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 + 1elements is greater than or equal to the sum of the lastn - i - 1elements. - 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
}
}