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Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: if len(s) == 1: return 1 def splitter(str): for i in range(1, len(str)): start = str[0:i] end = str[i:] yield start, end for split in splitter(end): result = [start] result.extend(split) yield result maxnum = 0 for split in splitter(s): if len(set(split)) == len(split): maxnum = max(len(split), maxnum) if maxnum == 0: return self.maxUniqueSplit(s[: len(s) - 1]) else: return maxnum
CLASS_DEF FUNC_DEF VAR IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER FUNC_DEF FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER VAR ASSIGN VAR VAR VAR EXPR VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR EXPR FUNC_CALL VAR VAR EXPR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER RETURN FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: max_len = [1] def helper(i, s, vis): if i == len(s) and i != s: max_len[0] = max(max_len[0], len(vis)) return j = i while j < len(s): if s[i : j + 1] not in vis and s[i : j + 1] != s: vis[s[i : j + 1]] = 1 helper(j + 1, s, vis) del vis[s[i : j + 1]] j += 1 helper(0, s, {}) return max_len[0]
CLASS_DEF FUNC_DEF VAR ASSIGN VAR LIST NUMBER FUNC_DEF IF VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR RETURN ASSIGN VAR VAR WHILE VAR FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR DICT RETURN VAR NUMBER VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: options = [] options = self.recursiveSplit(s) bestcount = 0 for option in options: count = 0 encountered = {} valid = True for substr in option: if substr not in encountered: encountered[substr] = 1 count += 1 else: valid = False break if valid and count > bestcount: bestcount = count return bestcount def recursiveSplit(self, s: str) -> str: if len(s) == 1: return [[s]] elif len(s) == 0: return [[]] options = [] for i in range(0, len(s)): recursiveOptions = self.recursiveSplit(s[:i]) for option in recursiveOptions: newoption = option.copy() newoption.append(s[i:]) options.append(newoption) return options
CLASS_DEF FUNC_DEF VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR DICT ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR ASSIGN VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR RETURN VAR VAR FUNC_DEF VAR IF FUNC_CALL VAR VAR NUMBER RETURN LIST LIST VAR IF FUNC_CALL VAR VAR NUMBER RETURN LIST LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str, mem=None, seen=None) -> int: if mem is None: mem = {} if seen is None: seen = set() res = 0 for i in range(0, len(s)): if s[0 : i + 1] in seen: continue seen.add(s[0 : i + 1]) res = max(res, 1 + self.maxUniqueSplit(s[i + 1 :], mem, seen)) seen.remove(s[0 : i + 1]) mem[s] = res return res
CLASS_DEF FUNC_DEF VAR NONE NONE IF VAR NONE ASSIGN VAR DICT IF VAR NONE ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR NUMBER BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: seen = set() def helper(start): nonlocal seen, s length = len(s) if start == length: return 0 max_split, cur_split = 0, 0 for split in range(start + 1, length): left, right = s[start:split], s[split:length] if left in seen or right in seen or left == right: continue seen.add(left) cur_split = 1 + helper(split) seen.remove(left) max_split = max(max_split, cur_split) return max_split return 1 + helper(0)
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR RETURN NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR IF VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR RETURN BIN_OP NUMBER FUNC_CALL VAR NUMBER VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def helper(start): if len(s[start:]) == 0: return 0 res = -math.inf for i in range(start + 1, len(s) + 1): if s[start:i] not in seen: seen.add(s[start:i]) followup = helper(i) res = max(res, 1 + followup) seen.remove(s[start:i]) return res seen = set() return helper(0)
CLASS_DEF FUNC_DEF VAR FUNC_DEF IF FUNC_CALL VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR RETURN FUNC_CALL VAR NUMBER VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: n = len(s) mx = 1 m = n - 1 total = 1 << m state = 0 for state in range(total): seen = set() lo = 0 for i in range(m + 1): if state & 1 << i: token = s[lo : i + 1] if token in seen: break else: seen.add(token) lo = i + 1 else: token = s[lo:n] if token not in seen: seen.add(token) mx = max(len(seen), mx) return mx
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: book = set() def maxUniqueSplit(self, s): result = 0 for i in range(1, len(s) + 1): curr = s[:i] if curr not in self.book: self.book.add(curr) result = max(result, 1 + self.maxUniqueSplit(s[i:])) self.book.remove(curr) return result
CLASS_DEF ASSIGN VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: prev = [[s[0]]] for elm in s[1:]: current = [] for sub in prev: copy = list(sub) copy[-1] = copy[-1] + elm current.append(copy) copy = list(sub) copy.append(elm) current.append(copy) prev = current return max(len(set(combo)) for combo in prev)
CLASS_DEF FUNC_DEF VAR ASSIGN VAR LIST LIST VAR NUMBER FOR VAR VAR NUMBER ASSIGN VAR LIST FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: return self.dp(s, 0, 0, []) def dp(self, s, start, pos, ans): if pos == len(s): return len(ans) a = 0 b = self.dp(s, start, pos + 1, list(ans)) if s[start : pos + 1] not in ans: ans.append(s[start : pos + 1]) a = self.dp(s, pos + 1, pos + 1, list(ans)) return a if a > b else b
CLASS_DEF FUNC_DEF VAR RETURN FUNC_CALL VAR VAR NUMBER NUMBER LIST VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR VAR RETURN VAR VAR VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: n = len(s) @lru_cache(None) def dp(i, cur): nonlocal ans if len(cur) + n - i < ans: return if i >= n: ans = max(ans, len(cur)) l = n - i for k in range(1, l + 1): cand = s[i : i + k] if cand not in cur: dp(i + k, cur + (cand,)) ans, cur = 0, () dp(0, cur) return ans
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF IF BIN_OP BIN_OP FUNC_CALL VAR VAR VAR VAR VAR RETURN IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR IF VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP VAR VAR FUNC_CALL VAR NONE ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: d = set() res = 1 n = len(s) def solve(p, ans): nonlocal res if p == n: res = max(res, ans) for i in range(p, n): ss = "".join(s[p : i + 1]) if ss in d: continue d.add(ss) solve(i + 1, ans + 1) d.remove(ss) solve(0, 0) return res
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL STRING VAR VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER NUMBER RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: best = -1 ans = [] for i in range(2 ** (len(s) - 1)): x = bin(i) x = x[2:] x = (len(s) - len(x) - 1) * "0" + x tmp = "" a = [] c = 0 for char in s: tmp += char if c < len(x): if x[c] == "1": a.append(tmp) tmp = "" c += 1 a.append(tmp) if len(a) == len(set(a)): if len(a) > best: best = len(a) ans = a return len(ans)
CLASS_DEF FUNC_DEF VAR ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER STRING VAR ASSIGN VAR STRING ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR VAR VAR IF VAR FUNC_CALL VAR VAR IF VAR VAR STRING EXPR FUNC_CALL VAR VAR ASSIGN VAR STRING VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR RETURN FUNC_CALL VAR VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def split(s, b): res = [] x, y = s[0], b[0] for i in range(1, len(s)): if b[i] == y: x += s[i] else: res.append(x) x, y = s[i], b[i] res.append(x) return res count = 0 for i in range(2 ** (len(s) - 1)): b = bin(i)[2:] bb = "0" * (len(s) - len(b)) + b res = split(s, bb) if len(res) == len(set(res)): count = max(count, len(res)) return count
CLASS_DEF FUNC_DEF VAR FUNC_DEF ASSIGN VAR LIST ASSIGN VAR VAR VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP STRING BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: self.res = 1 def backtrack(curr, idx): if len(curr) != len(set(curr)): return elif idx == len(s): self.res = max(self.res, len(curr)) return for i in range(idx + 1, len(s) + 1): backtrack(curr + [s[idx:i]], i) backtrack([], 0) return self.res
CLASS_DEF FUNC_DEF VAR ASSIGN VAR NUMBER FUNC_DEF IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR LIST VAR VAR VAR VAR EXPR FUNC_CALL VAR LIST NUMBER RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def dfs(self, s, req_l, segs, cur, hs): if cur >= len(s): return segs == req_l for pos in range(cur, len(s)): ts = s[cur : pos + 1] if ts in hs: continue hs.add(ts) if self.dfs(s, req_l, segs + 1, pos + 1, hs): return True hs.remove(ts) return False def maxUniqueSplit(self, s: str) -> int: for i in range(len(s), 0, -1): if self.dfs(s, i, 0, 0, set()): return i return 1
CLASS_DEF FUNC_DEF IF VAR FUNC_CALL VAR VAR RETURN VAR VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR RETURN NUMBER EXPR FUNC_CALL VAR VAR RETURN NUMBER FUNC_DEF VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER NUMBER IF FUNC_CALL VAR VAR VAR NUMBER NUMBER FUNC_CALL VAR RETURN VAR RETURN NUMBER VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def builder(s, seen): if not s: return 0 res = 0 for i, c in enumerate(s): candidate = s[: i + 1] if candidate not in seen: seen.add(candidate) res = max(res, 1 + builder(s[i + 1 :], seen)) seen.remove(candidate) return res return builder(s, set())
CLASS_DEF FUNC_DEF VAR FUNC_DEF IF VAR RETURN NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR RETURN VAR RETURN FUNC_CALL VAR VAR FUNC_CALL VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: max_ = 0 def maxUniqueSplit(self, s: str) -> int: b = len(s) - 1 ar = [] self.h(s, b, ar) return self.max_ def h(self, s, b, ar): ar2 = ar.copy() if b < 0: self.max_ = max(self.max_, len(ar2)) return 0 t = s[b:] self.h(s, b - 1, ar2) s = s[:b] if t not in ar2: ar2.append(t) self.h(s, b - 1, ar2)
