[leetcode] 1524. Number of Sub-arrays With Odd Sum

[leetcode] 1524. Number of sub-arrays With Odd Sum

Description

Given an array of integers arr. Return the number of sub-arrays with odd sum.

As the answer may grow large, the answer must be computed modulo 10^9 + 7.

Example 1:

Input: arr = [1,3,5]Output: 4Explanation: All sub-arrays are [[1],[1,3],[1,3,5],[3],[3,5],[5]]All sub-arrays sum are [1,4,9,3,8,5].Odd sums are [1,9,3,5] so the answer is 4.

Example 2:

Input: arr = [2,4,6]Output: 0Explanation: All sub-arrays are [[2],[2,4],[2,4,6],[4],[4,6],[6]]All sub-arrays sum are [2,6,12,4,10,6].All sub-arrays have even sum and the answer is 0.

Example 3:

Input: arr = [1,2,3,4,5,6,7]Output: 16

Example 4:

Input: arr = [100,100,99,99]Output: 4

Example 5:

Input: arr = [7]Output: 1

Constraints:

1 <= arr.length <= 10^51 <= arr[i] <= 100

分析

题目的意思是：给你一个数组，求其每个子数组的和为奇数的个数，这道题我暴力破解了一下，发现超时了。也没做出来，后面看了一下别人动态规划的思路，dp[i]表示的是以i为截止的新增子数组的和为奇数的数量，然后遍历数组，首先判断第0个位置的数是否为奇数，如果是dp[0]=0,否则为1，对于后面的遍历，如果是偶数，则不改变新增奇数数组的数量，dp[i]=dp[i-1]，如果是奇数，则新增了dp[i]=i-dp[i-1]+1个奇数。

代码

class Solution: def numOfSubarrays(self, arr: List[int]) -> int: n=len(arr) dp=[0]*n if(arr[0]%2==1): dp[0]=1 cnt=0 for i in range(0,len(arr)): if(i==0): cnt+=dp[0] continue elif(arr[i]%2==0): dp[i]=dp[i-1] else: dp[i]=i-dp[i-1]+1 cnt=(cnt+dp[i])%(10**9+7) return cnt

参考文献

​​[LeetCode] My Python, java Solutions DP​​

版权声明：本文内容由网络用户投稿，版权归原作者所有，本站不拥有其著作权，亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容，请联系我们jiasou666@gmail.com 处理，核实后本网站将在24小时内删除侵权内容。