LeetCode in Kotlin

3917. Count Indices With Opposite Parity

Easy

You are given an integer array nums of length n.

The score of an index i is defined as the number of indices j such that:

• i < j < n, and
• nums[i] and nums[j] have different parity (one is even and the other is odd).

Return an integer array answer of length n, where answer[i] is the score of index i.

Example 1:

Input: nums = [1,2,3,4]

Output: [2,1,1,0]

Explanation:

• nums[0] = 1, which is odd. Thus, the indices j = 1 and j = 3 satisfy the conditions, so the score of index 0 is 2.
• nums[1] = 2, which is even. Thus, the index j = 2 satisfies the conditions, so the score of index 1 is 1.
• nums[2] = 3, which is odd. Thus, the index j = 3 satisfies the conditions, so the score of index 2 is 1.
• nums[3] = 4, which is even. Thus, no index satisfies the conditions, so the score of index 3 is 0.

Thus, the answer = [2, 1, 1, 0].

Example 2:

Input: nums = [1]

Output: [0]

Explanation:

There is only one element in nums. Thus, the score of index 0 is 0.

Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 100

Solution

class Solution {
    fun countOppositeParity(nums: IntArray): IntArray {
        val n = nums.size
        var odd = 0
        var even = 0
        val result = IntArray(n)
        for (i in n - 1 downTo 0) {
            if ((nums[i] and 1) == 1) {
                result[i] = even
                odd++
            } else {
                result[i] = odd
                even++
            }
        }
        return result
    }
}

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