CLASS_DEF ASSIGN VAR NUMBER FUNC_DEF VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST EXPR FUNC_CALL VAR VAR VAR VAR RETURN VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN NUMBER ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def maxU(s, soFar=set()): if len(s) == 1: if s in soFar: return 0 maxSplit = len(soFar) + 1 for partition in range(1, len(s)): a = s[:partition] b = s[partition:] if a not in soFar: maxSplit = max(maxSplit, maxU(b, soFar | {a})) return maxSplit return maxU(s)
CLASS_DEF FUNC_DEF VAR FUNC_DEF FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER IF VAR VAR RETURN NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR RETURN VAR RETURN FUNC_CALL VAR VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: if s == None or len(s) == 0: return 0 self.res = 0 has = set() def dfs(cur, lasti, ct): if cur > len(s): self.res = max(self.res, ct) return for cur in range(cur, len(s) + 1): if s[lasti:cur] not in has: has.add(s[lasti:cur]) ct += 1 dfs(cur + 1, cur, ct) ct -= 1 has.remove(s[lasti:cur]) else: while cur <= len(s) and s[lasti:cur] in has: cur += 1 dfs(cur, lasti, ct) dfs(1, 0, 0) return self.res
CLASS_DEF FUNC_DEF VAR IF VAR NONE FUNC_CALL VAR VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN FOR VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR WHILE VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def allsplit(s): if len(s) == 1: return [[s]] res = [[s]] for i in range(1, len(s)): res += [([s[:i]] + x) for x in allsplit(s[i:])] return res ans = 0 for sp in allsplit(s): if len(sp) == len(set(sp)): ans = max(ans, len(sp)) return ans
CLASS_DEF FUNC_DEF VAR FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN LIST LIST VAR ASSIGN VAR LIST LIST VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR BIN_OP LIST VAR VAR VAR VAR FUNC_CALL VAR VAR VAR RETURN VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: result = [] def dfs(path, word): if not word: result.append(path) return for i in range(1, len(word) + 1): prefix, suffix = word[:i], word[i:] if prefix not in path: dfs(path | {prefix}, suffix) dfs(set(), s) l = 1 for item in result: l = max(len(item), l) return l
CLASS_DEF FUNC_DEF VAR ASSIGN VAR LIST FUNC_DEF IF VAR EXPR FUNC_CALL VAR VAR RETURN FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def check(st, cm): mx = 0 if st in cm else 1 for i in range(1, len(st)): a = st[:i] b = st[i:] if a not in cm: ncm = set(cm) ncm.add(a) c = check(b, ncm) + 1 if c > mx: mx = c return mx return check(s, set())
CLASS_DEF FUNC_DEF VAR FUNC_DEF ASSIGN VAR VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR RETURN VAR RETURN FUNC_CALL VAR VAR FUNC_CALL VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: n = len(s) self.res = 0 def helper(curIdx, curW, curSets): if curIdx == n: self.res = max(self.res, len(curSets)) return helper(curIdx + 1, curW + s[curIdx], curSets) if curW + s[curIdx] not in curSets: helper(curIdx + 1, "", curSets | {curW + s[curIdx]}) helper(0, "", set()) return self.res
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FUNC_DEF IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR IF BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER STRING BIN_OP VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER STRING FUNC_CALL VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
def get_sequences(symbols, length): if length == 0: return [[]] sequences = [] tail_sequences = get_sequences(symbols, length - 1) for tail_sequence in tail_sequences: for symbol in symbols: sequences.append(list(tail_sequence) + [symbol]) return sequences def is_unique(sequence, s): if s == "": return 0 words = set() letters = [s[0]] for i in range(len(sequence)): if sequence[i] == 1: word = "".join(letters) if word in words: return False words.add(word) letters = [s[i + 1]] else: letters.append(s[i + 1]) word = "".join(letters) if word in words: return False return True class Solution: def maxUniqueSplit(self, s: str) -> int: sequences = get_sequences([0, 1], len(s) - 1) max_split = 1 for sequence in sequences: if is_unique(sequence, s): max_split = max(max_split, sum(sequence) + 1) return max_split
FUNC_DEF IF VAR NUMBER RETURN LIST LIST ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER FOR VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR LIST VAR RETURN VAR FUNC_DEF IF VAR STRING RETURN NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER ASSIGN VAR FUNC_CALL STRING VAR IF VAR VAR RETURN NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL STRING VAR IF VAR VAR RETURN NUMBER RETURN NUMBER CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR LIST NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: N = len(s) res = 0 for mask in itertools.product("10", repeat=N - 1): mask = list(mask) + ["1"] substr = set() begin = 0 i = 0 valid = True while True: try: idx = mask.index("1", begin) begin = idx + 1 if s[i : idx + 1] in substr: valid = False break substr.add(s[i : idx + 1]) i = idx + 1 except ValueError: break if valid: res = max(res, mask.count("1")) return res
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR STRING BIN_OP VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR LIST STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE NUMBER ASSIGN VAR FUNC_CALL VAR STRING VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR IF VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR STRING RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: self.ans = 1 def rec(split, s): if not s: se = set(split) if len(se) == len(split): self.ans = max(self.ans, len(split)) return for i in range(len(s)): piece = s[0 : i + 1] spl = split[:] spl.append(piece) rec(spl, s[i + 1 :]) rec([], s) return self.ans
CLASS_DEF FUNC_DEF VAR ASSIGN VAR NUMBER FUNC_DEF IF VAR ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR LIST VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: best_ans = 1 cur_ans = set() def find_result(i): nonlocal best_ans if i >= len(s): return len(cur_ans) elif len(s) - i + len(cur_ans) < best_ans: return best_ans for j in range(i + 1, len(s) + 1): substr = s[i:j] if substr not in cur_ans: cur_ans.add(substr) best_ans = max(best_ans, find_result(j)) cur_ans.remove(substr) return max(best_ans, len(cur_ans)) return find_result(0)
CLASS_DEF FUNC_DEF VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR IF BIN_OP BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR NUMBER VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: if not s: return 0 cache = [None] * len(s) setSet = self.dp(s, 0, cache) return len(max(setSet, key=lambda x: len(x))) if setSet else 0 def dp(self, inputStr, i, cache): if i == len(inputStr): return [set()] if cache[i] is not None: return cache[i] end = len(inputStr) + 1 res = set() for j in range(i + 1, end): tempSol = self.dp(inputStr, j, cache) for substrArr in tempSol: if inputStr[i:j] in substrArr: continue combined = set(substrArr) combined.add(inputStr[i:j]) res.add(frozenset(combined)) cache[i] = frozenset(res) return cache[i]
CLASS_DEF FUNC_DEF VAR IF VAR RETURN NUMBER ASSIGN VAR BIN_OP LIST NONE FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR RETURN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR RETURN LIST FUNC_CALL VAR IF VAR VAR NONE RETURN VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FOR VAR VAR IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: ans = 1 def getans(self, s, start, n, temp): self.ans = max(self.ans, len(temp)) for x in range(start + 1, n + 1): boss = set(list(temp)) boss.add(s[start:x]) self.getans(s, x, n, boss) def maxUniqueSplit(self, s: str) -> int: temp = set() self.getans(s, 0, len(s), temp) return self.ans
CLASS_DEF ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: self.seen = set() self.result = 0 def backtrack(s, start): if start == len(s): return True for end in range(start + 1, len(s) + 1): if s[start:end] not in self.seen: self.seen.add(s[start:end]) if backtrack(s, end): self.result = max(self.result, len(self.seen)) self.seen.remove(s[start:end]) return False backtrack(s, 0) return self.result
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FUNC_DEF IF VAR FUNC_CALL VAR VAR RETURN NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR RETURN NUMBER EXPR FUNC_CALL VAR VAR NUMBER RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def search(s): ans = [[s]] for i in range(1, len(s)): ans += [([s[:i]] + comb) for comb in search(s[i:])] return ans a = search(s) r = 1 for comb in a: if len(set(comb)) == len(comb): r = max(r, len(comb)) return r
CLASS_DEF FUNC_DEF VAR FUNC_DEF ASSIGN VAR LIST LIST VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR BIN_OP LIST VAR VAR VAR VAR FUNC_CALL VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: if not s: return 0 return self.dfs(s, set()) def dfs(self, s: str, present: set) -> int: maximum = 0 for i in range(1, len(s) + 1): substring = s[:i] if substring not in present: sub_try = self.dfs(s[i:], {*present, substring}) maximum = max(maximum, 1 + sub_try) return maximum
CLASS_DEF FUNC_DEF VAR IF VAR RETURN NUMBER RETURN FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
def aux(s, lst, i): if i == len(s): return 0 mx = 0 for j in range(i, len(s)): if s[i : j + 1] not in lst: tmp = 1 + aux(s, lst + [s[i : j + 1]], j + 1) if tmp > mx: mx = tmp return mx class Solution: def maxUniqueSplit(self, s: str) -> int: lst = [] return aux(s, lst, 0)
