
dp.append(d[tp][mp[tp]])
mp[tp] += 1
LEN = n + 1
BIT = [0] * (LEN + 1)
def update(v, w):
while v <= LEN:
BIT[v] += w
v += v & -v
def getvalue(v):
ANS = 0
while v != 0:
ANS += BIT[v]
v -= v & -v
return ANS
ans = 0
for i in range(n - 1, -1, -1):
ans += getvalue(dp[i] + 1)
update(dp[i] + 1, 1)
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR ASSIGN VAR LIST ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FUNC_DEF WHILE VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR BIN_OP VAR VAR RETURN VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | N = int(input())
S = input()
dic = {}
for i, s in enumerate(S):
if s in dic:
dic[s].append(i)
else:
dic[s] = [i]
X = list(range(N - 1, -1, -1))
for c, idcs in dic.items():
for p, q in zip(idcs, reversed(idcs)):
X[p] = N - 1 - q
class BIT:
def __init__(self, N):
self.N = N
self.T = [0] * (N + 1)
def add(self, i, x):
i += 1
while i <= self.N:
self.T[i] += x
i += i & -i
def _sum(self, i):
ret = 0
while i > 0:
ret += self.T[i]
i ^= i & -i
return ret
def sum(self, l, r):
return self._sum(r) - self._sum(l)
def lower_bound(self, w):
if w <= 0:
return 0
x = 0
k = 1 << self.N.bit_length()
while k:
if x + k <= self.N and self.T[x + k] < w:
w -= self.t[x + k]
x += k
k >>= 1
return x + 1
b = BIT(N)
ans = 0
for i, x in enumerate(X):
ans += i - b._sum(x + 1)
b.add(x, 1)
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR DICT FOR VAR VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR LIST VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER FOR VAR VAR FUNC_CALL VAR FOR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP BIN_OP VAR NUMBER VAR CLASS_DEF FUNC_DEF ASSIGN VAR VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FUNC_DEF VAR NUMBER WHILE VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR BIN_OP VAR VAR RETURN VAR FUNC_DEF RETURN BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER FUNC_CALL VAR WHILE VAR IF BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER RETURN BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | class BIT:
def __init__(self, n):
self.BIT = [0] * (n + 1)
self.num = n
def query(self, idx):
res_sum = 0
while idx > 0:
res_sum += self.BIT[idx]
idx -= idx & -idx
return res_sum
def update(self, idx, x):
while idx <= self.num:
self.BIT[idx] += x
idx += idx & -idx
return
al = [chr(97 + i) for i in range(26)]
n = int(input())
s = input()
s = [s[i] for i in range(n)]
goal = s[::-1]
dic = {a: [] for a in al}
for i in range(n):
dic[goal[i]].append(i)
for a in dic:
dic[a] = dic[a][::-1]
seq = [(-1) for i in range(n)]
for i in range(n):
seq[i] = dic[s[i]].pop() + 1
res = 0
bit = BIT(n)
for i in range(n):
res += i - bit.query(seq[i])
bit.update(seq[i], 1)
print(res) | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_DEF ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR BIN_OP VAR VAR RETURN VAR FUNC_DEF WHILE VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR RETURN ASSIGN VAR FUNC_CALL VAR BIN_OP NUMBER VAR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR LIST VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR FOR VAR VAR ASSIGN VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | import sys
sys.setrecursionlimit(10**5)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II():
return int(sys.stdin.buffer.readline())
def MI():
return map(int, sys.stdin.buffer.readline().split())
def LI():
return list(map(int, sys.stdin.buffer.readline().split()))
def LLI(rows_number):
return [LI() for _ in range(rows_number)]
def BI():
return sys.stdin.buffer.readline().rstrip()
def SI():
return sys.stdin.buffer.readline().rstrip().decode()
class BitSum:
def __init__(self, n):
self.n = n + 1
self.table = [0] * self.n
def add(self, i, x):
i += 1
while i < self.n:
self.table[i] += x
i += i & -i
def sum(self, i):
i += 1
res = 0
while i > 0:
res += self.table[i]
i -= i & -i
return res
n = II()
s = BI()
rs = s[::-1]
pos = [[] for _ in range(26)]
for i in range(n - 1, -1, -1):
c = rs[i] - 97
pos[c].append(i)
aa = []
for c in s:
c -= 97
i = pos[c].pop()
aa.append(i)
bit = BitSum(n)
ans = 0
for i, a in enumerate(aa):
ans += i - bit.sum(a)
bit.add(a, 1)
print(ans) | IMPORT EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR STRING FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL FUNC_CALL FUNC_CALL VAR CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR FUNC_DEF VAR NUMBER WHILE VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR FUNC_DEF VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR BIN_OP VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST FOR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | def mergeSort(arr, n):
temp_arr = [0] * n
return _mergeSort(arr, temp_arr, 0, n - 1)
def _mergeSort(arr, temp_arr, left, right):
inv_count = 0
if left < right:
mid = (left + right) // 2
inv_count += _mergeSort(arr, temp_arr, left, mid)
inv_count += _mergeSort(arr, temp_arr, mid + 1, right)
inv_count += merge(arr, temp_arr, left, mid, right)
return inv_count
def merge(arr, temp_arr, left, mid, right):
i = left
j = mid + 1
k = left
inv_count = 0
while i <= mid and j <= right:
if arr[i] <= arr[j]:
temp_arr[k] = arr[i]
k += 1
i += 1
else:
temp_arr[k] = arr[j]
inv_count += mid - i + 1
k += 1
j += 1
while i <= mid:
temp_arr[k] = arr[i]
k += 1
i += 1
while j <= right:
temp_arr[k] = arr[j]
k += 1
j += 1
for loop_var in range(left, right + 1):
arr[loop_var] = temp_arr[loop_var]
return inv_count
n = int(input())
string = input()
occurences = [list() for _ in range(26)]
for i in range(n):
pos = ord(string[i]) - ord("a")
occurences[pos].append(i)
reversed = list()
for i in range(n - 1, -1, -1):
reversed.append(string[i])
inversion = [0] * n
for i in range(n - 1, -1, -1):
pos = ord(reversed[i]) - ord("a")
inversion[i] = occurences[pos].pop()
print(mergeSort(inversion, n)) | FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR RETURN FUNC_CALL VAR VAR VAR NUMBER BIN_OP VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR IF VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER VAR NUMBER VAR NUMBER WHILE VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER VAR NUMBER WHILE VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING ASSIGN VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | def build(v, l, r):
nonlocal tree
if r - l == 1:
tree[v] = u[l]
else:
m = (l + r) // 2
build(2 * v + 1, l, m)
build(2 * v + 2, m, r)
tree[v] = tree[2 * v + 1] + tree[2 * v + 2]
def rsq(v, l, r, a, b):
nonlocal tree
if l >= b or a >= r:
return 0
if l >= a and r <= b:
return tree[v]
m = (l + r) // 2
return rsq(2 * v + 1, l, m, a, b) + rsq(2 * v + 2, m, r, a, b)
def update(v, l, r, i, x):
nonlocal tree
if i < l or i >= r:
return
if r - l == 1:
tree[v] = x
else:
m = (l + r) // 2
update(2 * v + 1, l, m, i, x)
update(2 * v + 2, m, r, i, x)
tree[v] = tree[2 * v + 1] + tree[2 * v + 2]
n = int(input())
a = list(input())
sz = 1
while sz < n:
sz *= 2
tree = [0] * (2 * sz - 1)
u = [0] * n
while len(u) < sz:
u.append(0)
build(0, 0, sz)
ss = []
for i in range(26):
ss.append([])
for i in range(n):
a[i] = ord(a[i]) - ord("a")
ss[a[i]].append(i)
a.reverse()
ans = 0
for i in range(n - 1, -1, -1):
L, R = ss[a[i]][-1], i
ans += R - L + rsq(0, 0, sz, 0, L)
update(0, 0, sz, L, 1)
ss[a[i]].pop()
print(ans) | FUNC_DEF IF BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER FUNC_DEF IF VAR VAR VAR VAR RETURN NUMBER IF VAR VAR VAR VAR RETURN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER RETURN BIN_OP FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR VAR VAR VAR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR VAR VAR VAR FUNC_DEF IF VAR VAR VAR VAR RETURN IF BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR WHILE FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER VAR VAR BIN_OP BIN_OP VAR VAR FUNC_CALL VAR NUMBER NUMBER VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | n = int(input())
A = input()
flippedA = A[::-1]
revs = [(i + 1) for i in range(len(A))]
freqs = dict()
for i in range(len(flippedA)):
if flippedA[i] not in freqs:
freqs[flippedA[i]] = [i + 1]
else:
freqs[flippedA[i]].append(i + 1)
origs = []
ptr = [(0) for i in range(26)]
for i in range(len(A)):
origs.append(freqs[A[i]][ptr[ord(A[i]) - ord("a")]])
ptr[ord(A[i]) - ord("a")] += 1
def inversions(A):
ans = 0
for i in range(len(A)):
for j in range(i + 1, len(A)):
if A[i] > A[j]:
ans += 1
return ans
fenw = [(0) for i in range(n + 5)]
def update(idx, val):
while idx <= n + 1:
fenw[idx] += val
idx += idx & -idx
def query(idx):
ans = 0
while idx > 0:
ans += fenw[idx]
idx -= idx & -idx
return ans
def solve(l, r):
return query(r) - query(l - 1)
def inversions(A):
ans = 0
for i in range(n):
ans += solve(A[i] + 1, n)
update(A[i], 1)
return ans
print(inversions(origs)) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR VAR VAR LIST BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING NUMBER FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR NUMBER RETURN VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_DEF WHILE VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR BIN_OP VAR VAR RETURN VAR FUNC_DEF RETURN BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR NUMBER RETURN VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | import sys
mod = 1000000007
eps = 10**-9
def main():
import sys
input = sys.stdin.readline
class Bit:
def __init__(self, n):
self.size = n
self.tree = [0] * (n + 1)
def sum(self, i):
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, i, x):
while i <= self.size:
self.tree[i] += x
i += i & -i
def lower_bound(self, w):
if w <= 0:
return 0
x = 0
k = 1 << self.size.bit_length() - 1
while k:
if x + k <= self.size and self.tree[x + k] < w:
w -= self.tree[x + k]
x += k
k >>= 1
return x + 1
N = int(input())
S = input().rstrip("\n")
S_rev = S[::-1]
idx = [[] for _ in range(26)]
for i, s in enumerate(S):
idx[ord(s) - 97].append(i)
cnt = [0] * 26
A = [0] * N
for i, s in enumerate(S_rev):
si = ord(s) - 97
A[idx[si][cnt[si]]] = i
cnt[si] += 1
a2i = {a: (i + 1) for i, a in enumerate(A)}
bit = Bit(N)
ans = 0
for a in range(N - 1, -1, -1):
ans += bit.sum(a2i[a])
bit.add(a2i[a], 1)
print(ans)
main() | IMPORT ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER FUNC_DEF IMPORT ASSIGN VAR VAR CLASS_DEF FUNC_DEF ASSIGN VAR VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR BIN_OP VAR VAR RETURN VAR FUNC_DEF WHILE VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR FUNC_DEF IF VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER BIN_OP FUNC_CALL VAR NUMBER WHILE VAR IF BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR NUMBER RETURN BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | def merge(a, b):
i = 0
j = 0
c = 0
ans = []
while i < len(a) and j < len(b):
if a[i] < b[j]:
ans.append(a[i])
i += 1
else:
ans.append(b[j])
c += len(a) - i
j += 1
ans += a[i:]
ans += b[j:]
return ans, c
def mergesort(a):
if len(a) == 1:
return a, 0
mid = len(a) // 2
left, left_inversion = mergesort(a[:mid])
right, right_inversion = mergesort(a[mid:])
m, c = merge(left, right)
c += left_inversion + right_inversion
return m, c
n = int(input())
s = input()
a = [0] * n
ind = [[] for i in range(26)]
for i in range(n):
ind[ord(s[i]) - ord("a")].append(n - i)
for i in range(n):
a[i] = ind[ord(s[i]) - ord("a")].pop()
m, c = mergesort(a)
print(c) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR VAR VAR VAR VAR RETURN VAR VAR FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR RETURN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING ASSIGN VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | from sys import stdin, stdout
def string_reversal(n, s):
a = [[] for _ in range(26)]
bit_a = [0] * (n + 1)
t = n
res = 0
for i in range(len(s)):
bit_update(bit_a, i, 1)
a[ord(s[i]) - ord("a")].append(i)
for i in range(len(s)):
j = a[ord(s[i]) - ord("a")].pop()
res += t - bit_get(bit_a, j)
bit_update(bit_a, j, -1)
t -= 1
return res
def bit_update(bit_a, i, delta):
i += 1
while i < len(bit_a):
bit_a[i] += delta
i += i & -i
def bit_get(bit_a, i):
i += 1
v = 0
while i > 0:
