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Solutionap { Map<String, ArrayList<Pair<Integer, String>>> map; public TimeMap() { /

Solution: Binary search We remove nums[mid], resulting in left half arr and right half arr, the suba

Solution: sliding window O(∣S∣+∣T∣) t may include duplicates, so we need to count frequency of t Exp

Solution 1: check if loop exist Solution 2: Using graph property Check whether or not there are n -

Solution BFS we will perform a level-wise BFS iteration over the nodes level limited by k cost[] sto

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Solution Build graph and BFS The direction of edge indicates the order of division, and the weight o

Solution1 level order traversal O(N)O(N) Solution2: using established linkedlist O(N)O(1) since we d

Solution1: O(N) O(N) Level order traversal Solution2 : O(N) O(1) when we go over the nodes of a part