FUNC_DEF IF VAR FUNC_CALL VAR VAR RETURN NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP NUMBER FUNC_CALL VAR VAR BIN_OP VAR LIST VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR RETURN VAR CLASS_DEF FUNC_DEF VAR ASSIGN VAR LIST RETURN FUNC_CALL VAR VAR VAR NUMBER VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: n = len(s) ans = 0 for pat in range(1 << n - 1): start = 0 end = 1 split = [] for b in bin(pat)[2:].zfill(n - 1): if b == "1": split.append(s[start:end]) start = end end += 1 else: end += 1 split.append(s[start:end]) if len(set(split)) == len(split): ans = max(ans, len(split)) return ans
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER IF VAR STRING EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: for answer in range(len(s), 0, -1): num_splits = answer - 1 for split_points in combinations(range(1, len(s)), num_splits): current = 0 words = set() for split_point in split_points: words.add(s[current:split_point]) current = split_point words.add(s[current:]) if len(words) == answer: return answer
CLASS_DEF FUNC_DEF VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def dfs(index, visited): nonlocal ans if s[index:n] and s[index:n] not in visited: ans = max(ans, len(visited) + 1) for i in range(index, n): if ( s[index:i] and s[index:i] not in visited and len(visited) + 1 + n - i > ans ): visited.add(s[index:i]) dfs(i, visited) visited.remove(s[index:i]) n, ans = len(s), 0 dfs(0, set()) return ans
CLASS_DEF FUNC_DEF VAR FUNC_DEF IF VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR VAR VAR VAR BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER FUNC_CALL VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: n = len(s) def back(i, curr, addhere): if i == n: return len(addhere) - 1 if curr + s[i] in addhere: return back(i + 1, curr + s[i], addhere.copy()) else: return max( back(i + 1, "", addhere.union({curr + s[i]})), back(i + 1, curr + s[i], addhere.copy()), ) b = set() b.add("") return back(0, "", b)
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR VAR RETURN BIN_OP FUNC_CALL VAR VAR NUMBER IF BIN_OP VAR VAR VAR VAR RETURN FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR FUNC_CALL VAR RETURN FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER STRING FUNC_CALL VAR BIN_OP VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR STRING RETURN FUNC_CALL VAR NUMBER STRING VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: le = len(s) - 1 max = 1 for ma in range(pow(2, le) - 1, 0, -1): bi = format(ma, "0" + str(le) + "b") + "1" prev = 0 lis = [] valid = True for bit in range(len(bi)): if bi[bit] == "1": part = s[prev : bit + 1] if part in lis: valid = False break lis.append(part) prev = bit + 1 if valid and len(lis) > max: max = len(lis) return max
CLASS_DEF FUNC_DEF VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR NUMBER VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP STRING FUNC_CALL VAR VAR STRING STRING ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: N = len(s) possibilities = [] for k in range(1, N): comb = itertools.combinations(list(range(1, N)), k) possibilities += [ ( [s[0 : x[0]]] + [s[x[i] : x[i + 1]] for i in range(len(x) - 1)] + [s[x[-1] :]] ) for x in comb ] ansr = 1 for k in possibilities: if len(k) == len(set(k)): ansr = max(ansr, len(k)) return ansr
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR VAR VAR BIN_OP BIN_OP LIST VAR NUMBER VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER LIST VAR VAR NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def split(i): if i >= len(s): return len([v for v in t.values() if v]) best = 0 for j in range(i, len(s)): ss = s[i : j + 1] t[ss] += 1 best = max(best, split(j + 1)) t[ss] -= 1 return best t = collections.defaultdict(int) return split(0)
CLASS_DEF FUNC_DEF VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR NUMBER VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: @lru_cache(None) def helper(s): if not s: return [[]] splits = [] for i in range(1, len(s) + 1): word = s[:i] next_splits = helper(s[i:]) for split in next_splits: if word not in split: splits.append(split + [word]) return splits ret = helper(s) max_len = max([len(x) for x in ret]) return max_len
CLASS_DEF FUNC_DEF VAR FUNC_DEF IF VAR RETURN LIST LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR LIST VAR RETURN VAR FUNC_CALL VAR NONE ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: N, res = len(s), 0 for i in range(2 ** (N - 1)): start, seen = 0, set() for j in range(N): if i + (1 << N - 1) & 1 << j: seen.add(s[start : j + 1]) start = j + 1 res = max(res, len(seen)) return res
CLASS_DEF FUNC_DEF VAR ASSIGN VAR VAR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: self.ans = 0 n = len(s) def backtrack(start, have): if start >= n: self.ans = max(self.ans, len(have)) return for i in range(start + 1, n + 1): tmp = s[start:i] if tmp in have: continue else: have.add(tmp) backtrack(i, have) have.remove(tmp) backtrack(0, set()) return self.ans
CLASS_DEF FUNC_DEF VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER FUNC_CALL VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: n = len(s) currli = [] visiting = set() self.maxval = 1 def dfs(st, s): if len(s) <= 0: self.maxval = max(self.maxval, len(currli)) return True for i in range(1, len(s) + 1): if s[0:i] not in visiting: visiting.add(s[0:i]) currli.append(s[0:i]) dfs(i, s[i:]) currli.pop() visiting.remove(s[0:i]) return False for i in range(1, n + 1): visiting.add(s[0:i]) currli.append(s[0:i]) dfs(i, s[i:]) currli.pop() visiting.remove(s[0:i]) return self.maxval
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FUNC_DEF IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR NUMBER VAR RETURN NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR NUMBER VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: have = set() self.cur = [] def dfs(i): if i == len(s): return 0 if "".join(self.cur) not in have else -float("inf") self.cur.append(s[i]) ret = dfs(i + 1) st = "".join(self.cur) if st not in have: tmp, self.cur = self.cur, [] have.add(st) ret = max(dfs(i + 1) + 1, ret) have.remove(st) self.cur = tmp self.cur.pop() return ret return dfs(0)
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FUNC_DEF IF VAR FUNC_CALL VAR VAR RETURN FUNC_CALL STRING VAR VAR NUMBER FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL STRING VAR IF VAR VAR ASSIGN VAR VAR VAR LIST EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR RETURN VAR RETURN FUNC_CALL VAR NUMBER VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def all_partitions(string): for cutpoints in range(1 << len(string) - 1): result = [] lastcut = 0 for i in range(len(string) - 1): if 1 << i & cutpoints != 0: result.append(string[lastcut : i + 1]) lastcut = i + 1 result.append(string[lastcut:]) yield result ps = all_partitions(s) ret = 0 for p in ps: l = list(dict.fromkeys(p)) if len(p) == len(l): ret = max(ret, len(p)) return ret
CLASS_DEF FUNC_DEF VAR FUNC_DEF FOR VAR FUNC_CALL VAR BIN_OP NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF BIN_OP BIN_OP NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def numway(s): nw = 0 for i in range(1, len(s) + 1): if s[:i] not in nq: nq.add(s[:i]) nw = max(nw, 1 + numway(s[i:])) nq.remove(s[:i]) return nw nq = set() return numway(s)
CLASS_DEF FUNC_DEF VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR RETURN FUNC_CALL VAR VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: n = len(s) res = 1 if n == 1: return res for x in range(2 ** (n - 1)): m = 1 << n - 2 p0 = 0 p1 = 0 st = set() done = False while m: if x & m: sr = s[p0 : p1 + 1] if sr in st: done = True break st.add(sr) p0 = p1 + 1 p1 = p1 + 1 else: p1 += 1 m >>= 1 if not done: sr = s[p0 : p1 + 1] if sr not in st: res = max(res, len(st) + 1) return res
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER RETURN VAR FOR VAR FUNC_CALL VAR BIN_OP NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR IF BIN_OP VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER IF VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def backtrack(seen, sidx): if sidx == len(s): self.res = max(self.res, len(seen)) return if len(seen) + len(s) - sidx <= self.res: return for eidx in range(sidx, len(s)): if s[sidx : eidx + 1] not in seen: backtrack(seen | {s[sidx : eidx + 1]}, eidx + 1) self.res = 1 backtrack(set(), 0) return self.res
CLASS_DEF FUNC_DEF VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN IF BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR RETURN FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def func(self, s, idx, words): max_result = 0 for i in range(idx + 1, len(s)): max_result = max(max_result, self.func(s, i, words | {s[idx:i]})) max_result = max(max_result, len(words | {s[idx:]})) return max_result def maxUniqueSplit(self, s: str) -> int: return self.func(s, 0, set())
CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR RETURN VAR FUNC_DEF VAR RETURN FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: result = 1 for l in itertools.product([True, False], repeat=len(s)): if l[-1] == False: continue cache = list() now = "" for c, end in zip(s, l): now += c if end: cache.append(now) now = "" if len(cache) == len(set(cache)): result = max(result, len(cache)) return result
