2270. Number of Ways to Split Array

🟧 Medium

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 <= 10^5

• -10^5 <= nums[i] <= 10^5

Solution

My Solution

func waysToSplitArray(nums []int) int {
    sum := 0
    for _, num := range nums {
        sum += num
    }

    var (
        count   = 0
        leftSum = 0
    )
    for i := 0; i < len(nums)-1; i++ {
        leftSum += nums[i]
        if leftSum >= (sum - leftSum) {
            count++
        }
    }

    return count
}

result

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