Scribbling

Perhaps, the most thinkable solution would be DFS or BFS. Traverse all the '1' points and update mins and maxs. Time complexity would be O(M*N) in this case. class Solution: def minArea(self, image: List[List[str]], x: int, y: int) -> int: m, n = len(image), len(image[0]) ymin, ymax, xmin, xmax = m, -1, n, -1 dy, dx = [0, 1, 0, -1], [1, 0, -1, 0] stack, visited = [(x, y)], set() while stack: i, ..

Just a mere implementation problem. class Solution: def fullJustify(self, words: List[str], maxWidth: int) -> List[str]: ret = [] i = 0 while i

First solution is using regular expression. Below is some useful patterns. Integer: ^[+-]?\d+$ Flaot: ^[+-]?((\d+(\.\d*)?)|(\.\d+))$ Sicnetific notation: ^[+-]?((\d+(\.\d*)?)|(\.\d+))[eE][+-]?\d+$ class Solution: def isNumber(self, s: str) -> bool: import re pat = re.compile('(^[+-]?((\d+(\.\d*)?)|(\.\d+))$)|(^[+-]?((\d+(\.\d*)?)|(\.\d+))[eE][+-]?\d+$)') return re.match(pat, s) Second solution i..

A solution using sliding window. class Solution: def findSubstring(self, s: str, words: List[str]) -> List[int]: from collections import Counter ret = [] wordSet = set(words) n, m = len(words), len(words[0]) for i in range(m): wordCounter = Counter(words) k = j = i while j + m = n * m: if s[k:k+m] in wordSet: wordCounter[s[k:k+m]] += 1 k += m word = s[j:j+m] if word in wordSet: wordCounter[word]..

class Solution: def removeElement(self, nums: List[int], val: int) -> int: length = len(nums) i = 0 while i

First thing to try is iterating nums1, nums2, nums3 and looking for the corresponding element in nums4. This will result in O(N**3) time complexity, which raises TLE. class Solution: def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int: from collections import Counter c = Counter(nums4) ret = 0 for n1 in nums1: for n2 in nums2: for n3 in nums3: re..

Idea: Any characters that have frequencies less than k work as a breakpoints, meaning that the substring we're looking for can't include that particular character. So, recursively look for substrings while excluding those characters. Time complexity is O(N). from collections import Counter class Solution: def longestSubstring(self, s: str, k: int) -> int: return self.helper(s, k) def helper(self..

Refer to the below post for Knuth Shuffle. https://focalpoint.tistory.com/253 Shuffle Algorithm: Knuth Shuffle This post is about Knuth Shuffle, a shuffle algorithm. - This algorithm allows random shuffle, giving all permuations of array equal likelihood. - No need for additional arrays, so memory complexity.. focalpoint.tistory.com class Solution: def __init__(self, nums: List[int]): self.nums ..

Coming up with O(1) memory solution was very difficult for me. I refered to this post's last solution. https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/1322101/C%2B%2BJavaPython-MaxHeap-MinHeap-Binary-Search-Picture-Explain-Clean-and-Concise [C++/Java/Python] MaxHeap, MinHeap, Binary Search - Picture Explain - Clean & Concise - LeetCode Discuss Level up your coding s..

Making a new list is an easy solution to implement with recursion. class NestedIterator: def __init__(self, nestedList: [NestedInteger]): self.res = [] self._flattener(nestedList) self.ptr = 0 def _flattener(self, nestedList): for nl in nestedList: if nl.isInteger(): self.res.append(nl.getInteger()) else: self._flattener(nl.getList()) def next(self) -> int: ret = self.res[self.ptr] self.ptr += 1..