CLASS_DEF FUNC_DEF VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR LIST NUMBER NUMBER FUNC_CALL VAR VAR IF VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING FOR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR STRING IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: n = len(s) def dp(cur, ss): if cur == n - 1: return len(set(ss.split())) cur += 1 return max(dp(cur, ss + s[cur]), dp(cur, ss + " " + s[cur])) return dp(0, s[0])
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR BIN_OP VAR NUMBER RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR STRING VAR VAR RETURN FUNC_CALL VAR NUMBER VAR NUMBER VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: val = 0 if len(s) == 0: return val def helper(s, sett=()): return max( ( 1 + helper(s[i:], {candidate, *sett}) for i in range(1, len(s) + 1) if (candidate := s[:i]) not in sett ), default=0, ) val = helper(s) return val
CLASS_DEF FUNC_DEF VAR ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR NUMBER RETURN VAR FUNC_DEF RETURN FUNC_CALL VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR VAR VAR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: n = len(s) if n == 1: return 1 res = 1 seen = set() for m in range(1, 1 << n - 1): if bin(m).count("1") < res: continue valid = True p = 0 for i in range(1, n + 1): if valid: if m & 1 << i - 1 or i == n: if s[p:i] in seen: valid = False seen.add(s[p:i]) p = i if valid: res = max(res, len(seen)) seen = set() return res
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP NUMBER BIN_OP VAR NUMBER IF FUNC_CALL FUNC_CALL VAR VAR STRING VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR IF BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER VAR VAR IF VAR VAR VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR IF VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: res = 0 for count in range(len(s)): for partitions in combinations(range(1, len(s)), count): partitions = [0] + list(partitions) + [len(s)] parts = [s[a:b] for a, b in zip(partitions, partitions[1:])] if len(parts) == len(set(parts)): res = max(res, len(parts)) return res
CLASS_DEF FUNC_DEF VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP LIST NUMBER FUNC_CALL VAR VAR LIST FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def dfs(i, vis): nonlocal ans if len(vis) + n - i < ans: return if i >= n: ans = max(ans, len(vis)) l = n - i for k in range(1, l + 1): x = s[i : i + k] if x not in vis: vis.add(x) dfs(i + k, vis) vis.discard(x) ans, n, vis = 0, len(s), set() dfs(0, vis) return ans
CLASS_DEF FUNC_DEF VAR FUNC_DEF IF BIN_OP BIN_OP FUNC_CALL VAR VAR VAR VAR VAR RETURN IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER FUNC_CALL VAR VAR FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def f(self, s): xss = [] if len(s) == 0: temp = [] temp.append("") xss.append(temp) return xss for i in range(1, len(s) + 1): xs = self.f(s[i:]) for x in xs: if s[:i] not in x: x.append(s[:i]) xss.append(x) return xss def maxUniqueSplit(self, s: str) -> int: x = max([len(y) for y in self.f(s)]) return x - 1
CLASS_DEF FUNC_DEF ASSIGN VAR LIST IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR VAR IF VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR RETURN BIN_OP VAR NUMBER VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def helper(curr_set, s): if len(s) == 0: return len(s) max_split = 0 for i in range(len(s)): if s[: i + 1] not in curr_set: curr_set.add(s[: i + 1]) max_split = max(max_split, 1 + helper(curr_set, s[i + 1 :])) curr_set.remove(s[: i + 1]) return max_split return helper(set(), s)
CLASS_DEF FUNC_DEF VAR FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER RETURN VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: self.mx = 0 self.lim = len(s) def r(i, l, c): if i == self.lim - 1: if s[i] not in l: if c + 1 > self.mx: self.mx = c + 1 elif c > self.mx: self.mx = c else: for k in range(i + 1, self.lim + 1): if k == self.lim: if s[i:k] not in l: if c + 1 > self.mx: self.mx = c + 1 elif s[i:k] not in l: r(k, l | {s[i:k]}, c + 1) r(0, set(), 0) return self.mx
CLASS_DEF FUNC_DEF VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR BIN_OP VAR NUMBER IF VAR VAR VAR IF BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR IF VAR VAR VAR VAR IF BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER FUNC_CALL VAR NUMBER RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: self.s = s self.cache = {} count = self.maxUniqueSplitHelper(0, set()) return count def maxUniqueSplitHelper(self, pos: int, substrings: set) -> int: hashed = frozenset(substrings) if (pos, hashed) in self.cache: return self.cache[pos, hashed] if pos == len(self.s): return len(substrings) count = 0 for i in range(pos, len(self.s)): substring = self.s[pos : i + 1] if substring in substrings: continue substrings.add(substring) count = max(count, self.maxUniqueSplitHelper(i + 1, substrings)) substrings.remove(substring) self.cache[pos, hashed] = count return count
CLASS_DEF FUNC_DEF VAR ASSIGN VAR VAR ASSIGN VAR DICT ASSIGN VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR RETURN VAR VAR FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR VAR RETURN VAR VAR VAR IF VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
def partition(s): def backtrack(index, path): if index == len(s): res.append(list(path)) else: path.append(s[index]) backtrack(index + 1, path) path.pop() if path: path[-1] = path[-1] + s[index] backtrack(index + 1, path) res = [] backtrack(0, []) return res class Solution: def maxUniqueSplit(self, s: str) -> int: ans = 0 lst = partition(s) for prt in lst: if len(set(prt)) == len(prt): ans = max(ans, len(prt)) return ans
FUNC_DEF FUNC_DEF IF VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR IF VAR ASSIGN VAR NUMBER BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR NUMBER LIST RETURN VAR CLASS_DEF FUNC_DEF VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: self.maxsplits = 0 def DFS(i, substring_set): if i == len(s): self.maxsplits = max(len(substring_set), self.maxsplits) return for j in range(i + 1, len(s) + 1): DFS(j, substring_set | {s[i:j]}) DFS(0, set()) return self.maxsplits
CLASS_DEF FUNC_DEF VAR ASSIGN VAR NUMBER FUNC_DEF IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR RETURN FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER FUNC_CALL VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def helper(s, strings=set()): leng_s = len(s) leng_set = len(strings) if leng_s == 1 and s in strings: return 0 else: ans = leng_set + 1 for i in range(1, leng_s): substr = s[:i] if substr not in strings: ans = max(ans, helper(s[i:], strings | {substr})) return ans return helper(s)
CLASS_DEF FUNC_DEF VAR FUNC_DEF FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR VAR RETURN NUMBER ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR RETURN VAR RETURN FUNC_CALL VAR VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: ans = 0 queue = [(set([s[:i]]), s[i:]) for i in range(1, len(s) + 1)] while queue: curSet, curS = queue.pop(0) if curS == "": ans = max(ans, len(curSet)) for i in range(1, len(curS) + 1): if curS[:i] in curSet: continue else: tmpSet = curSet.copy() tmpSet.add(curS[:i]) queue.append((tmpSet, curS[i:])) return ans
CLASS_DEF FUNC_DEF VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR LIST VAR VAR VAR VAR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR ASSIGN VAR VAR FUNC_CALL VAR NUMBER IF VAR STRING ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: spl = list(s) n = len(spl) c = collections.Counter(spl) if all(v == 1 for v in c.values()): return n mx = 2 ** (n - 1) - 2 ans = 1 for cmb in range(mx, 0, -1): bs = list(bin(cmb)[2:].zfill(n - 1)) if bs.count("1") + 1 < ans: continue spl = [] sub = s[0] for i, b in enumerate(bs): if b == "0": sub += s[i + 1] else: spl.append(sub) sub = s[i + 1] spl.append(sub) c = collections.Counter(spl) if all(v == 1 for v in c.values()): if len(spl) > ans: ans = len(spl) return ans
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR RETURN VAR ASSIGN VAR BIN_OP BIN_OP NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER IF BIN_OP FUNC_CALL VAR STRING NUMBER VAR ASSIGN VAR LIST ASSIGN VAR VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR STRING VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
def count1(x): xx = x ans = 0 while x != 0: x &= x - 1 ans += 1 return ans def ok(mask, s): strs = [] prev = 0 for i in range(len(s) - 1): if 1 << i & mask: strs.append(s[prev : i + 1]) prev = i + 1 strs.append(s[prev:]) return len(set(strs)) == len(strs) class Solution: def maxUniqueSplit(self, s: str) -> int: n = len(s) ans = 0 for i in range(1, 1 << n - 1): if ok(i, s): ans = max(count1(i), ans) return ans + 1
FUNC_DEF ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF BIN_OP BIN_OP NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP NUMBER BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR RETURN BIN_OP VAR NUMBER VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: n = len(s) - 1 ans = 1 for i in range(2**n + 1): prev = 0 ok = True visited = set() for j in range(n): if i & 1 << j: tmp = s[prev : j + 1] prev = j + 1 if tmp in visited: ok = False break visited.add(tmp) if not ok: break if not ok: continue tmp = s[prev:] if tmp not in visited: ans = max(ans, len(visited) + 1) return ans
CLASS_DEF FUNC_DEF VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR IF VAR ASSIGN VAR VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: return get_max(s, set()) def get_max(s, set): maxi = 0 for i in range(1, len(s) + 1): if s[:i] not in set: set.add(s[:i]) maxi = max(maxi, 1 + get_max(s[i:], set)) set.remove(s[:i]) return maxi
