description
stringlengths
171
4k
code
stringlengths
94
3.98k
normalized_code
stringlengths
57
4.99k
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def is_unique(self, s): return len(set(s)) == len(s) def maxLength(self, arr: List[str]) -> int: dct = {} maxVal = 0 for i in range(0, len(arr)): v = frozenset(arr[i]) if self.is_unique(arr[i]): maxVal = max(maxVal, len(v)) for j in range(i + 1, len(arr)): t = arr[i] + arr[j] val = frozenset(t) if self.is_unique(t): arr.append(t) maxVal = max(maxVal, len(val)) return maxVal
CLASS_DEF FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF VAR VAR ASSIGN VAR DICT ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: res = 0 def dfs(n, s, l, cur): nonlocal res if n == l: if len(set(cur)) == len(cur): res = max(res, len(cur)) return for i in range(s, len(arr)): new_cur = cur + arr[i] if len(set(cur)) != len(cur): return dfs(n, i + 1, l + 1, new_cur) for i in range(len(arr) + 1): dfs(i, 0, 0, "") return res
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR NUMBER FUNC_DEF IF 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 FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER NUMBER STRING RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr) -> int: maxLen = 0 arr = sorted(arr, key=lambda x: len(x), reverse=True) self.compatible = {} for i in range(len(arr)): currString = arr[i] if self.noDupes(currString): if currString not in self.compatible: self.compatible[currString] = [] for j in range(i + 1, len(arr)): if self.noDupes(arr[j]) and self.uniqueChars(currString, arr[j]): self.compatible[currString].append(arr[j]) for key in self.compatible: if len(self.compatible[key]) == 0: length = len(key) else: length = self.findLength(key, key, 0) maxLen = max(maxLen, length) return maxLen def uniqueChars(self, s1, s2): for c in s2: if c in s1: return False return True def noDupes(self, string): counts = {} for c in string: counts[c] = counts.get(c, 0) + 1 if counts[c] > 1: return False return True def findLength(self, currString, totalString, maxLen): for string in self.compatible[currString]: if self.uniqueChars(totalString, string): maxLen = max( maxLen, self.findLength( string, totalString + string, len(totalString) + len(string) ), ) return maxLen
CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR DICT FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR FOR VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR VAR FUNC_DEF FOR VAR VAR IF VAR VAR RETURN NUMBER RETURN NUMBER FUNC_DEF ASSIGN VAR DICT FOR VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR NUMBER RETURN NUMBER RETURN NUMBER FUNC_DEF FOR VAR VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: newarr = [] for w in arr: if len(set(w)) == len(w): newarr.append(w) arr = newarr n = len(arr) res = 0 str_masks = [None] * n for i in range(n): wd = arr[i] mask = 0 for ch in wd: ch_pos = ord(ch) - ord("a") mask |= 1 << ch_pos str_masks[i] = mask def dfs(idx, mask, curlen): nonlocal res if idx == n: res = max(res, curlen) return if mask & str_masks[idx] == 0: dfs(idx + 1, mask | str_masks[idx], curlen + len(arr[idx])) dfs(idx + 1, mask, curlen) dfs(0, 0, 0) return res
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST FOR VAR VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NONE VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR FUNC_DEF IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN IF BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution(object): def maxLength(self, A): for i in range(len(A) - 1, -1, -1): if len(set(A[i])) != len(A[i]): A.pop(i) N = len(A) B = [] for word in A: ct = [0] * 26 for letter in word: ct[ord(letter) - ord("a")] += 1 B.append(ct) self.ans = 0 count = [0] * 26 def search(i): if i == N: cand = sum(count) if cand > self.ans: self.ans = cand return for letter, ct in enumerate(B[i]): if ct and count[letter]: search(i + 1) break else: search(i + 1) for letter, ct in enumerate(B[i]): if ct: count[letter] += 1 search(i + 1) for letter, ct in enumerate(B[i]): if ct: count[letter] -= 1 search(0) return self.ans
CLASS_DEF VAR FUNC_DEF FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER FUNC_DEF IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR VAR RETURN FOR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: letters_dict = {} def dfs(letter_set, arr, i): if i >= len(arr): return 0 if (tuple(letter_set), i) not in letters_dict: if len(set(arr[i]).intersection(letter_set)) == 0 and len( set(arr[i]) ) == len(arr[i]): letters_dict[tuple(letter_set), i] = max( dfs(letter_set.union(set(arr[i])), arr, i + 1) + len(set(arr[i])), dfs(letter_set, arr, i + 1), ) else: letters_dict[tuple(letter_set), i] = dfs(letter_set, arr, i + 1) return letters_dict[tuple(letter_set), i] return dfs(set(), arr, 0)
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR DICT FUNC_DEF IF VAR FUNC_CALL VAR VAR RETURN NUMBER IF FUNC_CALL VAR VAR VAR VAR IF FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR VAR VAR VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER RETURN VAR FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: arr.sort() self.arr = arr self.tmp = [] self.mlen = 0 self.len = 0 self.backtrack(0) return self.mlen def backtrack(self, pos): if pos == len(self.arr): if self.len > self.mlen: self.mlen = self.len return self.backtrack(pos + 1) lb = len(self.tmp) self.tmp = self.tmp + list(self.arr[pos]) l = len(self.arr[pos]) if l == len(self.arr[pos]) and lb + l == len(set(self.tmp)): self.len += l self.backtrack(pos + 1) self.len -= l for i in range(l): self.tmp.pop()
CLASS_DEF FUNC_DEF VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN VAR VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR VAR RETURN EXPR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: def dfs(arr, path, res): if self.checkUnique(path): res.append(path) for i in range(len(arr)): dfs(arr[i + 1 :], path + arr[i], res) concatenated_string, res = [], 0 dfs(arr, "", concatenated_string) for elem in concatenated_string: res = max(res, len(elem)) return res def checkUnique(self, string): dic = dict() for ch in string: if ch not in dic: dic[ch] = 1 else: return False return True
CLASS_DEF FUNC_DEF VAR VAR FUNC_DEF IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR LIST NUMBER EXPR FUNC_CALL VAR VAR STRING VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR ASSIGN VAR VAR NUMBER RETURN NUMBER RETURN NUMBER
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: def isUniqueChars(s): return len(set(s)) == len(s) arr = [set(w) for w in arr if isUniqueChars(w)] results = list(arr) for w in arr: temp = [] for r in results: if not w.intersection(r): temp.append(w.union(r)) results = results + temp max_length = 0 for w in results: max_length = max(max_length, len(w)) return max_length
CLASS_DEF FUNC_DEF VAR VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR ASSIGN VAR LIST FOR VAR VAR IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: res = 0 arr = list([x for x in arr if len(set(x)) == len(x)]) arr = [set(x) for x in arr] def dfs(n, s, l, cur): nonlocal res if n == l: res = max(res, len(cur)) return for i in range(s, len(arr)): if len(cur.intersection(arr[i])) == 0: dfs(n, i + 1, l + 1, cur.union(arr[i])) for i in range(len(arr) + 1): dfs(i, 0, 0, set()) return res
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_DEF IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER NUMBER FUNC_CALL VAR RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: dp = [set()] for word in arr: if len(word) != len(set(word)): continue curWord = set(word) for used in dp[:]: if len(used & curWord) == 0: dp.append(used | curWord) return len(max(dp, key=len))
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST FUNC_CALL VAR FOR VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR IF FUNC_CALL VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def recurse(self, letters, arr, csum): keys_as_str = "".join(sorted(letters)) if keys_as_str not in self.visited: self.visited.add(keys_as_str) self.max_len = max(self.max_len, csum) for i, new_letters in enumerate(arr): if len(new_letters.intersection(letters)) > 0: continue self.recurse( new_letters.union(letters), arr[:i] + arr[i + 1 :], csum + len(new_letters), ) def maxLength(self, arr: List[str]) -> int: new_arr = [] for word in arr: word_set = set(word) if len(word_set) == len(word): new_arr.append(word_set) self.max_len = 0 self.visited = set() for i, letters in enumerate(new_arr): self.recurse(letters, new_arr[:i] + new_arr[i + 1 :], len(letters)) return self.max_len
CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL STRING FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR FUNC_CALL VAR VAR FUNC_DEF VAR VAR ASSIGN VAR LIST FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: dp = [set()] length = 0 for word in arr: if len(word) != len(set(word)): continue curr = set(word) for used in dp[:]: if not len(used & curr): dp.append(used | curr) length = max(length, len(used | curr)) return length
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST FUNC_CALL 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 FOR VAR VAR IF FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR VAR RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def get_max_length(self, sets, idx, cur_set): if idx == len(sets): return len(cur_set) m = 0 for i in range(idx, len(sets)): if not cur_set & sets[i]: res = self.get_max_length(sets, i + 1, cur_set.union(sets[i])) else: res = self.get_max_length(sets, i + 1, cur_set) m = max(m, res) return m def maxLength(self, arr: List[str]) -> int: sets = [] for i, s in enumerate(arr): t = set() for ch in s: if ch in t: break else: t.add(ch) else: sets.append(t) print(sets) return self.get_max_length(sets, 0, set())
CLASS_DEF FUNC_DEF IF VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR FUNC_DEF VAR VAR ASSIGN VAR LIST FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
import itertools class Solution: def maxLength(self, arr): def powerset(iterable): import itertools s = list(iterable) return itertools.chain.from_iterable( itertools.combinations(s, r) for r in range(len(s) + 1) ) result = 0 for subset in powerset(arr): candidate = "".join(subset) if len(candidate) > result and len(candidate) == len(set(candidate)): result = len(candidate) return result
IMPORT CLASS_DEF FUNC_DEF FUNC_DEF IMPORT ASSIGN VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL STRING VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR RETURN VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: filtered = [] for string in arr: seen = {} good = True for letter in string: if letter in seen: good = False else: seen[letter] = True if good: filtered.append(string) self.memo = {} print(filtered) return self.getLength(0, filtered, {}) def getLength(self, current, arr, used): if not arr: return current if tuple(arr) in self.memo: return self.memo[tuple(arr)] maxi = current for index in range(len(arr)): string = arr[index] possible = True for char in string: if char in used: possible = False if not possible: continue for char in string: used[char] = True result = self.getLength( current + len(string), arr[:index] + arr[index + 1 :], used ) if result > maxi: maxi = result for char in string: del used[char] self.memo[tuple(arr)] = maxi return maxi
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST FOR VAR VAR ASSIGN VAR DICT ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER IF VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR DICT EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR NUMBER VAR DICT VAR FUNC_DEF IF VAR RETURN VAR IF FUNC_CALL VAR VAR VAR RETURN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR ASSIGN VAR NUMBER IF VAR FOR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR IF VAR VAR ASSIGN VAR VAR FOR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: ret = [set()] for string in arr: curr = set(string) if len(curr) == len(string): for seen in ret: if not seen & curr: ret.append(seen | curr) max_len = 0 for string in ret: max_len = max(max_len, len(string)) return max_len
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST FUNC_CALL VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR VAR IF BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: if len(arr) == 0: return 0 currRes = [] for el in arr: if self.isUnique(el): currRes.append(el) for el in arr: for subSeq in currRes: if self.isUnique(el + subSeq): currRes.append(el + subSeq) res = 0 for el in currRes: res = max(res, len(el)) return res def isUnique(self, s): chars = set() for c in s: if c not in chars: chars.add(c) else: return False return True
CLASS_DEF FUNC_DEF VAR VAR IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER ASSIGN VAR LIST FOR VAR VAR IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR FOR VAR VAR IF FUNC_CALL VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: self.maxLen = 0 def isUnique(s): return len(s) if len(s) == len(set(s)) else -1 def dfs(i, s): if i == len(candi) and isUnique(s) > self.maxLen: self.maxLen = len(s) return if i == len(candi): return dfs(i + 1, s) dfs(i + 1, s + candi[i]) candi = [] for ss in arr: if isUnique(ss) > 0: candi.append(ss) dfs(0, "") return self.maxLen
