哈哈哈个

solution: Sliding window + counter, O(N), O(K) simplest approach: sliding window + two counter hashm

solution: dp O(N^2 * k),O(NK) dp[i][j] represents the maximum length of subsequence ending at index

Intuitive: For each data center, can upgrade 1, 2, 3,... up to K servers, using O(N) loop over all t

Solution Build the Tree: Use an graph to represent the undirected tree. DFS: Bob going up, Alice goi

Solution: N-tree pre order traversal 「else return x's oldest child who's not in curOrder」-> pick a c

1.Keep record of previos of 1 2. Group by 0 s // 0000000001111 if (seats[prev] != 1)

solutiuon List<Integer> list1 = new ArrayList<>(); List<Integer> list2 = new ArrayList<>();

SOLUTION: DP dp[i][j] = number of ways to build the first i characters of target using the first j c

solution DFS int leftSum = resLeft[0], leftCount = resLeft[1]; int rightSum = resRight[0], r