LeetCode in Kotlin

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 <= 105
• -105 <= nums[i] <= 105

Solution

class Solution {
    fun waysToSplitArray(nums: IntArray): Int {
        var leftSum: Long = 0
        var rightSum: Long = 0
        for (i in nums) {
            rightSum += i.toLong()
        }
        var count = 0
        for (i in 0 until nums.size - 1) {
            rightSum -= nums[i].toLong()
            leftSum += nums[i].toLong()
            if (leftSum >= rightSum) {
                count++
            }
        }
        return count
    }
}

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