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR NUMBER FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER FUNC_DEF IF VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR RETURN IF VAR FUNC_CALL VAR VAR RETURN EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR ASSIGN VAR LIST FOR VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER STRING RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def is_unique(self, word): count = dict() for char in "".join(word): if char in count: return False else: count[char] = 1 return True def find_max(self, arr, output, current, start_index): if self.is_unique(current): output.append(len("".join(current))) for i in range(start_index, len(arr)): self.find_max(arr, output, current + [arr[i]], i + 1) def maxLength(self, arr: List[str]) -> int: output = [] self.find_max(arr, output, [], 0) return max(output)
CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL STRING VAR IF VAR VAR RETURN NUMBER ASSIGN VAR VAR NUMBER RETURN NUMBER FUNC_DEF IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL STRING VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR LIST VAR VAR BIN_OP VAR NUMBER FUNC_DEF VAR VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR VAR VAR LIST NUMBER RETURN FUNC_CALL VAR VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: d = {} def helper(banned, start): if start == len(arr): return len(banned) k = tuple(sorted(str(banned) + str(start))) if k in d: return d[k] string = arr[start] new_set = set() for i in string: if i in banned or i in new_set: return helper(banned, start + 1) new_set.add(i) new_banned = banned.union(new_set) val = max(helper(new_banned, start + 1), helper(banned, start + 1)) d[k] = val return val initial_banned = set() res = helper(initial_banned, 0) return res
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR DICT FUNC_DEF IF VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR RETURN VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: arr = [set(x) for x in arr if len(x) == len(set(x))] mx = 0 def util(start, cur): nonlocal mx if start == len(arr): mx = max(mx, len(cur)) for i in range(start, len(arr)): if cur & arr[i]: continue cur |= arr[i] util(i + 1, cur) cur ^= arr[i] mx = max(mx, len(cur)) util(0, set()) return mx
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FUNC_DEF IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER FUNC_CALL VAR RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: combinations = [set()] for a in arr: chars = set(a) if len(chars) != len(a): continue for c in combinations: if len(chars.intersection(c)) == 0: combinations.append(chars.union(c)) return max([len(c) for c in combinations])
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST FUNC_CALL VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
from itertools import combinations class Solution: def maxLength(self, arr: List[str]) -> int: max_count = 0 for i in range(0, len(arr) + 1): comb = combinations(arr, i) for c in comb: count = 0 bag = set() s = "".join(c) for k in s: if k in bag: count = 0 break else: count += 1 bag.add(k) max_count = max(max_count, count) return max_count
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL STRING VAR FOR VAR VAR IF VAR VAR ASSIGN VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def __init__(self): self._maxLength = 0 def maxLength(self, arr: List[str]) -> int: self.uniqueLengths(arr, []) return self._maxLength def uniqueLengths(self, arr, concat): result = "".join(concat) if not self.hasUniqueChars(result): return self._maxLength = max(self._maxLength, len(result)) for i in range(len(arr)): concat.append(arr[i]) self.uniqueLengths(arr[i + 1 :], concat) concat.pop() def hasUniqueChars(self, concat): counter = Counter(concat) return all([(x == 1) for x in counter.values()])
CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FUNC_DEF VAR VAR EXPR FUNC_CALL VAR VAR LIST RETURN VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL STRING VAR IF FUNC_CALL VAR VAR RETURN ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr): def isValid(chars): if len(chars) == len(set(chars)): return True else: return False result = 0 currentSubset = set() def iterate(idx): nonlocal result, currentSubset if idx == len(arr): candidateSolution = "".join(currentSubset) if isValid(candidateSolution) and len(candidateSolution) > result: result = len(candidateSolution) return currentSubset.add(arr[idx]) iterate(idx + 1) currentSubset.remove(arr[idx]) iterate(idx + 1) iterate(0) return result
CLASS_DEF FUNC_DEF FUNC_DEF IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL STRING VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR RETURN EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def uniquelen(self, cur): if len(cur) == len(set(cur)): return len(cur) else: return -1 def dfs(self, arr, index, cur): if index == len(arr) and self.uniquelen(cur) > self.ans: self.ans = self.uniquelen(cur) return if index == len(arr): return self.dfs(arr, index + 1, cur) self.dfs(arr, index + 1, cur + arr[index]) def maxLength(self, arr: List[str]) -> int: self.ans = -1 self.dfs(arr, 0, "") return self.ans
CLASS_DEF FUNC_DEF IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR RETURN NUMBER FUNC_DEF IF VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR RETURN IF VAR FUNC_CALL VAR VAR RETURN EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR FUNC_DEF VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER STRING RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: res = 0 unique = [""] def isvalid(s): return len(s) == len(set(s)) for word in arr: for u in unique: tmp = word + u if isvalid(tmp): unique.append(tmp) res = max(res, len(tmp)) return res
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST STRING FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: if arr == None: return path = [] result = [0] def hasUniqueLetters(word): hash_set = set() for c in word: if c in hash_set: return False else: hash_set.add(c) return True def generateCombinations(arr, i, result, path): if i == len(arr): combination = "".join(path) if path and hasUniqueLetters(combination): if result[0] < len(combination): result[0] = len(combination) return curr_path = [] curr_path.extend(path) curr_path.append(arr[i]) generateCombinations(arr, i + 1, result, path) generateCombinations(arr, i + 1, result, curr_path) generateCombinations(arr, 0, result, path) return result[0]
CLASS_DEF FUNC_DEF VAR VAR IF VAR NONE RETURN ASSIGN VAR LIST ASSIGN VAR LIST NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR RETURN NUMBER EXPR FUNC_CALL VAR VAR RETURN NUMBER FUNC_DEF IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL STRING VAR IF VAR FUNC_CALL VAR VAR IF VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR RETURN ASSIGN VAR LIST EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR NUMBER VAR VAR RETURN VAR NUMBER VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: self.output = [] self.res = 0 self.dfs(arr) return self.res def checkUnique(self, s): m = set() for i in s: if i not in m: m.add(i) else: return False return True def dfs(self, arr, first=0, curr=[]): st = "".join(curr) if self.checkUnique(st): self.output.append(st) self.res = max(len(st), self.res) for i in range(first, len(arr)): curr.append(arr[i]) self.dfs(arr, i + 1, curr) curr.pop()
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR RETURN NUMBER RETURN NUMBER FUNC_DEF NUMBER LIST ASSIGN VAR FUNC_CALL STRING VAR IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: if len(arr) == 1: if len(set(arr[0])) == len(list(arr[0])): return len(set(arr[0])) else: return 0 all_combinations = [] max_ = 0 for r in range(1, len(arr) + 1): combinations_object = itertools.combinations(arr, r) combinations_list = list(combinations_object) for st in combinations_list: comb = "".join(st) if len(set(comb)) == len(comb): max_ = max(max_, len(comb)) return max_
CLASS_DEF FUNC_DEF VAR VAR IF FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER RETURN NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR ASSIGN VAR FUNC_CALL STRING 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 an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution(object): def maxLength(self, arr): def get_max_len(arr): dp = [set(x) for x in arr if len(set(x)) == len(x)] for v in arr: if len((a := set(v))) == len(v): for b in dp: if a & b: continue dp.append(a | b) return max(len(x) for x in dp) if dp else 0 return get_max_len(arr)
CLASS_DEF VAR FUNC_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR VAR IF BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR RETURN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR NUMBER RETURN FUNC_CALL VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: def score(s): return sum(1 << ord(c) - ord("a") for c in s) def valid(s): return collections.Counter(s).most_common()[0][1] <= 1 arr = [s for s in arr if valid(s)] s = [score(s) for s in arr] res = 0 for i in range(2 ** len(s)): val = 0 cur = 0 for j in range(len(s)): if 1 << j & i == 0: continue if s[j] & val: break val |= s[j] cur += len(arr[j]) else: res = max(res, cur) return res
CLASS_DEF FUNC_DEF VAR VAR FUNC_DEF RETURN FUNC_CALL VAR BIN_OP NUMBER BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING VAR VAR FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP BIN_OP NUMBER VAR VAR NUMBER IF BIN_OP VAR VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: n = len(arr) self.max = 0 def helper(idx, chars): if idx >= n: if len(chars) == len(set(chars)): self.max = max(self.max, len(chars)) else: helper(idx + 1, chars) helper(idx + 1, chars + list(arr[idx])) helper(0, []) return self.max
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FUNC_DEF IF VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER LIST RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: myQueue = collections.deque([("", 0)]) maxLen = 0 while myQueue: word, start = myQueue.popleft() for i in range(start, len(arr)): newWord = word + arr[i] if self.isUnique(newWord): maxLen = max(maxLen, len(newWord)) myQueue.append((newWord, i + 1)) return maxLen def isUnique(self, s): return len(s) == len(set(s))
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR LIST STRING NUMBER ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR IF FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER RETURN VAR VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: return self.recursive(arr, "", 0) def recursive(self, arr, curr_str, pick_idx): if pick_idx == len(arr): table = {} for ch in curr_str: if ch in table: return 0 table[ch] = True return len(curr_str) return max( self.recursive(arr, curr_str + arr[pick_idx], pick_idx + 1), self.recursive(arr, curr_str, pick_idx + 1), )
CLASS_DEF FUNC_DEF VAR VAR RETURN FUNC_CALL VAR VAR STRING NUMBER VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR ASSIGN VAR DICT FOR VAR VAR IF VAR VAR RETURN NUMBER ASSIGN VAR VAR NUMBER RETURN FUNC_CALL VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: newarr = [] for cur in arr: newarr.append(collections.Counter(cur)) res = [0] def helper(cur, index): if index == len(arr): res[0] = max(res[0], len(cur)) else: if len(arr[index]) == len(newarr[index]): if not newarr[index] & cur: helper(newarr[index] + cur, index + 1) helper(cur, index + 1) helper(collections.Counter(""), 0) return res[0]
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR LIST NUMBER FUNC_DEF IF VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR STRING NUMBER RETURN VAR NUMBER VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: if not arr: return 0 def isUnique(s): return len(set(s)) == len(s) def dfs(arr, res, curr): if isUnique(curr): res[0] = max(res[0], len(curr)) for i in range(len(arr)): dfs(arr[i + 1 :], res, curr + arr[i]) res = [0] dfs(arr, res, "") return res[0]
CLASS_DEF FUNC_DEF VAR VAR IF VAR RETURN NUMBER FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_DEF IF FUNC_CALL VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR VAR VAR ASSIGN VAR LIST NUMBER EXPR FUNC_CALL VAR VAR VAR STRING RETURN VAR NUMBER VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: uniqELements = [""] maximum = 0 for i in range(len(arr)): sz = len(uniqELements) for j in range(sz): x = arr[i] + uniqELements[j] if len(x) == len(set(x)): uniqELements.append(x) maximum = max(maximum, len(x)) return maximum
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: def helper(arr, substring, i): unique_length = len(set(substring)) if len(substring) == unique_length: self.max_len = max(self.max_len, unique_length) if i == len(arr) - 1: return for j in range(i + 1, len(arr)): helper(arr, substring + arr[j], j) return self.max_len = 0 helper(arr, "", -1) return self.max_len
CLASS_DEF FUNC_DEF VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER RETURN FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR RETURN ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR STRING NUMBER RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: char_freq = {} for i in range(len(arr)): char_freq[i] = set() for c in arr[i]: if c not in char_freq[i]: char_freq[i].add(c) else: del char_freq[i] break items_count = len(char_freq) max_len = 0 for i in range(1, 2**items_count): num = i current_set = set() current_len = 0 for j in list(char_freq.keys()): intersect = current_set.intersection(char_freq[j]) if num & 1 and not intersect: current_set.update(char_freq[j]) current_len += len(arr[j]) max_len = max(max_len, current_len) elif num & 1 and intersect: break num >>= 1 return max_len