CLASS_DEF FUNC_DEF VAR RETURN FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR RETURN VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def helper(s, start, seen): ans = 0 for end in range(start + 1, len(s) + 1): added = s[start:end] print(added) if added not in seen: seen.add(added) ans = max(ans, 1 + helper(s, end, seen)) seen.remove(added) return ans seen = set() return helper(s, 0, seen)
CLASS_DEF FUNC_DEF VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR RETURN FUNC_CALL VAR VAR NUMBER VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def backtrack(m, s): if len(s) == 0: return 0 ans = 0 for i in range(1, len(s) + 1): candidate = s[:i] if candidate not in m: m.add(candidate) ans = max(ans, 1 + backtrack(m, s[i:])) m.remove(candidate) return ans m = set() return backtrack(m, s)
CLASS_DEF FUNC_DEF VAR FUNC_DEF IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR RETURN FUNC_CALL VAR VAR VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
import itertools class Solution: def maxUniqueSplit(self, s: str) -> int: maxres = 1 for sep in itertools.product([0, 1], repeat=len(s) - 1): sett = set() cur = s[0] for i, (se, c) in enumerate(zip(sep, s[1:]), start=1): if se: if cur in sett: break sett.add(cur) cur = c else: cur += c else: if cur in sett: continue sett.add(cur) maxres = max(maxres, len(sett)) return maxres
IMPORT CLASS_DEF FUNC_DEF VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR LIST NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR NUMBER NUMBER IF VAR IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
from itertools import combinations class Solution: def maxUniqueSplit(self, s: str) -> int: for num_splits in reversed(range(1, len(s))): num_strings = num_splits + 1 for split_indices in combinations(range(1, len(s)), num_splits): indices = [0] + list(split_indices) + [len(s)] split_strs = set() for start, end in zip(indices[:-1], indices[1:]): split_strs.add(s[start:end]) if len(split_strs) == num_strings: return num_strings return 1
CLASS_DEF FUNC_DEF VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP LIST NUMBER FUNC_CALL VAR VAR LIST FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR RETURN VAR RETURN NUMBER VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def helper(s1_set, s2): N = len(s2) if N == 1: if s2 in s1_set: return 0 return len(s1_set) + 1 res = 0 if s2 in s1_set else len(s1_set) + 1 for i in range(1, N): if s2[:i] in s1_set: continue s1_set.add(s2[:i]) res = max(res, helper(s1_set, s2[i:])) s1_set.discard(s2[:i]) return res res = helper(set(), s) return res
CLASS_DEF FUNC_DEF VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER IF VAR VAR RETURN NUMBER RETURN BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def split(s): if not s: return [set()] n = len(s) w = set() w.add(tuple([s])) ret = [set([s])] if n == 1: return ret if n == 2: if s[0] != s[1]: ret += [set([*s])] return ret for i in range(1, n): ls = split(s[0:i]) rs = split(s[i:n]) for l in ls: for r in rs: if len(l.intersection(r)) == 0: e = tuple(sorted(list(l | r))) if e not in w: w.add(e) ret += [l | r] return ret if not s: return 0 n = len(s) if n < 10: ws = split(s) return max(map(len, ws)) else: ans = 0 ls = split(s[:7]) rs = split(s[7:]) ws = [] w = set() for l in ls: for r in rs: if len(l.intersection(r)) == 0: e = tuple(sorted(list(l | r))) if e not in w: w.add(e) ws += [l | r] ans = max(ans, max(map(len, ws))) ls = split(s[:6]) rs = split(s[6:]) ws = [] w = set() for l in ls: for r in rs: if len(l.intersection(r)) == 0: e = tuple(sorted(list(l | r))) if e not in w: w.add(e) ws += [l | r] ans = max(ans, max(map(len, ws))) return ans
CLASS_DEF FUNC_DEF VAR FUNC_DEF IF VAR RETURN LIST FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR LIST VAR ASSIGN VAR LIST FUNC_CALL VAR LIST VAR IF VAR NUMBER RETURN VAR IF VAR NUMBER IF VAR NUMBER VAR NUMBER VAR LIST FUNC_CALL VAR LIST VAR RETURN VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FOR VAR VAR FOR VAR VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR LIST BIN_OP VAR VAR RETURN VAR IF VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR FOR VAR VAR FOR VAR VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR LIST BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR FOR VAR VAR FOR VAR VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR LIST BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, string: str) -> int: def split(s, splits): maxs = 1 if s not in splits else 0 for i in range(1, len(s)): if s[:i] in splits: continue rst = 1 + split(s[i:], splits + [s[:i]]) maxs = max(rst, maxs) return maxs return split(string, [])
CLASS_DEF FUNC_DEF VAR FUNC_DEF ASSIGN VAR VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR BIN_OP VAR LIST VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR RETURN FUNC_CALL VAR VAR LIST VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: n = len(s) ans = -1 for bit in range(1 << n - 1): sep = [] for i in range(n - 1): if bit >> i & 1: sep.append(i + 1) sep.append(n) ww = set() l = r = 0 ng = False for r in sep: w = s[l:r] if w in ww: ng = True break ww.add(w) l = r if ng: continue ans = max(ans, len(ww)) return ans
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP NUMBER BIN_OP VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR VAR VAR VAR IF VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR IF VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: if len(s) == 1: return 1 if len(s) == 2: if s[0] == s[1]: return 1 return 2 ans = 0 for i in range(2 ** (len(s) - 1)): ss = format(i, "0" + str(len(s) - 1) + "b") + "1" st = set() count = 0 prev = 0 for j in range(len(ss)): lth = len(st) if ss[j] == "1": st.add(s[prev : j + 1]) if len(st) == lth: break prev = j + 1 count += 1 ans = max(ans, count) return ans
CLASS_DEF FUNC_DEF VAR IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER IF FUNC_CALL VAR VAR NUMBER IF VAR NUMBER VAR NUMBER RETURN NUMBER RETURN NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP STRING FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER STRING STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR STRING EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def dfs(i): nonlocal used, ret if i == n: ret = max(ret, len(used)) return for j in range(i + 1, n + 1): w = s[i:j] if w in used: continue used.add(w) dfs(j) used.discard(w) n = len(s) used = set() ret = 0 dfs(0) return ret
CLASS_DEF FUNC_DEF VAR FUNC_DEF IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def __init__(self): self.ans = 0 self.set = set() def maxUniqueSplit(self, s: str) -> int: self.solve(s, 0) return self.ans def solve(self, s: str, pos: int) -> None: if pos > len(s): return self.ans = max(self.ans, len(self.set)) if self.ans - len(self.set) >= len(s) - pos: return for i in range(pos, len(s)): v = s[pos : i + 1] if v in self.set: continue self.set.add(v) self.solve(s, i + 1) self.set.remove(v)
CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_DEF VAR EXPR FUNC_CALL VAR VAR NUMBER RETURN VAR VAR FUNC_DEF VAR VAR IF VAR FUNC_CALL VAR VAR RETURN ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR RETURN FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR NONE
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: i = 0 self.res = 0 def sub(s, x): if len(s) == 0: if sorted(x) == sorted(list(set(x))): self.res = max(self.res, len(set(x))) return for i in range(1, len(s) + 1): sub(s[i:], x + [s[:i]]) res = sub(s, []) return self.res
CLASS_DEF FUNC_DEF VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FUNC_DEF IF FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR LIST VAR VAR ASSIGN VAR FUNC_CALL VAR VAR LIST RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: n = len(s) - 1 ans = 1 for i in range(1 << n): subs = set() start = 0 flag = False for j in range(n): if 1 << j & i: tmp = s[start : j + 1] if tmp in subs: flag = True break subs.add(tmp) start = j + 1 if flag: continue tmp = s[start:] if tmp in subs: continue ans = max(ans, len(subs) + 1) return ans
CLASS_DEF FUNC_DEF VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP NUMBER VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP BIN_OP NUMBER VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR ASSIGN VAR VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER RETURN VAR VAR
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: lookup = set() return self.traverse(s, 0, lookup) def traverse(self, s, index, lookup): if index > len(s): return 0 array = [] for idx in range(index, len(s) + 1): if len(s[index:idx]) > 0 and s[index:idx] not in lookup: lookup.add(s[index:idx]) array.append(1 + self.traverse(s, idx, lookup)) lookup.remove(s[index:idx]) return max(array) if array else 0
CLASS_DEF FUNC_DEF VAR ASSIGN VAR FUNC_CALL VAR RETURN FUNC_CALL VAR VAR NUMBER VAR VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR RETURN NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR RETURN VAR FUNC_CALL VAR VAR NUMBER
Given a string s, return the maximum number of unique substrings that the given string can be split into. You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique. A substring is a contiguous sequence of characters within a string.   Example 1: Input: s = "ababccc" Output: 5 Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times. Example 2: Input: s = "aba" Output: 2 Explanation: One way to split maximally is ['a', 'ba']. Example 3: Input: s = "aa" Output: 1 Explanation: It is impossible to split the string any further.   Constraints: 1 <= s.length <= 16 s contains only lower case English letters.
