[Leetcode] 823. Binary Trees With Factors

方法1

Dynamic Programming
dp[k] += dp[i] * dp[j], if arr[i] * arr[j] == arr[k]
initialize all dp[k] = 1
we need to use hashmap or two pointers to get possible arr[i] and arr[j], otherwise the total time complexity is O(N**3), exceed time limit

time complexity: O(n**2), space complexity: O(n)

class Solution:
    def numFactoredBinaryTrees(self, arr: List[int]) -> int:
        arr.sort()
        dp = [1] * len(arr)
        for k in range(1, len(arr)):
            i = 0
            j = k - 1
            while i <= k - 1 and j >= 0:
                if arr[i] * arr[j] < arr[k]:
                    i += 1
                elif arr[i] * arr[j] > arr[k]:
                    j -= 1
                else:
                    dp[k] += (dp[i] * dp[j]) % (10 ** 9 + 7)
                    i += 1
                    j -= 1

        return sum(dp) % (10 ** 9 + 7)

标签：Binary,arr,int,Trees,complexity,len,time,Leetcode,dp
来源： https://blog.csdn.net/weixin_44473990/article/details/112059507