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR DICT FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR FOR VAR VAR VAR IF VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP NUMBER VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER VAR VAR NUMBER RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: def valid(s, baskets): return all([(c not in baskets) for c in s]) def find_max_len_recursive(baskets, currIdx): if currIdx == len(arr): return len(baskets) res1 = float("-inf") if valid(arr[currIdx], baskets): res1 = find_max_len_recursive( baskets.union(set(arr[currIdx])), currIdx + 1 ) res2 = find_max_len_recursive(baskets, currIdx + 1) return max(res1, res2) arr = [s for s in arr if len(s) == len(set(s))] return find_max_len_recursive(set(), 0)
CLASS_DEF FUNC_DEF VAR VAR FUNC_DEF RETURN FUNC_CALL VAR VAR VAR VAR VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR STRING IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER RETURN FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: N = len(arr) charSets = [set()] for s in arr: currSet = set(s) if len(currSet) != len(s): continue currSize = len(charSets) for idx in range(currSize): charSet = charSets[idx] if charSet & currSet: continue charSets.append(charSet | currSet) return max(len(charSet) for charSet in charSets)
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FUNC_CALL VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: self.res = 0 def walk(i, temp): self.res = max(self.res, len(temp)) for j in range(i, len(arr)): if all( v <= 1 for v in list(collections.Counter(temp + arr[j]).values()) ): walk(j + 1, temp + arr[j]) walk(0, "") return self.res
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER STRING RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: arr_sets = [set(x) for x in arr] def buildSet(bitlist): output = set() for i, bit in enumerate(bitlist): if bit: output = output.union(arr_sets[i]) return output def recurse(bitlist, index): if index == len(arr): return len(buildSet(bitlist)) new_bitlist = bitlist[:] new_bitlist[index] = 0 not_included = recurse(new_bitlist, index + 1) if len(arr_sets[index]) != len(arr[index]) or buildSet( bitlist ).intersection(arr_sets[index]): return not_included else: new_bitlist[index] = 1 return max(not_included, recurse(new_bitlist, index + 1)) return recurse([(0) for _ in arr], 0)
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FOR VAR VAR FUNC_CALL VAR VAR IF VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR VAR VAR VAR RETURN VAR ASSIGN VAR VAR NUMBER RETURN FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER RETURN FUNC_CALL VAR NUMBER VAR VAR NUMBER VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
def bo(i): return ord(i) - ord("a") class Solution: def maxLength(self, arr: List[str]) -> int: n = len(arr) freq = [] ans = 0 for i in range(n): s1 = set() for j in arr[i]: s1.add(bo(j)) if len(s1) == len(arr[i]): freq.append(s1) n = len(freq) for i in range(1 << n): s = set() cnt = 0 for j in range(n): if i & 1 << j: for k in freq[j]: s.add(k) cnt += len(freq[j]) if cnt != len(s): break if cnt == len(s): ans = max(ans, cnt) if ans == 26: break return ans
FUNC_DEF RETURN BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP NUMBER VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR BIN_OP NUMBER VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr): prev = {(0): 0} for word in arr: seen = 0 duplicate = False for char in word: binaryC = 1 << ord(char) - ord("a") if binaryC & seen != 0: duplicate = True break else: seen |= binaryC if duplicate: continue toAdd = dict() for k in prev: if k & seen == 0: toAdd[k | seen] = prev[k] + len(word) prev.update(toAdd) return max(prev.values()) class Solution: def maxLength(self, arr): cands = {(0): 0} for word in arr: if len(word) != len(set(word)): continue currW = sum(1 << ord(char) - ord("a") for char in word) toAdd = dict() for prevW in cands: if prevW & currW == 0: toAdd[prevW + currW] = cands[prevW] + len(word) cands.update(toAdd) return max(cands.values())
CLASS_DEF FUNC_DEF ASSIGN VAR DICT NUMBER NUMBER FOR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR BIN_OP NUMBER BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING IF BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER VAR VAR IF VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR CLASS_DEF FUNC_DEF ASSIGN VAR DICT NUMBER NUMBER FOR VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP NUMBER BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR BIN_OP VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def is_unique(self, word): letters = set() for letter in word: if letter in letters: return False letters.add(letter) return True def maxLength(self, arr: List[str]) -> int: uniques = set() for piece in arr: next_uniques = set() for unique in uniques: concat = piece + unique if self.is_unique(concat): next_uniques.add(concat) if self.is_unique(piece): next_uniques.add(piece) for u in next_uniques: uniques.add(u) if uniques: return max([len(x) for x in uniques]) return 0
CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR RETURN NUMBER EXPR FUNC_CALL VAR VAR RETURN NUMBER FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR RETURN NUMBER VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: if len(arr) == 1: return len(arr[0]) self.res = 0 def backtrack(start: int, path: set): self.res = max(self.res, len(path)) if start == len(arr): return for i in range(start, len(arr)): new_chars = set(arr[i]) if len(new_chars) != len(arr[i]): continue if len(new_chars & path) == 0: new_path = new_chars | path backtrack(i + 1, new_path) backtrack(0, set()) return self.res
CLASS_DEF FUNC_DEF VAR VAR IF FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR RETURN FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER FUNC_CALL VAR RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: def digit_representation(s): ans = 0 for c in s: ans |= 1 << ord(c) - ord("a") return ans A = sorted( [ (len(s), digit_representation(s)) for s in set(arr) if len(set(s)) == len(s) ], reverse=True, ) if not A: return 0 R = [sum(t[0] for t in A)] for i in range(1, len(A)): R.append(R[-1] - A[i][0]) self.ans = A[0][0] def helper(i, b, k): if i == len(A): self.ans = max(self.ans, k) elif k + R[i] > self.ans: if not b & A[i][1]: helper(i + 1, b | A[i][1], k + A[i][0]) helper(i + 1, b, k) helper(0, 0, 0) return self.ans
CLASS_DEF FUNC_DEF VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR BIN_OP NUMBER BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER IF VAR RETURN NUMBER ASSIGN VAR LIST FUNC_CALL VAR VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FUNC_DEF IF VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR VAR VAR VAR IF BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR NUMBER BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR NUMBER NUMBER NUMBER RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: self.ans = 0 vis = set() def help(s, i): if len(s) == len(set(s)): self.ans = max(self.ans, len(s)) if i >= len(arr): return for j in range(i, len(arr)): if j not in vis: vis.add(j) help(s + arr[j], j + 1) vis.remove(j) help("", 0) return self.ans
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_DEF IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR RETURN FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING NUMBER RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: self.out = set() self.visited = set() opts = [] for sub in arr: tmp = set() omit = False for ch in sub: if ch in tmp: omit = True tmp.add(ch) if not omit: opts.append(tmp) self.dfs(set(), opts) return len(self.out) def dfs(self, curr, opts): if tuple(sorted(curr)) in self.visited: return self.visited.add(tuple(sorted(curr))) if len(curr) > len(self.out): self.out = curr if not opts: return for j, opt in enumerate(opts): if not curr.intersection(opt): self.dfs(curr.union(opt), opts[:j] + opts[j + 1 :])
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR FUNC_DEF IF FUNC_CALL VAR FUNC_CALL VAR VAR VAR RETURN EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR IF VAR RETURN FOR VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: def isUnique(concatString): concatSet = set() for char in concatString: if char in concatSet: return False concatSet.add(char) return True length = len(arr) self.result = 0 def dfs(index, curr): if isUnique(curr): self.result = max(self.result, len(curr)) if index == length - 1: return for i in range(index + 1, length): dfs(i, curr + arr[i]) for i in range(length): dfs(i, arr[i]) return self.result
CLASS_DEF FUNC_DEF VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR RETURN NUMBER EXPR FUNC_CALL VAR VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FUNC_DEF IF FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER RETURN FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: def maxLength(i, string): nonlocal ans if i == len(arr): s, invalid = set(), False for ch in string: if ch in s: invalid = True break s.add(ch) if not invalid: ans = max(ans, len(string)) return maxLength(i + 1, string) maxLength(i + 1, string + arr[i]) ans = 0 maxLength(0, "") return ans
CLASS_DEF FUNC_DEF VAR VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR NUMBER FOR VAR VAR IF VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR NUMBER STRING RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: def unique_char_count(s): chars = set() for char in s: if char in chars: return -1 else: chars.add(char) return len(s) def max_unique(arr, index, cur_str, result): if index >= len(arr): if unique_char_count(cur_str) > result[0]: result[0] = len(cur_str) return max_unique(arr, index + 1, cur_str + arr[index], result) max_unique(arr, index + 1, cur_str, result) result = [0] max_unique(arr, 0, "", result) return result[0]
CLASS_DEF FUNC_DEF VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR RETURN NUMBER EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR RETURN EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR LIST NUMBER EXPR FUNC_CALL VAR VAR NUMBER STRING VAR RETURN VAR NUMBER VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: count = 0 def dfs(self, arr, i, visited): for index in range(i + 1, len(arr)): visit = True for j in range(0, len(arr[index])): if arr[index][j] in visited: for k in range(0, j): visited.remove(arr[index][k]) visit = False break else: visited.add(arr[index][j]) if visit: self.count = max(self.count, len(visited)) self.dfs(arr, index, visited) for char in arr[index]: if char in visited and visit: visited.remove(char) def maxLength(self, arr: List[str]) -> int: if len(arr) == 0: return 0 visited = set() self.count = 0 self.dfs(arr, -1, visited) return self.count
CLASS_DEF ASSIGN VAR NUMBER FUNC_DEF FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR IF VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR FOR VAR VAR VAR IF VAR VAR VAR EXPR FUNC_CALL VAR VAR FUNC_DEF VAR VAR IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER VAR RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: a = [set(i) for i in arr if len(set(i)) == len(i)] n = len(a) an = 0 for v in range(1, 2**n): s = set() i = 0 while v > 0: if v & 1 == 1: if s & a[i]: break s = s ^ a[i] v = v >> 1 i += 1 an = max(an, len(s)) return an
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP NUMBER VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: d = [""] for a in arr: if len(set(a)) < len(a): continue else: for x in d: if set(x) & set(a): continue else: t = set(a) | set(x) d.append(t) max = 0 for i in d: if len(i) > max: max = len(i) return max
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST STRING FOR VAR VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR VAR IF BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR RETURN VAR VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: dp = defaultdict(lambda: set()) def dfs(index, cur): if index == len(arr) or cur in dp[index]: return len(cur) dp[index].add(cur) best = len(cur) for x in range(index, len(arr)): count = Counter(cur) xCount = Counter(arr[x]) canTake = True for k, v in xCount.items(): if k in count or v > 1: canTake = False if canTake: best = max(best, dfs(x + 1, cur + arr[x])) best = max(best, dfs(x + 1, cur)) return best return dfs(0, "")
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR IF VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR RETURN VAR RETURN FUNC_CALL VAR NUMBER STRING VAR
Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possible length of s.   Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique". Maximum length is 4. Example 2: Input: arr = ["cha","r","act","ers"] Output: 6 Explanation: Possible solutions are "chaers" and "acters". Example 3: Input: arr = ["abcdefghijklmnopqrstuvwxyz"] Output: 26   Constraints: 1 <= arr.length <= 16 1 <= arr[i].length <= 26 arr[i] contains only lower case English letters.