class Solution: def maxUniqueSplit(self, s: str) -> int: def dfs(s, n): if n < 0: return [[s]] else: res = [] for i in range(1, len(s) - n): for group in dfs(s[i:], n - 1): res.append([s[:i]] + group) return res res = 1 for i in range(len(s)): for group in dfs(s, i): res = max(res, len(set(group))) return res
CLASS_DEF FUNC_DEF VAR FUNC_DEF IF VAR NUMBER RETURN LIST LIST VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP LIST VAR VAR VAR RETURN VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1. Example 1: Input: {xs,ys} = {0,0} {xd,yd} = {1,7} matrix = 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 Output: 24 Explanation: Example 2: Input: {xs,ys} = {0,3} {xd,yd} = {2,2} matrix = 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 Output: -1 Explanation: We can see that it is impossible to reach the cell (2,2) from (0,3). Your Task: You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path. Expected Time Complexity: O(2^(N*M)) Expected Auxiliary Space: O(N*M) Constraints: 1 <= N,M <= 10
class Solution: def longestPath( self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int ) -> int: if mat[xs][ys] == 0 or mat[xd][yd] == 0: return -1 def dfs(r, c, pathlen, res): if r == xd and c == yd: res[0] = max(res[0], pathlen) return for x, y in directions: row = r + x col = c + y if ( -1 < row < rows and -1 < col < cols and (r, c) not in visited and mat[r][c] == 1 ): visited.add((r, c)) dfs(row, col, pathlen + 1, res) visited.remove((r, c)) directions = [[1, 0], [-1, 0], [0, 1], [0, -1]] rows = n cols = m res = [-1] pathlen = 0 visited = set() dfs(xs, ys, pathlen, res) return res[0]
CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER FUNC_DEF IF VAR VAR VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR RETURN FOR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF NUMBER VAR VAR NUMBER VAR VAR VAR VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR LIST NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR RETURN VAR NUMBER VAR
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1. Example 1: Input: {xs,ys} = {0,0} {xd,yd} = {1,7} matrix = 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 Output: 24 Explanation: Example 2: Input: {xs,ys} = {0,3} {xd,yd} = {2,2} matrix = 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 Output: -1 Explanation: We can see that it is impossible to reach the cell (2,2) from (0,3). Your Task: You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path. Expected Time Complexity: O(2^(N*M)) Expected Auxiliary Space: O(N*M) Constraints: 1 <= N,M <= 10
class Solution: def longestPath( self, matrix, n: int, m: int, xs: int, ys: int, xd: int, yd: int ) -> int: currlen = 0 def safe(matrix, xs, ys): if ( xs < 0 or ys < 0 or xs >= len(matrix) or ys >= len(matrix[0]) or matrix[xs][ys] != 1 ): return False else: return True res = 0 def solve(matrix, xs, ys, xd, yd, currlen, length): currlen = max(currlen, length) nonlocal res if xs == xd and ys == yd: res = max(res, currlen) return if safe(matrix, xs + 1, ys): matrix[xs][ys] = 2 solve(matrix, xs + 1, ys, xd, yd, currlen + 1, length) matrix[xs][ys] = 1 if safe(matrix, xs - 1, ys): matrix[xs][ys] = 2 solve(matrix, xs - 1, ys, xd, yd, currlen + 1, length) matrix[xs][ys] = 1 if safe(matrix, xs, ys + 1): matrix[xs][ys] = 2 solve(matrix, xs, ys + 1, xd, yd, currlen + 1, length) matrix[xs][ys] = 1 if safe(matrix, xs, ys - 1): matrix[xs][ys] = 2 solve(matrix, xs, ys - 1, xd, yd, currlen + 1, length) matrix[xs][ys] = 1 if matrix[xs][ys] == 0 or matrix[xd][yd] == 0: return -1 solve(matrix, xs, ys, xd, yd, currlen, 0) return res
CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER FUNC_DEF IF VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER RETURN NUMBER ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN IF FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR NUMBER RETURN VAR VAR
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1. Example 1: Input: {xs,ys} = {0,0} {xd,yd} = {1,7} matrix = 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 Output: 24 Explanation: Example 2: Input: {xs,ys} = {0,3} {xd,yd} = {2,2} matrix = 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 Output: -1 Explanation: We can see that it is impossible to reach the cell (2,2) from (0,3). Your Task: You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path. Expected Time Complexity: O(2^(N*M)) Expected Auxiliary Space: O(N*M) Constraints: 1 <= N,M <= 10
class Solution: def longestPath( self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int ) -> int: if mat[xs][ys] == 0 or mat[xd][yd] == 0: return -1 if xs == xd and ys == yd: return 0 vis = [[(0) for i in range(m)] for j in range(n)] def dfs(grid, sr, sc, dr, dc, vis, n, m, ans, cnt): if sr == dr and sc == dc: ans[0] = max(ans[0], cnt) return if ( sr < 0 or sc < 0 or sr >= n or sc >= m or vis[sr][sc] == 1 or grid[sr][sc] == 0 ): return vis[sr][sc] = 1 dfs(grid, sr + 1, sc, dr, dc, vis, n, m, ans, cnt + 1) dfs(grid, sr - 1, sc, dr, dc, vis, n, m, ans, cnt + 1) dfs(grid, sr, sc + 1, dr, dc, vis, n, m, ans, cnt + 1) dfs(grid, sr, sc - 1, dr, dc, vis, n, m, ans, cnt + 1) vis[sr][sc] = 0 ans = [-1] cnt = 0 dfs(mat, xs, ys, xd, yd, vis, n, m, ans, cnt) return ans[0]
CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR VAR VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR RETURN IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR RETURN VAR NUMBER VAR
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1. Example 1: Input: {xs,ys} = {0,0} {xd,yd} = {1,7} matrix = 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 Output: 24 Explanation: Example 2: Input: {xs,ys} = {0,3} {xd,yd} = {2,2} matrix = 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 Output: -1 Explanation: We can see that it is impossible to reach the cell (2,2) from (0,3). Your Task: You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path. Expected Time Complexity: O(2^(N*M)) Expected Auxiliary Space: O(N*M) Constraints: 1 <= N,M <= 10
class Solution: def longestPath( self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int ) -> int: def solve(mat, n, m, xs, ys, xd, yd, cnt): if xs == xd and yd == ys: ans[0] = max(cnt, ans[0]) return dir = [[0, 1], [0, -1], [1, 0], [-1, 0]] for i, j in dir: dx = xs + i dy = ys + j if ( dx >= 0 and dy >= 0 and dx <= n - 1 and dy <= m - 1 and mat[dx][dy] == 1 ): mat[dx][dy] = 0 solve(mat, n, m, dx, dy, xd, yd, cnt + 1) mat[dx][dy] = 1 if mat[xs][ys] == 0 or mat[xd][yd] == 0: return -1 ans = [-1] mat[xs][ys] = 0 solve(mat, n, m, xs, ys, xd, yd, 0) return ans[0]
CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR FUNC_DEF IF VAR VAR VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER RETURN ASSIGN VAR LIST LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER FOR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR NUMBER RETURN VAR NUMBER VAR
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1. Example 1: Input: {xs,ys} = {0,0} {xd,yd} = {1,7} matrix = 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 Output: 24 Explanation: Example 2: Input: {xs,ys} = {0,3} {xd,yd} = {2,2} matrix = 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 Output: -1 Explanation: We can see that it is impossible to reach the cell (2,2) from (0,3). Your Task: You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path. Expected Time Complexity: O(2^(N*M)) Expected Auxiliary Space: O(N*M) Constraints: 1 <= N,M <= 10
class Solution: def longestPath(self, M, R, C, sc, sr, dc, dr): if M[sc][sr] == 0: return -1 ans = dfs(sc, sr, R, C, M, set(), dr, dc) return -1 if ans == 0 else ans def dfs(r, c, R, C, M, v, dc, dr): if (r, c) == (dr, dc): return len(v) v.add((r, c)) l = 0 for nr, nc in ((r + 1, c), (r - 1, c), (r, c + 1), (r, c - 1)): if 0 <= nr < R and 0 <= nc < C and (nr, nc) not in v and M[nr][nc] == 1: l = max(l, dfs(nr, nc, R, C, M, v, dc, dr)) v.remove((r, c)) return l
CLASS_DEF FUNC_DEF IF VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR RETURN VAR NUMBER NUMBER VAR FUNC_DEF IF VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF NUMBER VAR VAR NUMBER VAR VAR VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR RETURN VAR
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1. Example 1: Input: {xs,ys} = {0,0} {xd,yd} = {1,7} matrix = 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 Output: 24 Explanation: Example 2: Input: {xs,ys} = {0,3} {xd,yd} = {2,2} matrix = 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 Output: -1 Explanation: We can see that it is impossible to reach the cell (2,2) from (0,3). Your Task: You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path. Expected Time Complexity: O(2^(N*M)) Expected Auxiliary Space: O(N*M) Constraints: 1 <= N,M <= 10
