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2270. Number of Ways to Split Array

Description

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

Solutions

Solution 1

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class Solution:
    def waysToSplitArray(self, nums: List[int]) -> int:
        s = sum(nums)
        ans = t = 0
        for v in nums[:-1]:
            t += v
            if t >= s - t:
                ans += 1
        return ans

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class Solution {
    public int waysToSplitArray(int[] nums) {
        long s = 0;
        for (int v : nums) {
            s += v;
        }
        int ans = 0;
        long t = 0;
        for (int i = 0; i < nums.length - 1; ++i) {
            t += nums[i];
            if (t >= s - t) {
                ++ans;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int waysToSplitArray(vector<int>& nums) {
        long long s = accumulate(nums.begin(), nums.end(), 0ll);
        long long t = 0;
        int ans = 0;
        for (int i = 0; i < nums.size() - 1; ++i) {
            t += nums[i];
            ans += t >= s - t;
        }
        return ans;
    }
};

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func waysToSplitArray(nums []int) int {
    s := 0
    for _, v := range nums {
        s += v
    }
    ans, t := 0, 0
    for _, v := range nums[:len(nums)-1] {
        t += v
        if t >= s-t {
            ans++
        }
    }
    return ans
}

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