class Solution: def maxLength(self, arr: List[str]) -> int: self.array = arr return max(self.recurse("", set(), 0, True), self.recurse("", set(), 0, False)) def recurse(self, prefix: str, letters: set, index: int, include: bool): if index >= len(self.array): return len(prefix) if include: if len(self.array[index]) != len(set(self.array[index])): return len(prefix) if any(letter in letters for letter in self.array[index]): return len(prefix) return max( self.recurse( "".join((prefix, self.array[index])), letters | set(self.array[index]), index + 1, True, ), self.recurse( "".join((prefix, self.array[index])), letters | set(self.array[index]), index + 1, False, ), ) return max( self.recurse(prefix, letters, index + 1, True), self.recurse(prefix, letters, index + 1, False), )
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR STRING FUNC_CALL VAR NUMBER NUMBER FUNC_CALL VAR STRING FUNC_CALL VAR NUMBER NUMBER VAR FUNC_DEF VAR VAR VAR VAR IF VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR IF VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL STRING VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR FUNC_CALL STRING VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER
Childan is making up a legendary story and trying to sell his forgery — a necklace with a strong sense of "Wu" to the Kasouras. But Mr. Kasoura is challenging the truth of Childan's story. So he is going to ask a few questions about Childan's so-called "personal treasure" necklace. This "personal treasure" is a multiset S of m "01-strings". A "01-string" is a string that contains n characters "0" and "1". For example, if n=4, strings "0110", "0000", and "1110" are "01-strings", but "00110" (there are 5 characters, not 4) and "zero" (unallowed characters) are not. Note that the multiset S can contain equal elements. Frequently, Mr. Kasoura will provide a "01-string" t and ask Childan how many strings s are in the multiset S such that the "Wu" value of the pair (s, t) is not greater than k. Mrs. Kasoura and Mr. Kasoura think that if s_i = t_i (1≤ i≤ n) then the "Wu" value of the character pair equals to w_i, otherwise 0. The "Wu" value of the "01-string" pair is the sum of the "Wu" values of every character pair. Note that the length of every "01-string" is equal to n. For example, if w=[4, 5, 3, 6], "Wu" of ("1001", "1100") is 7 because these strings have equal characters only on the first and third positions, so w_1+w_3=4+3=7. You need to help Childan to answer Mr. Kasoura's queries. That is to find the number of strings in the multiset S such that the "Wu" value of the pair is not greater than k. Input The first line contains three integers n, m, and q (1≤ n≤ 12, 1≤ q, m≤ 5⋅ 10^5) — the length of the "01-strings", the size of the multiset S, and the number of queries. The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_i ≤ 100) — the value of the i-th caracter. Each of the next m lines contains the "01-string" s of length n — the string in the multiset S. Each of the next q lines contains the "01-string" t of length n and integer k (0≤ k≤ 100) — the query. Output For each query, print the answer for this query. Examples Input 2 4 5 40 20 01 01 10 11 00 20 00 40 11 20 11 40 11 60 Output 2 4 2 3 4 Input 1 2 4 100 0 1 0 0 0 100 1 0 1 100 Output 1 2 1 2 Note In the first example, we can get: "Wu" of ("01", "00") is 40. "Wu" of ("10", "00") is 20. "Wu" of ("11", "00") is 0. "Wu" of ("01", "11") is 20. "Wu" of ("10", "11") is 40. "Wu" of ("11", "11") is 60. In the first query, pairs ("11", "00") and ("10", "00") satisfy the condition since their "Wu" is not greater than 20. In the second query, all strings satisfy the condition. In the third query, pairs ("01", "11") and ("01", "11") satisfy the condition. Note that since there are two "01" strings in the multiset, the answer is 2, not 1. In the fourth query, since k was increased, pair ("10", "11") satisfies the condition too. In the fifth query, since k was increased, pair ("11", "11") satisfies the condition too.
from sys import stdin def main(): n, m, q = map(int, input().split()) c = sz = 1 << n costs, cnt = [0] * sz, [0] * sz for w in map(int, input().split()): c //= 2 for i in range(c, sz, c * 2): for j in range(i, i + c): costs[j] += w for i, c in enumerate(costs): if c > 101: costs[i] = 101 costs.reverse() l = stdin.read().splitlines() for s in l[:m]: cnt[int(s, 2)] += 1 del l[:m] cache = [] for i in range(sz): row = [0] * 102 cache.append(row) for j, c in enumerate(cnt): row[costs[i ^ j]] += c a = 0 for j, b in enumerate(row): a += b row[j] = a for i, s in enumerate(l): row = s.split() l[i] = str(cache[int(row[0], 2)][int(row[1])]) print("\n".join(l)) main()
FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR BIN_OP NUMBER VAR ASSIGN VAR VAR BIN_OP LIST NUMBER VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR VAR FOR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR FOR VAR VAR VAR VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR VAR FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR
Childan is making up a legendary story and trying to sell his forgery — a necklace with a strong sense of "Wu" to the Kasouras. But Mr. Kasoura is challenging the truth of Childan's story. So he is going to ask a few questions about Childan's so-called "personal treasure" necklace. This "personal treasure" is a multiset S of m "01-strings". A "01-string" is a string that contains n characters "0" and "1". For example, if n=4, strings "0110", "0000", and "1110" are "01-strings", but "00110" (there are 5 characters, not 4) and "zero" (unallowed characters) are not. Note that the multiset S can contain equal elements. Frequently, Mr. Kasoura will provide a "01-string" t and ask Childan how many strings s are in the multiset S such that the "Wu" value of the pair (s, t) is not greater than k. Mrs. Kasoura and Mr. Kasoura think that if s_i = t_i (1≤ i≤ n) then the "Wu" value of the character pair equals to w_i, otherwise 0. The "Wu" value of the "01-string" pair is the sum of the "Wu" values of every character pair. Note that the length of every "01-string" is equal to n. For example, if w=[4, 5, 3, 6], "Wu" of ("1001", "1100") is 7 because these strings have equal characters only on the first and third positions, so w_1+w_3=4+3=7. You need to help Childan to answer Mr. Kasoura's queries. That is to find the number of strings in the multiset S such that the "Wu" value of the pair is not greater than k. Input The first line contains three integers n, m, and q (1≤ n≤ 12, 1≤ q, m≤ 5⋅ 10^5) — the length of the "01-strings", the size of the multiset S, and the number of queries. The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_i ≤ 100) — the value of the i-th caracter. Each of the next m lines contains the "01-string" s of length n — the string in the multiset S. Each of the next q lines contains the "01-string" t of length n and integer k (0≤ k≤ 100) — the query. Output For each query, print the answer for this query. Examples Input 2 4 5 40 20 01 01 10 11 00 20 00 40 11 20 11 40 11 60 Output 2 4 2 3 4 Input 1 2 4 100 0 1 0 0 0 100 1 0 1 100 Output 1 2 1 2 Note In the first example, we can get: "Wu" of ("01", "00") is 40. "Wu" of ("10", "00") is 20. "Wu" of ("11", "00") is 0. "Wu" of ("01", "11") is 20. "Wu" of ("10", "11") is 40. "Wu" of ("11", "11") is 60. In the first query, pairs ("11", "00") and ("10", "00") satisfy the condition since their "Wu" is not greater than 20. In the second query, all strings satisfy the condition. In the third query, pairs ("01", "11") and ("01", "11") satisfy the condition. Note that since there are two "01" strings in the multiset, the answer is 2, not 1. In the fourth query, since k was increased, pair ("10", "11") satisfies the condition too. In the fifth query, since k was increased, pair ("11", "11") satisfies the condition too.