class Solution: def longestPath(self, mat, n, m, xs, ys, xd, yd) -> int: res = [0] vis = [[(0) for i in range(m)] for i in range(n)] if mat[xs][ys] == 0 or mat[xd][yd] == 0: return -1 def recsol(row, col, temp): if row == xd and col == yd: res[0] = max(res[0], temp) return if ( row < 0 or col < 0 or row >= n or col >= m or vis[row][col] == 1 or mat[row][col] == 0 ): return vis[row][col] = 1 recsol(row + 1, col, temp + 1) recsol(row, col + 1, temp + 1) recsol(row - 1, col, temp + 1) recsol(row, col - 1, temp + 1) vis[row][col] = 0 return recsol(xs, ys, 0) return res[0]
CLASS_DEF FUNC_DEF ASSIGN VAR LIST NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER FUNC_DEF IF VAR VAR VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR RETURN IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER RETURN EXPR FUNC_CALL VAR VAR VAR NUMBER RETURN VAR NUMBER VAR
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1. Example 1: Input: {xs,ys} = {0,0} {xd,yd} = {1,7} matrix = 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 Output: 24 Explanation: Example 2: Input: {xs,ys} = {0,3} {xd,yd} = {2,2} matrix = 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 Output: -1 Explanation: We can see that it is impossible to reach the cell (2,2) from (0,3). Your Task: You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path. Expected Time Complexity: O(2^(N*M)) Expected Auxiliary Space: O(N*M) Constraints: 1 <= N,M <= 10
class Solution: def longestPath(self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int): visited = [[(0) for i in range(m)] for j in range(n)] directions = [(0, 1), (0, -1), (1, 0), (-1, 0)] def dfs(x, y): if x == xd and y == yd: return 0 mn = -float("inf") visited[x][y] = 1 for i, j in directions: nx = x + i ny = y + j if ( nx >= n or nx < 0 or ny >= m or ny < 0 or visited[nx][ny] or mat[nx][ny] == 0 ): continue mn = max(mn, 1 + dfs(nx, ny)) visited[x][y] = 0 return mn if mat[xs][ys] == 0: return -1 val = dfs(xs, ys) if val == -float("inf"): return -1 return val
CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER FUNC_DEF IF VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR VAR VAR NUMBER FOR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR NUMBER RETURN VAR IF VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR FUNC_CALL VAR STRING RETURN NUMBER RETURN VAR
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1. Example 1: Input: {xs,ys} = {0,0} {xd,yd} = {1,7} matrix = 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 Output: 24 Explanation: Example 2: Input: {xs,ys} = {0,3} {xd,yd} = {2,2} matrix = 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 Output: -1 Explanation: We can see that it is impossible to reach the cell (2,2) from (0,3). Your Task: You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path. Expected Time Complexity: O(2^(N*M)) Expected Auxiliary Space: O(N*M) Constraints: 1 <= N,M <= 10
class Solution: def longestPath(self, l, n, m, xs, ys, xd, yd): if l[xs][ys] == 0 or l[xd][yd] == 0: return -1 self.ans = -1 def ss(i, j, t): if i == xd and j == yd: self.ans = max(self.ans, t) return if i < 0 or j < 0 or i >= n or j >= m or l[i][j] == 0: return l[i][j] = 0 ss(i + 1, j, t + 1) ss(i - 1, j, t + 1) ss(i, j + 1, t + 1) ss(i, j - 1, t + 1) l[i][j] = 1 ss(xs, ys, 0) return self.ans
CLASS_DEF FUNC_DEF IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER FUNC_DEF IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR VAR VAR NUMBER RETURN ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER RETURN VAR
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1. Example 1: Input: {xs,ys} = {0,0} {xd,yd} = {1,7} matrix = 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 Output: 24 Explanation: Example 2: Input: {xs,ys} = {0,3} {xd,yd} = {2,2} matrix = 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 Output: -1 Explanation: We can see that it is impossible to reach the cell (2,2) from (0,3). Your Task: You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path. Expected Time Complexity: O(2^(N*M)) Expected Auxiliary Space: O(N*M) Constraints: 1 <= N,M <= 10
class Solution: def longestPath(self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int): visited = [[(0) for i in range(m)] for j in range(n)] directions = [(0, 1), (0, -1), (1, 0), (-1, 0)] def dfs(x, y): if x == xd and y == yd: return 0 mn = -float("inf") visited[x][y] = 1 for i, j in directions: nx = x + i ny = y + j if ( nx >= n or nx < 0 or ny >= m or ny < 0 or visited[nx][ny] or mat[nx][ny] == 0 ): continue mn = max(mn, 1 + dfs(nx, ny)) visited[x][y] = 0 return mn if mat[xs][ys] == 0: return -1 val = dfs(xs, ys) if val == -float("inf"): return -1 return val class IntArray: def __init__(self) -> None: pass def Input(self, n): arr = [int(i) for i in input().strip().split()] return arr def Print(self, arr): for i in arr: print(i, end=" ") print() class IntMatrix: def __init__(self) -> None: pass def Input(self, n, m): matrix = [] for _ in range(n): matrix.append([int(i) for i in input().strip().split()]) return matrix def Print(self, arr): for i in arr: for j in i: print(j, end=" ") print() if __name__ == "__main__": t = int(input()) for _ in range(t): a = IntArray().Input(2) b = IntArray().Input(4) mat = IntMatrix().Input(a[0], a[0]) obj = Solution() res = obj.longestPath(mat, a[0], a[1], b[0], b[1], b[2], b[3]) print(res)
CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER FUNC_DEF IF VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR VAR VAR NUMBER FOR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR NUMBER RETURN VAR IF VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR FUNC_CALL VAR STRING RETURN NUMBER RETURN VAR CLASS_DEF FUNC_DEF NONE FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR RETURN VAR FUNC_DEF FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR CLASS_DEF FUNC_DEF NONE FUNC_DEF ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR RETURN VAR FUNC_DEF FOR VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR IF VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1. Example 1: Input: {xs,ys} = {0,0} {xd,yd} = {1,7} matrix = 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 Output: 24 Explanation: Example 2: Input: {xs,ys} = {0,3} {xd,yd} = {2,2} matrix = 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 Output: -1 Explanation: We can see that it is impossible to reach the cell (2,2) from (0,3). Your Task: You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path. Expected Time Complexity: O(2^(N*M)) Expected Auxiliary Space: O(N*M) Constraints: 1 <= N,M <= 10
class Solution: def rec(self, mat, i, j, n, m, temp, ans, vis, xd, yd): if i < 0 or i >= n or j < 0 or j >= m or vis[i][j] == 1 or mat[i][j] == 0: return vis[i][j] = 1 if i == xd and j == yd: ans[0] = max(ans[0], temp) paths = [[0, 0, 1, -1], [1, -1, 0, 0]] for k in range(4): self.rec( mat, paths[0][k] + i, paths[1][k] + j, n, m, temp + 1, ans, vis, xd, yd ) vis[i][j] = 0 def longestPath(self, mat, n, m, xs, ys, xd, yd): ans = [-1] vis = [([0] * m) for i in range(n)] self.rec(mat, xs, ys, n, m, 0, ans, vis, xd, yd) return ans[0]
CLASS_DEF FUNC_DEF IF VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN ASSIGN VAR VAR VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR LIST LIST NUMBER NUMBER NUMBER NUMBER LIST NUMBER NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER FUNC_DEF ASSIGN VAR LIST NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR VAR RETURN VAR NUMBER
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1. Example 1: Input: {xs,ys} = {0,0} {xd,yd} = {1,7} matrix = 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 Output: 24 Explanation: Example 2: Input: {xs,ys} = {0,3} {xd,yd} = {2,2} matrix = 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 Output: -1 Explanation: We can see that it is impossible to reach the cell (2,2) from (0,3). Your Task: You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path. Expected Time Complexity: O(2^(N*M)) Expected Auxiliary Space: O(N*M) Constraints: 1 <= N,M <= 10
class Solution: def solve(self, mat, i, j, m, n, dp, c, ans): if ( i < 0 or j < 0 or i >= len(mat) or j >= len(mat[0]) or mat[i][j] == 0 or dp[i][j] == False ): return -100000000 if i == n and j == m: return 0 dp[i][j] = False ans = 1 + max( self.solve(mat, i + 1, j, m, n, dp, c + 1, ans), self.solve(mat, i, j + 1, m, n, dp, c + 1, ans), self.solve(mat, i - 1, j, m, n, dp, c + 1, ans), self.solve(mat, i, j - 1, m, n, dp, c + 1, ans), ) dp[i][j] = True return ans def longestPath(self, mat, n, m, xs, ys, xd, yd): if mat[xs][ys] == 0 or mat[xd][yd] == 0: return -1 dp = [[(True) for i in range(m)] for j in range(n)] ans = self.solve(mat, xs, ys, yd, xd, dp, 0, 0) if ans < 0: return -1 return ans
CLASS_DEF FUNC_DEF IF VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR BIN_OP NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER RETURN VAR FUNC_DEF IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR NUMBER NUMBER IF VAR NUMBER RETURN NUMBER RETURN VAR