v += bit_a[i]
i -= i & -i
return v
n = int(stdin.readline())
s = stdin.readline().strip()
r = string_reversal(n, s)
print(r) | FUNC_DEF ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING VAR BIN_OP VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER RETURN VAR FUNC_DEF VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR FUNC_DEF VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR BIN_OP VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | n = int(input())
s = list(str(input()))
t = list(reversed(s))
X = [[] for _ in range(26)]
Y = [[] for _ in range(26)]
for i, c in enumerate(s):
j = ord(c) - ord("a")
X[j].append(i)
for i, c in enumerate(t):
j = ord(c) - ord("a")
Y[j].append(i)
A = [-1] * n
for i in range(26):
for j, k in zip(X[i], Y[i]):
A[j] = k
class BIT:
def __init__(self, n):
self.n = n
self.bit = [0] * (self.n + 1)
def init(self, init_val):
for i, v in enumerate(init_val):
self.add(i, v)
def add(self, i, x):
i += 1
while i <= self.n:
self.bit[i] += x
i += i & -i
def sum(self, i, j):
return self._sum(j) - self._sum(i)
def _sum(self, i):
res = 0
while i > 0:
res += self.bit[i]
i -= i & -i
return res
def lower_bound(self, x):
s = 0
pos = 0
depth = self.n.bit_length()
v = 1 << depth
for i in range(depth, -1, -1):
k = pos + v
if k <= self.n and s + self.bit[k] < x:
s += self.bit[k]
pos += v
v >>= 1
return pos
bit = BIT(max(A) + 1)
ans = 0
for a in A:
ans += bit.sum(a + 1, bit.n)
bit.add(a, 1)
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR VAR FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR CLASS_DEF FUNC_DEF ASSIGN VAR VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FUNC_DEF FOR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR FUNC_DEF VAR NUMBER WHILE VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR FUNC_DEF RETURN BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR BIN_OP VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP NUMBER VAR FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | n = int(input())
s = input().strip()
character_indexes = [[] for indx in range(26)]
for indx in range(n):
char_to_append = ord(s[indx]) - 97
character_indexes[char_to_append].append(indx)
arr = [(0) for i in range(n + 2)]
result = 0
freqs = [(0) for i in range(26)]
for i in range(n - 1, -1, -1):
char_to_append = ord(s[i]) - 97
tmp = character_indexes[char_to_append][freqs[char_to_append]]
res = 0
first_tmp = tmp + 1
while first_tmp != 0:
res += arr[first_tmp]
first_tmp -= first_tmp & -first_tmp
result += tmp - res
freqs[char_to_append] += 1
second_tmp = tmp + 1
while second_tmp <= n:
arr[second_tmp] += 1
second_tmp += second_tmp & -second_tmp
print(result) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR VAR VAR NUMBER VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | import sys
input = sys.stdin.readline
n = int(input())
s = input().rstrip()
class FenwickTree:
def __init__(self, n):
self.t = [0] * (n + 1)
def update(self, i, x):
i += 1
while i < len(self.t):
self.t[i] += x
i += i & -i
def query(self, r):
ans = 0
while r:
ans += self.t[r]
r -= r & -r
return ans
rev = s[::-1]
indices = [[] for _ in range(26)]
def ind(x):
return ord(x) - ord("a")
for i, x in enumerate(s):
indices[ind(x)].append(i)
for x in indices:
x.reverse()
ft = FenwickTree(n)
for i in range(n):
ft.update(i, 1)
ans = 0
for x in rev:
ans += ft.query(indices[ind(x)][-1])
ft.update(indices[ind(x)][-1], -1)
indices[ind(x)].pop()
print(ans) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FUNC_DEF VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR VAR BIN_OP VAR VAR RETURN VAR ASSIGN VAR VAR NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER FUNC_DEF RETURN BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING FOR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | import sys
__lines = list(sys.stdin.readlines())
__cur_line = 0
def input():
global __lines, __cur_line
ans = __lines[__cur_line]
__cur_line += 1
return ans
class SegmentTree:
def __init__(self, n):
self.size = 1
while self.size < n:
self.size *= 2
self.vals = [0] * (2 * self.size)
def set(self, i, v):
return self._set(i, v, 0, 0, self.size)
def _set(self, i, v, x, lx, rx):
if rx - lx == 1:
self.vals[x] = v
else:
m = (lx + rx) // 2
if i < m:
self._set(i, v, x * 2 + 1, lx, m)
else:
self._set(i, v, x * 2 + 2, m, rx)
self.vals[x] = self.vals[2 * x + 1] + self.vals[2 * x + 2]
def range(self, l, r):
return self._range(l, r, 0, 0, self.size)
def _range(self, l, r, x, lx, rx):
if r <= lx or rx <= l:
return 0
if l <= lx and rx <= r:
return self.vals[x]
m = (lx + rx) // 2
s = 0
if l < m:
s += self._range(l, r, x * 2 + 1, lx, m)
if m < r:
s += self._range(l, r, x * 2 + 2, m, rx)
return s
n = int(input())
s = input().strip()
poss = [[] for _ in range(26)]
for i, c in enumerate(s):
poss[ord(c) - ord("a")].append(i)
target_pos = [0] * n
for l in poss:
l2 = [(n - 1 - i) for i in l][::-1]
for x1, x2 in zip(l, l2):
target_pos[x1] = x2
total = 0
st = SegmentTree(n)
for pos in target_pos:
total += st.range(pos + 1, n)
st.set(pos, 1)
print(total) | IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR VAR VAR VAR NUMBER RETURN VAR CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER WHILE VAR VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP NUMBER VAR FUNC_DEF RETURN FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR FUNC_DEF IF BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER FUNC_DEF RETURN FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR FUNC_DEF IF VAR VAR VAR VAR RETURN NUMBER IF VAR VAR VAR VAR RETURN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR IF VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR VAR VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | count = 0
def merge_count(arr):
if len(arr) == 1:
return arr
global count
mid = len(arr) // 2
l = merge_count(arr[:mid])
r = merge_count(arr[mid:])
new_arr = []
lp, rp = 0, 0
while lp < len(l) and rp < len(r):
if l[lp] <= r[rp]:
new_arr.append(l[lp])
lp += 1
else:
count += len(l) - lp
new_arr.append(r[rp])
rp += 1
while lp < len(l):
new_arr.append(l[lp])
lp += 1
while rp < len(r):
new_arr.append(r[rp])
rp += 1
return new_arr
def solve(n, s):
arr = [(0) for _ in s]
for chi in range(26):
indexes = []
ch = chr(ord("a") + chi)
for i in range(n):
if s[i] == ch:
indexes.append(i)
m = len(indexes)
for i in range(m):
x, y = indexes[i], n - 1 - indexes[m - 1 - i]
arr[x] = y
global count
count = 0
merge_count(arr)
return count
print(solve(int(input()), input())) | ASSIGN VAR NUMBER FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR VAR NUMBER NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR STRING VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | import sys
input = sys.stdin.readline
class Bit:
def __init__(self, n):
self.size = n
self.tree = [0] * (n + 1)
def sum(self, i):
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, i, x):
while i <= self.size:
self.tree[i] += x
i += i & -i
n = int(input())
s = list(input().rstrip())
rev_s = s[::-1]
p = [0] * n
cnt = [0] * 26
pos = [[] for i in range(26)]
for i in range(n):
pos[ord(s[i]) - ord("a")].append(i)
for i in range(n):
x = ord(rev_s[i]) - ord("a")
cur = pos[x][cnt[x]]
p[i] = cur
cnt[x] += 1
for i in range(n):
p[i] += 1
bit = Bit(n)
ans = 0
for i, x in enumerate(p):
bit.add(x, 1)
ans += i + 1 - bit.sum(x)
print(ans) | IMPORT ASSIGN VAR VAR CLASS_DEF FUNC_DEF ASSIGN VAR VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR BIN_OP VAR VAR RETURN VAR FUNC_DEF WHILE VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING ASSIGN VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | ans = 0
def count_inversion(arr):
global ans
if len(arr) <= 1:
return arr
mid = len(arr) // 2
left = count_inversion(arr[:mid])
right = count_inversion(arr[mid:])
final = []
i = 0
j = 0
while i < len(left) and j < len(right):
if left[i] <= right[j]:
final.append(left[i])
i += 1
else:
final.append(right[j])
ans += len(left) - i
j += 1
while i < len(left):
final.append(left[i])
i += 1
while j < len(right):
final.append(right[j])
j += 1
return final
n = int(input())
s = input()
position = [0] * n
for i in range(26):
temp = []
for j in range(len(s)):
if s[j] == chr(ord("a") + i):
temp.append(j)
for j in range(len(temp)):
curr = temp[j]
dest = n - (temp[len(temp) - j - 1] + 1)
position[curr] = dest
count_inversion(position)
print(ans) | ASSIGN VAR NUMBER FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR STRING VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | import sys
input = sys.stdin.readline
n = int(input())
S = input().strip()
LIST = [[] for i in range(26)]
for i in range(n):
LIST[ord(S[i]) - 97].append(i)
LEN = n + 1
BIT = [0] * (LEN + 1)
def update(v, w):
while v <= LEN:
BIT[v] += w
v += v & -v
def getvalue(v):
ANS = 0
while v != 0:
ANS += BIT[v]
v -= v & -v
return ANS
ANS = 0
moji = [0] * 26
for i in range(n - 1, -1, -1):
s = ord(S[i]) - 97
x = LIST[s][moji[s]]
ANS += x - getvalue(x + 1)
moji[s] += 1
update(x + 1, 1)
print(ANS) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FUNC_DEF WHILE VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR BIN_OP VAR VAR RETURN VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR BIN_OP VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | import sys
input = sys.stdin.readline
def segment_tree(ar, n):
data = [0] * n + ar.copy()
for i in range(n - 1, 0, -1):
data[i] += data[i * 2] + data[i * 2 + 1]
return data
def update(data, idx, n, value):
idx += n
data[idx] += value
while idx > 1:
idx //= 2
data[idx] += value
def summation(data, l, r, n):
l += n
r += n
maxi = 0
while l < r:
if l % 2 != 0:
maxi += data[l]
l += 1
if r % 2 != 0:
r -= 1
maxi += data[r]
l //= 2
r //= 2
return maxi
n = int(input())
st = list(input())[:-1]
ar = []
dic = {}
for i in range(n):
ar.append(1)
if st[i] in dic:
dic[st[i]].append(i)
else:
dic[st[i]] = [i]
ar[0] = 0
for i in dic:
dic[i] = dic[i][::-1]
rev = st[::-1]
data = segment_tree(ar, n)
ans = 0
for i in range(n):
cur = rev[i]
ind = dic[cur].pop()
ind1 = summation(data, 0, ind + 1, n)
tem = ind1 - i
ans += tem
if tem:
update(data, 0, n, 1)
update(data, ind, n, -ar[ind])
print(ans) | IMPORT ASSIGN VAR VAR FUNC_DEF ASSIGN VAR BIN_OP BIN_OP LIST NUMBER VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER RETURN VAR FUNC_DEF VAR VAR VAR VAR VAR WHILE VAR NUMBER VAR NUMBER VAR VAR VAR FUNC_DEF VAR VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR VAR VAR VAR NUMBER VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER ASSIGN VAR LIST ASSIGN VAR DICT FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR VAR LIST VAR ASSIGN VAR NUMBER NUMBER FOR VAR VAR ASSIGN VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR IF VAR EXPR FUNC_CALL VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
-----Input-----
The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.
The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.
-----Output-----
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
-----Examples-----
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
-----Note-----
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it. | import sys
input = sys.stdin.readline
inp, ip = lambda: int(input()), lambda: [int(w) for w in input().split()]
def update(i, val):
i += 1
while i <= n:
BIT[i] += val
i += i & -i
def getsum(i):
sm = 0
i += 1
while i:
sm += BIT[i]
i -= i & -i
return sm
n = inp()
s = input().strip()
if s == s[::-1]:
print(0)
exit()
dt = [[] for i in range(26)]
for i in range(n - 1, -1, -1):
dt[ord(s[i]) - 97].append(i)
BIT = [0] * (n + 1)
ans = 0
for i in range(n):
ind = dt[ord(s[n - i - 1]) - 97].pop(-1)
put = getsum(ind)
ans += ind - put
update(ind, 1)
print(ans) | IMPORT ASSIGN VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF VAR NUMBER WHILE VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR NUMBER VAR NUMBER WHILE VAR VAR VAR VAR VAR BIN_OP VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR LIST VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
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