import sys n, m, q = map(int, sys.stdin.readline().strip().split()) w = [int(x) for x in sys.stdin.readline().strip().split()] mx = pow(2, n) mask = mx - 1 sn = [0] * mx for i in range(m): s = sys.stdin.readline().strip() si = int(s, 2) sn[si] += 1 wu = [0] * mx for i in range(mx): for j in range(n): if 1 << j & i != 0: wu[i] += w[n - j - 1] c = [] for i in range(mx): c.append([0] * 102) for j in range(mx): join = (~i ^ j) & mask val = wu[join] if val > 100: val = 101 c[i][val] += sn[j] for j in range(1, 102): c[i][j] += c[i][j - 1] for i in range(q): row = sys.stdin.readline().strip().split() t = int(row[0], 2) k = int(row[1]) sys.stdout.write(str(c[t][k]) + "\n")
IMPORT ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP BIN_OP NUMBER VAR VAR NUMBER VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER NUMBER VAR VAR VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR VAR STRING
Given a set of distinct integers, nums, return all possible subsets (the power set). Note: The solution set must not contain duplicate subsets. Example: Input: nums = [1,2,3] Output: [ [3],   [1],   [2],   [1,2,3],   [1,3],   [2,3],   [1,2],   [] ]
class Solution: def subsets(self, nums): subsets = [[]] for v in nums: extra = [] for s in subsets: extra.append(s + [v]) subsets += extra return subsets
CLASS_DEF FUNC_DEF ASSIGN VAR LIST LIST FOR VAR VAR ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR LIST VAR VAR VAR RETURN VAR
Given a set of distinct integers, nums, return all possible subsets (the power set). Note: The solution set must not contain duplicate subsets. Example: Input: nums = [1,2,3] Output: [ [3],   [1],   [2],   [1,2,3],   [1,3],   [2,3],   [1,2],   [] ]
class Solution: def subsets(self, nums): result = [[]] for num in nums: result += self.get_subset(result, num) return result def get_subset(self, subsets, item): new_subset_with_item = [] for subset in subsets: new_subset_with_item.append(subset + [item]) return new_subset_with_item
CLASS_DEF FUNC_DEF ASSIGN VAR LIST LIST FOR VAR VAR VAR FUNC_CALL VAR VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR LIST VAR RETURN VAR
Given a set of distinct integers, nums, return all possible subsets (the power set). Note: The solution set must not contain duplicate subsets. Example: Input: nums = [1,2,3] Output: [ [3],   [1],   [2],   [1,2,3],   [1,3],   [2,3],   [1,2],   [] ]
class Solution: def subsets(self, nums): def mask(x): return 1 << x exp = 1 << len(nums) ans = [] for count in range(exp): new = [] i = 0 while mask(i) <= count: if mask(i) & count: new.append(nums[i]) i += 1 ans.append(new) return ans
CLASS_DEF FUNC_DEF FUNC_DEF RETURN BIN_OP NUMBER VAR ASSIGN VAR BIN_OP NUMBER FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE FUNC_CALL VAR VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN VAR
On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 represents obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.   Example 1: Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.   Note: 1 <= grid.length * grid[0].length <= 20
class Solution: def uniquePathsIII(self, grid: List[List[int]]) -> int: self.ans, empty = 0, 0 m, n = len(grid[0]), len(grid) def dfs(x, y, rest): if x < 0 or x >= n or y < 0 or y >= m or grid[x][y] < 0: return if grid[x][y] == 2 and rest == 0: self.ans += 1 return neibs = [[0, 1], [0, -1], [1, 0], [-1, 0]] for dx, dy in neibs: tmp = grid[x][y] grid[x][y] = -2 dfs(x + dx, y + dy, rest - 1) grid[x][y] = tmp for i, j in product(list(range(n)), list(range(m))): if grid[i][j] == 1: sx, sy = i, j empty += grid[i][j] != -1 dfs(sx, sy, empty - 1) return self.ans
CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR FUNC_DEF IF VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR VAR VAR NUMBER RETURN IF VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER RETURN ASSIGN VAR LIST LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER FOR VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR FOR VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER RETURN VAR VAR
On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 represents obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.   Example 1: Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.   Note: 1 <= grid.length * grid[0].length <= 20
class Solution: def uniquePathsIII(self, grid: List[List[int]]) -> int: rows, cols = len(grid), len(grid[0]) non_obstacles = 0 start_row, start_col = 0, 0 for row in range(0, rows): for col in range(0, cols): cell = grid[row][col] if cell >= 0: non_obstacles += 1 if cell == 1: start_row, start_col = row, col path_count = 0 def backtrack(row, col, remain): nonlocal path_count if grid[row][col] == 2 and remain == 1: path_count += 1 return temp = grid[row][col] grid[row][col] = -4 remain -= 1 for ro, co in [(0, 1), (0, -1), (-1, 0), (1, 0)]: next_row, next_col = row + ro, col + co if not (0 <= next_row < rows and 0 <= next_col < cols): continue if grid[next_row][next_col] < 0: continue backtrack(next_row, next_col, remain) grid[row][col] = temp backtrack(start_row, start_col, non_obstacles) return path_count
CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER FUNC_DEF IF VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER RETURN ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER VAR NUMBER FOR VAR VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR IF NUMBER VAR VAR NUMBER VAR VAR IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR RETURN VAR VAR
On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 represents obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.   Example 1: Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.   Note: 1 <= grid.length * grid[0].length <= 20
class Solution: def uniquePathsIII(self, grid: List[List[int]]) -> int: def traverse(row, col, spacesLeft): if spacesLeft == 1 and grid[row][col] == 2: self.results += 1 return temp = grid[row][col] grid[row][col] = "#" spacesLeft -= 1 for rowOffset, colOffset in [(1, 0), (0, 1), (-1, 0), (0, -1)]: nextRow, nextCol = row + rowOffset, col + colOffset if nextRow < 0 or nextRow == rows or nextCol < 0 or nextCol == cols: continue if grid[nextRow][nextCol] == -1 or grid[nextRow][nextCol] == "#": continue traverse(nextRow, nextCol, spacesLeft) grid[row][col] = temp rows, cols = len(grid), len(grid[0]) spaces = 0 for row in range(rows): for col in range(cols): if grid[row][col] >= 0: spaces += 1 if grid[row][col] == 1: start = row, col self.results = 0 traverse(start[0], start[1], spaces) return self.results
CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF IF VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER RETURN ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR STRING VAR NUMBER FOR VAR VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR IF VAR NUMBER VAR VAR VAR NUMBER VAR VAR IF VAR VAR VAR NUMBER VAR VAR VAR STRING EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR RETURN VAR VAR
On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 represents obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.   Example 1: Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.   Note: 1 <= grid.length * grid[0].length <= 20
class Solution: def uniquePathsIII(self, grid: List[List[int]]) -> int: def dfs(i, j, out, to_visit): if ( i < 0 or j < 0 or i >= len(grid) or j >= len(grid[0]) or grid[i][j] == "#" or grid[i][j] == -1 ): return if grid[i][j] == 2 and to_visit == 0: out[0] += 1 else: tmp = grid[i][j] grid[i][j] = "#" dfs(i, j + 1, out, to_visit - 1) dfs(i, j - 1, out, to_visit - 1) dfs(i + 1, j, out, to_visit - 1) dfs(i - 1, j, out, to_visit - 1) grid[i][j] = tmp res = [0] ori, orj = 0, 0 to_visit = 0 for i in range(len(grid)): for j in range(len(grid[0])): if grid[i][j] == 1: ori, orj = i, j if grid[i][j] != -1: to_visit += 1 dfs(ori, orj, res, to_visit - 1) return res[0]
CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF IF VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR STRING VAR VAR VAR NUMBER RETURN IF VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR STRING EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR LIST NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER RETURN VAR NUMBER VAR
On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 represents obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.   Example 1: Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.   Note: 1 <= grid.length * grid[0].length <= 20
class Solution: def uniquePathsIII(self, grid: List[List[int]]) -> int: def fun(g, i0, i1, n1, t1, f2, row, col): if n1 == t1: if abs(f2[1] - i1) + abs(f2[0] - i0) == 1: self.ans += 1 else: if i1 + 1 != col and g[i0][i1 + 1] == 0: g[i0][i1 + 1] = 1 fun(g, i0, i1 + 1, n1 + 1, t1, f2, row, col) g[i0][i1 + 1] = 0 if i1 != 0 and g[i0][i1 - 1] == 0: g[i0][i1 - 1] = 1 fun(g, i0, i1 - 1, n1 + 1, t1, f2, row, col) g[i0][i1 - 1] = 0 if i0 + 1 != row and g[i0 + 1][i1] == 0: g[i0 + 1][i1] = 1 fun(g, i0 + 1, i1, n1 + 1, t1, f2, row, col) g[i0 + 1][i1] = 0 if i0 != 0 and g[i0 - 1][i1] == 0: g[i0 - 1][i1] = 1 fun(g, i0 - 1, i1, n1 + 1, t1, f2, row, col) g[i0 - 1][i1] = 0 self.ans = 0 i0, i1, t1 = 0, 0, 0 f2 = [0, 0] row = len(grid) col = len(grid[0]) for i in range(row): for j in range(col): if grid[i][j] == 0: t1 += 1 elif grid[i][j] == 1: i0, i1 = i, j elif grid[i][j] == 2: f2 = [i, j] fun(grid, i0, i1, 0, t1, f2, row, col) return self.ans
CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF IF VAR VAR IF BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER IF BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER NUMBER IF VAR NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER NUMBER IF BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR LIST VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR VAR VAR RETURN VAR VAR
On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 represents obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.   Example 1: Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.   Note: 1 <= grid.length * grid[0].length <= 20
class Solution: def uniquePathsIII(self, grid: List[List[int]]) -> int: def dfs(r, c, n): if r < 0 or r >= R or c < 0 or c >= C or grid[r][c] == -1: return 0 if grid[r][c] == 2: return n == 0 grid[r][c] = -1 paths = ( dfs(r + 1, c, n - 1) + dfs(r - 1, c, n - 1) + dfs(r, c + 1, n - 1) + dfs(r, c - 1, n - 1) ) grid[r][c] = 0 return paths R = len(grid) C = len(grid[0]) n = 1 sr, sc = -1, -1 for r in range(R): for c in range(C): if grid[r][c] == 0: n += 1 elif grid[r][c] == 1: sr, sc = r, c return dfs(sr, sc, n)
CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF IF VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR VAR VAR NUMBER RETURN NUMBER IF VAR VAR VAR NUMBER RETURN VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR VAR
On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 represents obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.   Example 1: Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.   Note: 1 <= grid.length * grid[0].length <= 20
class Solution: def uniquePathsIII(self, grid: List[List[int]]) -> int: self.ans, self.count = 0, 0 def go(cur, seen): i, j = cur if grid[i][j] == 2: if len(seen) == self.count: self.ans += 1 else: return for r, c in [(i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)]: if ( 0 <= r < len(grid) and 0 <= c < len(grid[0]) and (r, c) not in seen and grid[r][c] != -1 ): seen.add((r, c)) go((r, c), seen) seen.remove((r, c)) for i in range(len(grid)): for j in range(len(grid[0])): if grid[i][j] in [0, 1, 2]: self.count += 1 for i in range(len(grid)): for j in range(len(grid[0])): if grid[i][j] == 1: go((i, j), {(i, j)}) return self.ans return ans
CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR VAR NUMBER NUMBER FUNC_DEF ASSIGN VAR VAR VAR IF VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR NUMBER RETURN FOR VAR VAR LIST BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER IF NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR LIST NUMBER NUMBER NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR RETURN VAR RETURN VAR VAR
On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 represents obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.   Example 1: Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.   Note: 1 <= grid.length * grid[0].length <= 20
class Solution: def uniquePathsIII(self, grid: List[List[int]]) -> int: nr = len(grid) nc = len(grid[0]) all_cells = 0 start = 0, 0 res = 0 direction = [(0, 1), (0, -1), (1, 0), (-1, 0)] for r in range(nr): for c in range(nc): if grid[r][c] >= 0: all_cells += 1 if grid[r][c] == 1: start = r, c def dfs(r, c, cell_nums): nonlocal res nonlocal direction nonlocal grid nonlocal nr nonlocal nc if grid[r][c] == 2 and cell_nums == 1: res += 1 return temp = grid[r][c] grid[r][c] = -2 cell_nums -= 1 for d in direction: rr = r + d[0] cc = c + d[1] if 0 <= rr < nr and 0 <= cc < nc and grid[rr][cc] >= 0: dfs(rr, cc, cell_nums) grid[r][c] = temp dfs(*start, all_cells) return res
CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR FUNC_DEF IF VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER RETURN ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER VAR NUMBER FOR VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER IF NUMBER VAR VAR NUMBER VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR RETURN VAR VAR
On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 represents obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.   Example 1: Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.   Note: 1 <= grid.length * grid[0].length <= 20