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1. Example 1: Input: {xs,ys} = {0,0} {xd,yd} = {1,7} matrix = 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 Output: 24 Explanation: Example 2: Input: {xs,ys} = {0,3} {xd,yd} = {2,2} matrix = 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 Output: -1 Explanation: We can see that it is impossible to reach the cell (2,2) from (0,3). Your Task: You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path. Expected Time Complexity: O(2^(N*M)) Expected Auxiliary Space: O(N*M) Constraints: 1 <= N,M <= 10
class Solution: def longestPath( self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int ) -> int: if mat[xs][ys] == 0 or mat[xd][yd] == 0: return -1 def sol(mat, i, j, xd, yd, length): if i == xd and j == yd: if length > self.ans: self.ans = length return dirs = [[0, 1], [0, -1], [1, 0], [-1, 0]] for idx in range(len(dirs)): newx, newy = i + dirs[idx][0], j + dirs[idx][1] if ( 0 <= newx < len(mat) and 0 <= newy < len(mat[0]) and mat[newx][newy] == 1 ): mat[i][j] = 2 sol(mat, newx, newy, xd, yd, length + 1) mat[i][j] = 1 return self.ans = -1 sol(mat, xs, ys, xd, yd, 0) return self.ans
CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER FUNC_DEF IF VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR RETURN ASSIGN VAR LIST LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR NUMBER BIN_OP VAR VAR VAR NUMBER IF NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER RETURN ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER RETURN VAR VAR
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1. Example 1: Input: {xs,ys} = {0,0} {xd,yd} = {1,7} matrix = 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 Output: 24 Explanation: Example 2: Input: {xs,ys} = {0,3} {xd,yd} = {2,2} matrix = 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 Output: -1 Explanation: We can see that it is impossible to reach the cell (2,2) from (0,3). Your Task: You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path. Expected Time Complexity: O(2^(N*M)) Expected Auxiliary Space: O(N*M) Constraints: 1 <= N,M <= 10
class Solution: def isBound(self, x, y): if 0 <= x < self.N and 0 <= y < self.M: return True return False def dfs(self, x, y, dist): if x == self.targetX and y == self.targetY: self.maxLength = max(self.maxLength, dist) return self.mat[x][y] = 0 for delX, delY in zip(self.dx, self.dy): newX = x + delX newY = y + delY if not self.isBound(newX, newY): continue if self.mat[newX][newY] == 0: continue self.dfs(newX, newY, dist + 1) self.mat[x][y] = 1 def longestPath(self, mat, n, m, xs, ys, xd, yd): if mat[xs][ys] == 0 or mat[xd][yd] == 0: return -1 self.dx = [1, -1, 0, 0] self.dy = [0, 0, 1, -1] self.N, self.M = n, m self.maxLength = -1 self.mat = mat self.targetX, self.targetY = xd, yd self.dfs(xs, ys, 0) return self.maxLength
CLASS_DEF FUNC_DEF IF NUMBER VAR VAR NUMBER VAR VAR RETURN NUMBER RETURN NUMBER FUNC_DEF IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN ASSIGN VAR VAR VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR VAR IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER FUNC_DEF IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER RETURN VAR
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1. Example 1: Input: {xs,ys} = {0,0} {xd,yd} = {1,7} matrix = 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 Output: 24 Explanation: Example 2: Input: {xs,ys} = {0,3} {xd,yd} = {2,2} matrix = 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 Output: -1 Explanation: We can see that it is impossible to reach the cell (2,2) from (0,3). Your Task: You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path. Expected Time Complexity: O(2^(N*M)) Expected Auxiliary Space: O(N*M) Constraints: 1 <= N,M <= 10
class Solution: def longestPath( self, mat, n: int, m: int, xs: int, ys: int, xd: int, yd: int ) -> int: row = len(mat) col = len(mat[0]) ans = [] visited = [([False] * col) for i in range(row)] def solve(i, j, c, visited): if ( i < 0 or j < 0 or j >= col or i >= row or mat[i][j] == 0 or visited[i][j] == True ): return if i == xd and j == yd: ans.append(c) return visited[i][j] = True solve(i + 1, j, c + 1, visited) solve(i - 1, j, c + 1, visited) solve(i, j - 1, c + 1, visited) solve(i, j + 1, c + 1, visited) visited[i][j] = False solve(xs, ys, 0, visited) if len(ans) == 0: return -1 return max(ans)
CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR BIN_OP LIST NUMBER VAR VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER VAR IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER RETURN FUNC_CALL VAR VAR VAR
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1. Example 1: Input: {xs,ys} = {0,0} {xd,yd} = {1,7} matrix = 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 Output: 24 Explanation: Example 2: Input: {xs,ys} = {0,3} {xd,yd} = {2,2} matrix = 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 Output: -1 Explanation: We can see that it is impossible to reach the cell (2,2) from (0,3). Your Task: You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path. Expected Time Complexity: O(2^(N*M)) Expected Auxiliary Space: O(N*M) Constraints: 1 <= N,M <= 10
class Solution: def check(self, i, j, n, m, mat): if i < n and 0 <= i and j < m and j >= 0 and mat[i][j] == 1: return True return False def find(self, i, j, x, y, n, m, mat, ans, visited, cnt): if i == x and j == y: ans[0] = max(ans[0], cnt) return if self.check(i, j, n, m, mat) == False or visited[i][j] == 1: return visited[i][j] = 1 if self.check(i + 1, j, n, m, mat): self.find(i + 1, j, x, y, n, m, mat, ans, visited, cnt + 1) if self.check(i - 1, j, n, m, mat): self.find(i - 1, j, x, y, n, m, mat, ans, visited, cnt + 1) if self.check(i, j + 1, n, m, mat): self.find(i, j + 1, x, y, n, m, mat, ans, visited, cnt + 1) if self.check(i, j - 1, n, m, mat): self.find(i, j - 1, x, y, n, m, mat, ans, visited, cnt + 1) visited[i][j] = 0 def longestPath(self, mat, n, m, xs, ys, xd, yd): ans = [-1] visited = [[(0) for i in range(100)] for j in range(100)] self.find(xs, ys, xd, yd, n, m, mat, ans, visited, 0) return ans[0]
CLASS_DEF FUNC_DEF IF VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER RETURN NUMBER FUNC_DEF IF VAR VAR VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR RETURN IF FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER RETURN ASSIGN VAR VAR VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER FUNC_DEF ASSIGN VAR LIST NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR NUMBER RETURN VAR NUMBER
Given an N x M matrix, with a few hurdles(denoted by 0) arbitrarily placed, calculate the length of the longest possible route possible from source(xs,ys) to a destination(xd,yd) within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.If it is impossible to reach the destination from the source return -1. Example 1: Input: {xs,ys} = {0,0} {xd,yd} = {1,7} matrix = 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 Output: 24 Explanation: Example 2: Input: {xs,ys} = {0,3} {xd,yd} = {2,2} matrix = 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 Output: -1 Explanation: We can see that it is impossible to reach the cell (2,2) from (0,3). Your Task: You don't need to read input or print anything. Your task is to complete the function longestPath() which takes matrix ,source and destination as input parameters and returns an integer denoting the longest path. Expected Time Complexity: O(2^(N*M)) Expected Auxiliary Space: O(N*M) Constraints: 1 <= N,M <= 10
class Solution: def longestPath(self, arr, n, m, xs, ys, xd, yd) -> int: if arr[xd][yd] == 0: return -1 def helper(i, j): if i == xd and j == yd: return 0 if i < 0 or i >= n or j < 0 or j >= m or arr[i][j] == 0: return -100000 arr[i][j] = 0 val = 1 + max( helper(i + 1, j), helper(i, j + 1), helper(i - 1, j), helper(i, j - 1) ) arr[i][j] = 1 return val res = helper(xs, ys) if res < 0: return -1 return res class IntArray: def __init__(self) -> None: pass def Input(self, n): arr = [int(i) for i in input().strip().split()] return arr def Print(self, arr): for i in arr: print(i, end=" ") print() class IntMatrix: def __init__(self) -> None: pass def Input(self, n, m): matrix = [] for _ in range(n): matrix.append([int(i) for i in input().strip().split()]) return matrix def Print(self, arr): for i in arr: for j in i: print(j, end=" ") print() if __name__ == "__main__": t = int(input()) for _ in range(t): a = IntArray().Input(2) b = IntArray().Input(4) mat = IntMatrix().Input(a[0], a[0]) obj = Solution() res = obj.longestPath(mat, a[0], a[1], b[0], b[1], b[2], b[3]) print(res)
CLASS_DEF FUNC_DEF IF VAR VAR VAR NUMBER RETURN NUMBER FUNC_DEF IF VAR VAR VAR VAR RETURN NUMBER IF VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR BIN_OP NUMBER FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER RETURN NUMBER RETURN VAR VAR CLASS_DEF FUNC_DEF NONE FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR RETURN VAR FUNC_DEF FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR CLASS_DEF FUNC_DEF NONE FUNC_DEF ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR RETURN VAR FUNC_DEF FOR VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR IF VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR