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Main idea: current_presum - old_presum = target , where old_presum is in hmap, so we can check if current_presum - target in hmap. if yes, res + 1; then update current_presum in hmap before entereing subtree. Note, each time when current node and its subtrees are traversed, hmap need to... [Read More]

class Solution: def threeSum(self, nums: List[int]) -> List[List[int]]: # main idea: sort + two pointers # time O(N^2) space O(1) nums.sort() res = [] size = len(nums) for i in range(size-2): # optimization: since nums is sorted, it is impossible to have a sum of 0 if nums[i] > 0... [Read More]

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class Solution: def nextGreaterElement(self, n: int) -> int: # time O(NlgN) worst 2N, space O(N), where N is # digits ? # notice, 32bit max interger is 2^31 -1 # step1, reverse digits, find target number with, idx smaller than lower digit # step2, reverve its lower digit then loop... [Read More]