class Solution: def uniquePathsIII(self, grid: List[List[int]]) -> int: self.ret = 0 empty = 0 m = len(grid) n = len(grid[0]) for i in range(m): for j in range(n): if grid[i][j] == 1: sx = i sy = j if grid[i][j] >= 0: empty += 1 def backtrack(grid, x, y, num_left): if x < 0 or y < 0 or x >= m or y >= n or grid[x][y] < 0: return if grid[x][y] == 2 and num_left == 0: self.ret += 1 return neighs = [[0, 1], [1, 0], [0, -1], [-1, 0]] for dx, dy in neighs: orig_value = grid[x][y] grid[x][y] = -1 backtrack(grid, x + dx, y + dy, num_left - 1) grid[x][y] = orig_value backtrack(grid, sx, sy, empty - 1) return self.ret
CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR IF VAR VAR VAR NUMBER VAR NUMBER FUNC_DEF IF VAR NUMBER VAR NUMBER VAR VAR VAR VAR VAR VAR VAR NUMBER RETURN IF VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER RETURN ASSIGN VAR LIST LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER LIST NUMBER NUMBER FOR VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER RETURN VAR VAR
On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 represents obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.   Example 1: Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.   Note: 1 <= grid.length * grid[0].length <= 20
class Solution: def uniquePathsIII(self, grid: List[List[int]]) -> int: rows, cols = len(grid), len(grid[0]) directions = [(1, 0), (0, 1), (-1, 0), (0, -1)] start_cell, end_cell = None, None empty_cells = {} for i in range(rows): for j in range(cols): if grid[i][j] in [0, 1]: empty_cells[i, j] = set() if grid[i][j] == 1: start_cell = i, j elif grid[i][j] == 2: end_cell = i, j for x, y in empty_cells: for xch, ych in directions: neighbor = x + xch, y + ych if neighbor in empty_cells: empty_cells[x, y].add(neighbor) next_to_finish = set( (end_cell[0] + xch, end_cell[1] + ych) for xch, ych in directions ) visited = set([start_cell]) def count_routes(cell=start_cell): left_to_visit = len(empty_cells) - len(visited) if not left_to_visit: return int(cell in next_to_finish) routes_to_end = 0 for neighbor in empty_cells[cell] - visited: visited.add(neighbor) routes_to_end += count_routes(neighbor) visited.remove(neighbor) return routes_to_end return count_routes()
CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER ASSIGN VAR VAR NONE NONE ASSIGN VAR DICT FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR LIST NUMBER NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR FOR VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR BIN_OP VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR LIST VAR FUNC_DEF VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR RETURN FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR RETURN FUNC_CALL VAR VAR
On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 represents obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.   Example 1: Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.   Note: 1 <= grid.length * grid[0].length <= 20
class Solution: def uniquePathsIII(self, grid: List[List[int]]) -> int: self.ans = 0 m, n = len(grid), len(grid[0]) candidates = set() for i in range(m): for j in range(n): if grid[i][j] == 1: si, sj = i, j elif grid[i][j] == 0 or grid[i][j] == 2: candidates.add((i, j)) def helper(i, j): if grid[i][j] == 2: if not candidates: self.ans += 1 return if grid[i][j] == -1: return if i < m - 1 and (i + 1, j) in candidates: candidates.remove((i + 1, j)) helper(i + 1, j) candidates.add((i + 1, j)) if i > 0 and (i - 1, j) in candidates: candidates.remove((i - 1, j)) helper(i - 1, j) candidates.add((i - 1, j)) if j < n - 1 and (i, j + 1) in candidates: candidates.remove((i, j + 1)) helper(i, j + 1) candidates.add((i, j + 1)) if j > 0 and (i, j - 1) in candidates: candidates.remove((i, j - 1)) helper(i, j - 1) candidates.add((i, j - 1)) helper(si, sj) return self.ans
CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR FUNC_DEF IF VAR VAR VAR NUMBER IF VAR VAR NUMBER RETURN IF VAR VAR VAR NUMBER RETURN IF VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR RETURN VAR VAR
On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 represents obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.   Example 1: Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.   Note: 1 <= grid.length * grid[0].length <= 20
class Solution: def uniquePathsIII(self, grid: List[List[int]]) -> int: count = 0 visited = set() def dfs(i: int, j: int): nonlocal count nonlocal num_empty if grid[i][j] == -1 or (i, j) in visited: return elif grid[i][j] == 2 and len(visited) == num_empty + 1: count += 1 return else: visited.add((i, j)) if i > 0: dfs(i - 1, j) if j > 0: dfs(i, j - 1) if i < len(grid) - 1: dfs(i + 1, j) if j < len(grid[0]) - 1: dfs(i, j + 1) visited.remove((i, j)) start_i = start_j = None num_empty = 0 for i in range(len(grid)): for j in range(len(grid[0])): if grid[i][j] == 1: start_i, start_j = i, j elif grid[i][j] == 0: num_empty += 1 dfs(start_i, start_j) return count
CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_DEF VAR VAR IF VAR VAR VAR NUMBER VAR VAR VAR RETURN IF VAR VAR VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR NUMBER RETURN EXPR FUNC_CALL VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR NONE ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR RETURN VAR VAR
On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 represents obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.   Example 1: Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.   Note: 1 <= grid.length * grid[0].length <= 20
class Solution: def uniquePathsIII(self, grid: List[List[int]]) -> int: start = end = None cnt = ans = 0 m, n = len(grid), len(grid[0]) for i in range(m): for j in range(n): if grid[i][j] == 1: start = i, j elif grid[i][j] == 1: end = i, j if grid[i][j] != -1: cnt += 1 def dfs(i, j, cnt): nonlocal m, n, ans if grid[i][j] == 2 and cnt == 1: ans += 1 return val = grid[i][j] grid[i][j] = -1 for x, y in map( lambda t: (t[0] + i, t[1] + j), [(-1, 0), (1, 0), (0, -1), (0, 1)] ): if 0 <= x < m and 0 <= y < n and grid[x][y] != -1: dfs(x, y, cnt - 1) grid[i][j] = val dfs(*start, cnt) return ans
CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR VAR NONE ASSIGN VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR IF VAR VAR VAR NUMBER VAR NUMBER FUNC_DEF IF VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER RETURN ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER FOR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER IF NUMBER VAR VAR NUMBER VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR RETURN VAR VAR
On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 represents obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.   Example 1: Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.   Note: 1 <= grid.length * grid[0].length <= 20
class Solution: def uniquePathsIII(self, grid: List[List[int]]) -> int: start = end = None remain = 0 for m in range(len(grid)): for n in range(len(grid[0])): if grid[m][n] == 1: start = m, n elif grid[m][n] == 2: end = m, n if grid[m][n] == 0: remain += 1 if start is None or end is None: return 0 return self._helper(grid, start[0], start[1], end[0], end[1], remain + 1) def _helper( self, grid: List[List[int]], start_m: int, start_n: int, end_m: int, end_n: int, remain: int, ) -> int: if ( 0 <= start_m < len(grid) and 0 <= start_n < len(grid[0]) and grid[start_m][start_n] > -1 ): if remain < 0: return 0 elif remain == 0 and start_m == end_m and start_n == end_n: return 1 count = 0 grid[start_m][start_n] = -2 count += self._helper(grid, start_m + 1, start_n, end_m, end_n, remain - 1) count += self._helper(grid, start_m, start_n + 1, end_m, end_n, remain - 1) count += self._helper(grid, start_m - 1, start_n, end_m, end_n, remain - 1) count += self._helper(grid, start_m, start_n - 1, end_m, end_n, remain - 1) grid[start_m][start_n] = 0 return count return 0
CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR VAR NONE ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR IF VAR VAR VAR NUMBER VAR NUMBER IF VAR NONE VAR NONE RETURN NUMBER RETURN FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR IF NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR NUMBER IF VAR NUMBER RETURN NUMBER IF VAR NUMBER VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER RETURN VAR RETURN NUMBER VAR
On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 represents obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.   Example 1: Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.   Note: 1 <= grid.length * grid[0].length <= 20
class Solution: def uniquePathsIII(self, grid): maxPath = 2 neighbors = [None] * (len(grid) * len(grid[0])) visited = [False] * (len(grid) * len(grid[0])) for x in range(len(grid)): for y in range(len(grid[0])): node = x * len(grid[0]) + y if grid[x][y] == 0: neighbors[node] = getNeighbors(grid, x, y) visited[node] = False maxPath += 1 elif grid[x][y] == 1: start = node neighbors[node] = getNeighbors(grid, x, y) elif grid[x][y] == 2: end = node visited[node] = False return count(neighbors, visited, start, maxPath, 1, end) def getNeighbors(grid, x, y): output = [] if x > 0 and grid[x - 1][y] % 2 == 0: output.append((x - 1) * len(grid[0]) + y) if x < len(grid) - 1 and grid[x + 1][y] % 2 == 0: output.append((x + 1) * len(grid[0]) + y) if y > 0 and grid[x][y - 1] % 2 == 0: output.append(x * len(grid[0]) + y - 1) if y < len(grid[0]) - 1 and grid[x][y + 1] % 2 == 0: output.append(x * len(grid[0]) + y + 1) return output def count(neighbors, visited, node, maxPath, pathLength, destination): if pathLength == maxPath and node == destination: return 1 elif pathLength == maxPath or node == destination: return 0 visited[node] = True sum = 0 for k in neighbors[node]: if not visited[k]: sum += count(neighbors, visited, k, maxPath, pathLength + 1, destination) visited[node] = False return sum
CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NONE BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR NUMBER VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR NUMBER VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR NUMBER RETURN FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR FUNC_DEF ASSIGN VAR LIST IF VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR IF VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER RETURN VAR FUNC_DEF IF VAR VAR VAR VAR RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR IF VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR NUMBER RETURN VAR
On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 represents obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.   Example 1: Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.   Note: 1 <= grid.length * grid[0].length <= 20
class Solution: def uniquePathsIII(self, A: List[List[int]]) -> int: self.res = 0 rows = len(A) cols = len(A[0]) empty = 1 for i in range(rows): for j in range(cols): if A[i][j] == 1: x, y = i, j elif A[i][j] == 2: end = i, j elif A[i][j] == 0: empty += 1 self.dfs(A, x, y, empty, end) return self.res def dfs(self, A, x, y, empty, end): if x < 0 or x >= len(A) or y < 0 or y >= len(A[0]) or A[x][y] < 0: return if (x, y) == end: self.res += empty == 0 return A[x][y] = -2 self.dfs(A, x + 1, y, empty - 1, end) self.dfs(A, x - 1, y, empty - 1, end) self.dfs(A, x, y + 1, empty - 1, end) self.dfs(A, x, y - 1, empty - 1, end) A[x][y] = 0
CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR IF VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR RETURN VAR VAR FUNC_DEF IF VAR NUMBER VAR FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR NUMBER RETURN IF VAR VAR VAR VAR VAR NUMBER RETURN ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER
On a 2-dimensional grid, there are 4 types of squares: 1 represents the starting square.  There is exactly one starting square. 2 represents the ending square.  There is exactly one ending square. 0 represents empty squares we can walk over. -1 represents obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.   Example 1: Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.   Note: 1 <= grid.length * grid[0].length <= 20
class Solution: def uniquePathsIII(self, grid: List[List[int]]) -> int: def dfs(i, j, N): if N == -1 and grid[i][j] == 2: return 1 grid[i][j] = -1 res = sum( [ dfs(row, col, N - 1) for x, y in dirs for row, col in [(i + x, j + y)] if 0 <= row < m and 0 <= col < n and (N > 0 and not grid[row][col] or N == 0 and grid[row][col] == 2) ] ) grid[i][j] = 0 return res m, n = len(grid), len(grid[0]) dirs = [(0, 1), (0, -1), (1, 0), (-1, 0)] N = 0 for i in range(m): for j in range(n): if grid[i][j] == 1: start = [i, j] elif not grid[i][j]: N += 1 return dfs(start[0], start[1], N)
CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF IF VAR NUMBER VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR LIST BIN_OP VAR VAR BIN_OP VAR VAR NUMBER VAR VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER RETURN VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR LIST VAR VAR IF VAR VAR VAR VAR NUMBER RETURN FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR VAR
Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into k non-empty subsets whose sums are all equal. Example 1: Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4 Output: True Explanation: It's possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums. Note: 1 . 0 < nums[i] < 10000.
