LeetCode in Kotlin

3618. Split Array by Prime Indices

Medium

You are given an integer array nums.

Split nums into two arrays A and B using the following rule:

• Elements at prime indices in nums must go into array A.
• All other elements must go into array B.

Return the absolute difference between the sums of the two arrays: |sum(A) - sum(B)|.

Note: An empty array has a sum of 0.

Example 1:

Input: nums = [2,3,4]

Output: 1

Explanation:

• The only prime index in the array is 2, so nums[2] = 4 is placed in array A.
• The remaining elements, nums[0] = 2 and nums[1] = 3 are placed in array B.
• sum(A) = 4, sum(B) = 2 + 3 = 5.
• The absolute difference is |4 - 5| = 1.

Example 2:

Input: nums = [-1,5,7,0]

Output: 3

Explanation:

• The prime indices in the array are 2 and 3, so nums[2] = 7 and nums[3] = 0 are placed in array A.
• The remaining elements, nums[0] = -1 and nums[1] = 5 are placed in array B.
• sum(A) = 7 + 0 = 7, sum(B) = -1 + 5 = 4.
• The absolute difference is |7 - 4| = 3.

Constraints:

• 1 <= nums.length <= 105
• -109 <= nums[i] <= 109

Solution

import kotlin.math.abs

class Solution {
    fun splitArray(nums: IntArray): Long {
        val n = nums.size
        val isPrime = sieve(n)
        var sumA: Long = 0
        var sumB: Long = 0
        for (i in 0..<n) {
            if (isPrime[i]) {
                sumA += nums[i].toLong()
            } else {
                sumB += nums[i].toLong()
            }
        }
        return abs(sumA - sumB)
    }

    // Sieve of Eratosthenes to find all prime indices up to n
    private fun sieve(n: Int): BooleanArray {
        val isPrime = BooleanArray(n)
        if (n > 2) {
            isPrime[2] = true
        }
        run {
            var i = 3
            while (i < n) {
                isPrime[i] = true
                i += 2
            }
        }
        if (n > 2) {
            isPrime[2] = true
        }
        var i = 3
        while (i * i < n) {
            if (isPrime[i]) {
                var j = i * i
                while (j < n) {
                    isPrime[j] = false
                    j += i * 2
                }
            }
            i += 2
        }
        isPrime[0] = false
        if (n > 1) {
            isPrime[1] = false
        }
        return isPrime
    }
}

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