[LeetCode] 2270. Number of Ways to Split Array

You are given a 0-indexed integer array nums of length n.

nums contains a valid split at index i if the following are true:

• The sum of the first i + 1 elements is greater than or equal to the sum of the last n - i - 1 elements.
• There is at least one element to the right of i. That is, 0 <= i < n - 1.

Return the number of valid splits in nums.

Example 1:

Input: nums = [10,4,-8,7]
Output: 2
Explanation: 
There are three ways of splitting nums into two non-empty parts:
- Split nums at index 0. Then, the first part is [10], and its sum is 10. The second part is [4,-8,7], and its sum is 3. Since 10 >= 3, i = 0 is a valid split.
- Split nums at index 1. Then, the first part is [10,4], and its sum is 14. The second part is [-8,7], and its sum is -1. Since 14 >= -1, i = 1 is a valid split.
- Split nums at index 2. Then, the first part is [10,4,-8], and its sum is 6. The second part is [7], and its sum is 7. Since 6 < 7, i = 2 is not a valid split.
Thus, the number of valid splits in nums is 2.

Example 2:

Input: nums = [2,3,1,0]
Output: 2
Explanation: 
There are two valid splits in nums:
- Split nums at index 1. Then, the first part is [2,3], and its sum is 5. The second part is [1,0], and its sum is 1. Since 5 >= 1, i = 1 is a valid split. 
- Split nums at index 2. Then, the first part is [2,3,1], and its sum is 6. The second part is [0], and its sum is 0. Since 6 >= 0, i = 2 is a valid split.

Constraints:

• 2 <= nums.length <= 105
• -105 <= nums[i] <= 105

分割数组的方案数。

给你一个下标从 0 开始长度为 n 的整数数组 nums 。
如果以下描述为真，那么 nums 在下标 i 处有一个 合法的分割 ：

前 i + 1 个元素的和 大于等于 剩下的 n - i - 1 个元素的和。
下标 i 的右边 至少有一个 元素，也就是说下标 i 满足 0 <= i < n - 1 。
请你返回 nums 中的 合法分割 方案数。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/number-of-ways-to-split-array
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这道题的思路跟2256类似。2256求的是最小的满足条件的 index，这道题其实更简单，就是求符合题意的 index 一共有几个，而且只是单纯的求子数组的和，用前缀和的思路解决即可。

时间O(n)

空间O(1)

Java实现

class Solution {
    public int waysToSplitArray(int[] nums) {
        long sum = 0;
        for (int num : nums) {
            sum += num;
        }
        
        int count = 0;
        int len = nums.length;
        long left = 0;
        long right = 0;
        for (int i = 0; i < len - 1; i++) {
            left += nums[i];
            right = sum - left;
            if (left >= right) {
                count++;
            }
        }
        return count;
    }
}

 

前缀和prefix sum题目总结

LeetCode 题目总结