class Solution: def canPartitionKSubsets(self, nums, k): tot = sum(nums) if k <= 0 or tot % k != 0: return False visited = [0] * len(nums) def canPart(nums, visited, idx, curSum, k, target): if k == 1: return True if curSum == target: return canPart(nums, visited, 0, 0, k - 1, target) if curSum > target: return False for i in range(idx, len(nums)): if not visited[i]: visited[i] = 1 if canPart(nums, visited, i + 1, curSum + nums[i], k, target): return True visited[i] = 0 return False return canPart(nums, visited, 0, 0, k, tot / k)
CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER BIN_OP VAR VAR NUMBER RETURN NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FUNC_DEF IF VAR NUMBER RETURN NUMBER IF VAR VAR RETURN FUNC_CALL VAR VAR VAR NUMBER NUMBER BIN_OP VAR NUMBER VAR IF VAR VAR RETURN NUMBER FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR VAR NUMBER RETURN NUMBER RETURN FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR BIN_OP VAR VAR
Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into k non-empty subsets whose sums are all equal. Example 1: Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4 Output: True Explanation: It's possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums. Note: 1 . 0 < nums[i] < 10000.
class Solution: def canPartitionKSubsets(self, nums, k): if not nums: return True if sum(nums) % k != 0: return False target = sum(nums) / k nums.sort() if nums[-1] > target: return False while nums and nums[-1] == target: nums.pop() k -= 1 def partition(nums, subsets, target): if not nums: return True selected = nums.pop() for i in range(len(subsets)): if subsets[i] + selected <= target: subsets[i] += selected if partition(nums, subsets, target): return True subsets[i] -= selected if subsets[i] == 0: break nums.append(selected) return False return partition(nums, subsets=[0] * k, target=target)
CLASS_DEF FUNC_DEF IF VAR RETURN NUMBER IF BIN_OP FUNC_CALL VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR IF VAR NUMBER VAR RETURN NUMBER WHILE VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR NUMBER FUNC_DEF IF VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR VAR RETURN NUMBER VAR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN NUMBER RETURN FUNC_CALL VAR VAR BIN_OP LIST NUMBER VAR VAR
Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into k non-empty subsets whose sums are all equal. Example 1: Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4 Output: True Explanation: It's possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums. Note: 1 . 0 < nums[i] < 10000.
class Solution: def canPartitionKSubsets(self, nums, k): quotient = sum(nums) / k if quotient % 1 != 0: return False for num in nums: if num > quotient: return False if num == quotient: nums.remove(quotient) k -= 1 nums.sort(reverse=True) self.nums = nums self.quotient = quotient print(self.nums) answer = self.dfs([0] * len(nums), 0, k) return answer def dfs(self, visit, accu, k): if k == 1: return True for i in range(len(self.nums)): if not visit[i]: if self.nums[i] + accu < self.quotient: visit[i] = 1 if self.dfs(visit, self.nums[i] + accu, k): return True visit[i] = 0 if self.nums[i] + accu == self.quotient: visit[i] = 1 accu = self.nums[i] + accu return self.dfs(visit, 0, k - 1) return False
CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER RETURN NUMBER FOR VAR VAR IF VAR VAR RETURN NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR NUMBER VAR RETURN VAR FUNC_DEF IF VAR NUMBER RETURN NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR IF BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR VAR NUMBER IF BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR RETURN FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER RETURN NUMBER
Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into k non-empty subsets whose sums are all equal. Example 1: Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4 Output: True Explanation: It's possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums. Note: 1 . 0 < nums[i] < 10000.
class Solution: def canPartitionKSubsets(self, nums, k): total = sum(nums) if total % k != 0: return False target = total / k used = [False] * len(nums) nums = sorted(nums) def check(nums, k, cur, pos): if k == 1: return True for i in range(pos, -1, -1): if not used[i]: if nums[i] + cur < target: used[i] = True if check(nums, k, cur + nums[i], i - 1): return True used[i] = False elif nums[i] + cur == target: used[i] = True if check(nums, k - 1, 0, len(nums) - 1): return True used[i] = False return False return check(nums, k, 0, len(nums) - 1)
CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER RETURN NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR NUMBER RETURN NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER IF VAR VAR IF BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER RETURN NUMBER ASSIGN VAR VAR NUMBER IF BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER RETURN NUMBER ASSIGN VAR VAR NUMBER RETURN NUMBER RETURN FUNC_CALL VAR VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER
Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into k non-empty subsets whose sums are all equal. Example 1: Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4 Output: True Explanation: It's possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums. Note: 1 . 0 < nums[i] < 10000.
class Solution: def canPartitionKSubsets(self, nums, k): if k <= 0: return False nums_sum = sum(nums) if nums_sum % k != 0: return False visited = [False] * len(nums) nums.sort(reverse=True) return self.canPartition(nums, visited, k, 0, 0, nums_sum / k) def canPartition(self, nums, visited, k, currentIndex, currentSum, target): if k == 1: return True if currentSum == target: return self.canPartition(nums, visited, k - 1, 0, 0, target) for i in range(currentIndex, len(nums)): if visited[i] is False and currentSum + nums[i] <= target: visited[i] = True if self.canPartition( nums, visited, k, i + 1, currentSum + nums[i], target ): return True visited[i] = False return False
CLASS_DEF FUNC_DEF IF VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER RETURN NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER BIN_OP VAR VAR FUNC_DEF IF VAR NUMBER RETURN NUMBER IF VAR VAR RETURN FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR VAR NUMBER RETURN NUMBER
Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into k non-empty subsets whose sums are all equal. Example 1: Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4 Output: True Explanation: It's possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums. Note: 1 . 0 < nums[i] < 10000.
class Solution(object): def canPartitionKSubsets(self, nums, k): he = sum(nums) if he % k != 0: return False target = he // k visit = [(0) for _ in range(len(nums))] def dfs(k, ind, cur, cnt): if k == 0: return True if cur == target and cnt > 0: return dfs(k - 1, 0, 0, 0) for i in range(ind, len(nums)): if not visit[i] and cur + nums[i] <= target: visit[i] = 1 if dfs(k, i + 1, cur + nums[i], cnt + 1): return True visit[i] = 0 return False return dfs(k, 0, 0, 0)
CLASS_DEF VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER RETURN NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR NUMBER RETURN NUMBER IF VAR VAR VAR NUMBER RETURN FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR BIN_OP VAR NUMBER RETURN NUMBER ASSIGN VAR VAR NUMBER RETURN NUMBER RETURN FUNC_CALL VAR VAR NUMBER NUMBER NUMBER
Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into k non-empty subsets whose sums are all equal. Example 1: Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4 Output: True Explanation: It's possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums. Note: 1 . 0 < nums[i] < 10000.
class Solution: def canPartitionKSubsets(self, nums, k): target, rem = divmod(sum(nums), k) if rem or max(nums) > target: return False n = len(nums) seen = [0] * n nums.sort(reverse=True) def dfs(k, index, current_sum): if k == 1: return True if current_sum == target: return dfs(k - 1, 0, 0) for i in range(index, n): if not seen[i] and current_sum + nums[i] <= target: seen[i] = 1 if dfs(k, i + 1, current_sum + nums[i]): return True seen[i] = 0 return False return dfs(k, 0, 0)
CLASS_DEF FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF VAR FUNC_CALL VAR VAR VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR EXPR FUNC_CALL VAR NUMBER FUNC_DEF IF VAR NUMBER RETURN NUMBER IF VAR VAR RETURN FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR VAR IF VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR RETURN NUMBER ASSIGN VAR VAR NUMBER RETURN NUMBER RETURN FUNC_CALL VAR VAR NUMBER NUMBER
Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into k non-empty subsets whose sums are all equal. Example 1: Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4 Output: True Explanation: It's possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums. Note: 1 . 0 < nums[i] < 10000.
class Solution: def canPartitionKSubsets(self, nums, k): if sum(nums) % k or len(nums) < k: return False if k == 1: return True target, used = sum(nums) / k, [(False) for i in range(len(nums))] def dfs(start, sum_, k): if k == 1: return True if sum_ == target: return dfs(0, 0, k - 1) for i in range(start, len(nums)): if not used[i] and sum_ < target: used[i] = True if dfs(i + 1, sum_ + nums[i], k): return True used[i] = False return False return dfs(0, 0, k)
CLASS_DEF FUNC_DEF IF BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR RETURN NUMBER IF VAR NUMBER RETURN NUMBER ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR NUMBER RETURN NUMBER IF VAR VAR RETURN FUNC_CALL VAR NUMBER NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR VAR NUMBER RETURN NUMBER RETURN FUNC_CALL VAR NUMBER NUMBER VAR
Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into k non-empty subsets whose sums are all equal. Example 1: Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4 Output: True Explanation: It's possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums. Note: 1 . 0 < nums[i] < 10000.
class Solution: def canPartitionKSubsets(self, nums, k): target, rem = divmod(sum(nums), k) if rem: return False def dfs(groups): if not nums: return True v = nums.pop() for i, group in enumerate(groups): if group + v <= target: groups[i] += v if dfs(groups): return True groups[i] -= v if not group: break nums.append(v) return False nums.sort() if nums[-1] > target: return False while nums and nums[-1] == target: k -= 1 nums.pop() return dfs([0] * k)
CLASS_DEF FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF VAR RETURN NUMBER FUNC_DEF IF VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR VAR VAR VAR IF FUNC_CALL VAR VAR RETURN NUMBER VAR VAR VAR IF VAR EXPR FUNC_CALL VAR VAR RETURN NUMBER EXPR FUNC_CALL VAR IF VAR NUMBER VAR RETURN NUMBER WHILE VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR RETURN FUNC_CALL VAR BIN_OP LIST NUMBER VAR
Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into k non-empty subsets whose sums are all equal. Example 1: Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4 Output: True Explanation: It's possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums. Note: 1 . 0 < nums[i] < 10000.
class Solution: def canPartitionKSubsets(self, nums, k): total = sum(nums) if total % k != 0: return False target = total // k nums.sort() if nums[-1] > target: return False while nums and nums[-1] == target: nums.pop() k -= 1 return self.helper(nums, target, [0] * k) def helper(self, nums, target, dp): if not nums: return True num = nums.pop() for i in range(len(dp)): if dp[i] + num <= target: dp[i] += num if self.helper(nums, target, dp): return True dp[i] -= num if dp[i] == 0: break nums.append(num) return False
CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER RETURN NUMBER ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR IF VAR NUMBER VAR RETURN NUMBER WHILE VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR VAR VAR BIN_OP LIST NUMBER VAR FUNC_DEF IF VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR VAR RETURN NUMBER VAR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN NUMBER
Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into k non-empty subsets whose sums are all equal. Example 1: Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4 Output: True Explanation: It's possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums. Note: 1 . 0 < nums[i] < 10000.
class Solution: def canPartitionKSubsets(self, nums, k): total = sum(nums) if total % k != 0: return False subsetTotal = total // k visited = [0] * len(nums) return self.helper(k, 0, 0, visited, nums, k, subsetTotal) def helper(self, remainingSets, index, s, visited, nums, k, subsetTotal): if remainingSets == 1: return True if s == subsetTotal: return self.helper(remainingSets - 1, 0, 0, visited, nums, k, subsetTotal) for i in range(index, len(nums)): if visited[i] == 0 and s + nums[i] <= subsetTotal: visited[i] = 1 if self.helper( remainingSets, i + 1, s + nums[i], visited, nums, k, subsetTotal ): return True visited[i] = 0 return False
CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER RETURN NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR NUMBER NUMBER VAR VAR VAR VAR FUNC_DEF IF VAR NUMBER RETURN NUMBER IF VAR VAR RETURN FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR VAR VAR VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR NUMBER IF FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR VAR NUMBER RETURN NUMBER
Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into k non-empty subsets whose sums are all equal. Example 1: Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4 Output: True Explanation: It's possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums. Note: 1 . 0 < nums[i] < 10000.
class Solution: def partition(self, nums, target, current, pos, fill): if fill == 0: return True for i in range(pos, len(nums)): next = nums[:i] + nums[i + 1 :] if current + nums[i] == target: if self.partition(next, target, 0, 0, fill - 1): return True elif current + nums[i] < target: if self.partition(next, target, current + nums[i], i, fill): return True elif current == 0: return False return False def canPartitionKSubsets(self, nums, k): s = sum(nums) if s % k > 0: return False target = s // k return self.partition(nums, target, 0, 0, k)
CLASS_DEF FUNC_DEF IF VAR NUMBER RETURN NUMBER FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER NUMBER BIN_OP VAR NUMBER RETURN NUMBER IF BIN_OP VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR RETURN NUMBER IF VAR NUMBER RETURN NUMBER RETURN NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER RETURN NUMBER ASSIGN VAR BIN_OP VAR VAR RETURN FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR
Given an array A of non-negative integers, the array is squareful if for every pair of adjacent elements, their sum is a perfect square. Return the number of permutations of A that are squareful.  Two permutations A1 and A2 differ if and only if there is some index i such that A1[i] != A2[i].   Example 1: Input: [1,17,8] Output: 2 Explanation: [1,8,17] and [17,8,1] are the valid permutations. Example 2: Input: [2,2,2] Output: 1   Note: 1 <= A.length <= 12 0 <= A[i] <= 1e9
class Solution: def numSquarefulPerms(self, A: List[int]) -> int: visited = collections.Counter(A) ans = [0] self.calc(A, visited, [], ans) return ans[0] def calc(self, A, visited, tmp, ans): if len(tmp) == len(A): ans[0] += 1 for i in list(visited.keys()): if visited[i] == 0: continue visited[i] -= 1 tmp.append(i) if len(tmp) < 2 or (tmp[-1] + tmp[-2]) ** 0.5 == int( (tmp[-1] + tmp[-2]) ** 0.5 ): self.calc(A, visited, tmp, ans) tmp.pop() visited[i] += 1
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST NUMBER EXPR FUNC_CALL VAR VAR VAR LIST VAR RETURN VAR NUMBER VAR FUNC_DEF IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR IF VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR NUMBER NUMBER FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER
Given an array A of non-negative integers, the array is squareful if for every pair of adjacent elements, their sum is a perfect square. Return the number of permutations of A that are squareful.  Two permutations A1 and A2 differ if and only if there is some index i such that A1[i] != A2[i].   Example 1: Input: [1,17,8] Output: 2 Explanation: [1,8,17] and [17,8,1] are the valid permutations. Example 2: Input: [2,2,2] Output: 1   Note: 1 <= A.length <= 12 0 <= A[i] <= 1e9
class Solution: def numSquarefulPerms(self, nums: List[int]) -> int: self.ans = 0 def isSquare(v): x = int(v**0.5) return x * x == v def dfs(pos): if pos >= len(nums): self.ans += 1 return used = set() for i in range(pos, len(nums)): if nums[i] in used: continue used.add(nums[i]) nums[pos], nums[i] = nums[i], nums[pos] if pos == 0 or pos > 0 and isSquare(nums[pos] + nums[pos - 1]): dfs(pos + 1) nums[pos], nums[i] = nums[i], nums[pos] dfs(0) return self.ans
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER RETURN BIN_OP VAR VAR VAR FUNC_DEF IF VAR FUNC_CALL VAR VAR VAR NUMBER RETURN ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN VAR VAR
Given an array A of non-negative integers, the array is squareful if for every pair of adjacent elements, their sum is a perfect square. Return the number of permutations of A that are squareful.  Two permutations A1 and A2 differ if and only if there is some index i such that A1[i] != A2[i].   Example 1: Input: [1,17,8] Output: 2 Explanation: [1,8,17] and [17,8,1] are the valid permutations. Example 2: Input: [2,2,2] Output: 1   Note: 1 <= A.length <= 12 0 <= A[i] <= 1e9
class Solution: def numSquarefulPerms(self, A: List[int]) -> int: n = len(A) A = sorted(A) g = [([0] * n) for _ in range(n)] dp = [([0] * n) for _ in range(1 << n)] for i in range(n): for j in range(n): if i == j: continue r = int((A[i] + A[j]) ** 0.5) if r**2 == A[i] + A[j]: g[i][j] = 1 for i in range(n): if i == 0 or A[i] != A[i - 1]: dp[1 << i][i] = 1 ans = 0 for s in range(1 << n): for i in range(n): if not dp[s][i]: continue for j in range(n): if not g[i][j]: continue if s & 1 << j: continue if j > 0 and not s & 1 << j - 1 and A[j - 1] == A[j]: continue dp[s | 1 << j][j] += dp[s][i] for i in range(n): ans += dp[(1 << n) - 1][i] return ans
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR VAR FUNC_CALL VAR BIN_OP NUMBER VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP NUMBER VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP NUMBER VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR IF BIN_OP VAR BIN_OP NUMBER VAR IF VAR NUMBER BIN_OP VAR BIN_OP NUMBER BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR BIN_OP NUMBER VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR RETURN VAR VAR
Given an array A of non-negative integers, the array is squareful if for every pair of adjacent elements, their sum is a perfect square. Return the number of permutations of A that are squareful.  Two permutations A1 and A2 differ if and only if there is some index i such that A1[i] != A2[i].   Example 1: Input: [1,17,8] Output: 2 Explanation: [1,8,17] and [17,8,1] are the valid permutations. Example 2: Input: [2,2,2] Output: 1   Note: 1 <= A.length <= 12 0 <= A[i] <= 1e9
class Solution: def numSquarefulPerms(self, A: List[int]) -> int: c = collections.Counter(A) cand = {i: {j for j in c if int((i + j) ** 0.5) ** 2 == i + j} for i in c} def dfs(x, left=len(A) - 1): c[x] -= 1 count = sum(dfs(y, left - 1) for y in cand[x] if c[y]) if left else 1 c[x] += 1 return count return sum(map(dfs, c))
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER BIN_OP VAR VAR VAR VAR FUNC_DEF BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR NUMBER VAR VAR NUMBER RETURN VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR
Given an array A of non-negative integers, the array is squareful if for every pair of adjacent elements, their sum is a perfect square. Return the number of permutations of A that are squareful.  Two permutations A1 and A2 differ if and only if there is some index i such that A1[i] != A2[i].   Example 1: Input: [1,17,8] Output: 2 Explanation: [1,8,17] and [17,8,1] are the valid permutations. Example 2: Input: [2,2,2] Output: 1   Note: 1 <= A.length <= 12 0 <= A[i] <= 1e9
class Solution: def numSquarefulPerms(self, A: List[int]) -> int: def get_valid(): ans = 0 seen = set() visited = set() while q: cur, remaining = q.popleft() if not remaining and cur not in seen: ans += 1 for i, n in enumerate(remaining): sq = (cur[-1] + n) ** 0.5 if float(int(sq)) == sq: add = cur + (n,) remain = remaining[:i] + remaining[i + 1 :] if (add, remain) not in visited: q.append((add, remain)) visited.add((add, remain)) return ans q = deque() for i, num in enumerate(A): q.append(((num,), tuple(A[:i] + A[i + 1 :]))) return get_valid()
CLASS_DEF FUNC_DEF VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR WHILE VAR ASSIGN VAR VAR FUNC_CALL VAR IF VAR VAR VAR VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER IF FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER RETURN FUNC_CALL VAR VAR
Given an array A of non-negative integers, the array is squareful if for every pair of adjacent elements, their sum is a perfect square. Return the number of permutations of A that are squareful.  Two permutations A1 and A2 differ if and only if there is some index i such that A1[i] != A2[i].   Example 1: Input: [1,17,8] Output: 2 Explanation: [1,8,17] and [17,8,1] are the valid permutations. Example 2: Input: [2,2,2] Output: 1   Note: 1 <= A.length <= 12 0 <= A[i] <= 1e9
class Solution: def numSquarefulPerms(self, A: List[int]) -> int: candidates = collections.Counter(A) graph = { x: [y for y in candidates if int((x + y) ** 0.5) ** 2 == x + y] for x in candidates } graph[-1] = [x for x in candidates] def dfs(node, node_left): if node_left == 0: return 1 res = 0 for n in graph[node]: if candidates[n] == 0: continue candidates[n] -= 1 res += dfs(n, node_left - 1) candidates[n] += 1 return res return dfs(-1, len(A))
CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER BIN_OP VAR VAR VAR VAR ASSIGN VAR NUMBER VAR VAR VAR FUNC_DEF IF VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR IF VAR VAR NUMBER VAR VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR NUMBER RETURN VAR RETURN FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR