LeetCode in Kotlin

3876. Construct Uniform Parity Array II

Medium

You are given an array nums1 of n distinct integers.

You want to construct another array nums2 of length n such that the elements in nums2 are either all odd or all even.

For each index i, you must choose exactly one of the following (in any order):

• nums2[i] = nums1[i]
• nums2[i] = nums1[i] - nums1[j], for an index j != i, such that nums1[i] - nums1[j] >= 1

Return true if it is possible to construct such an array, otherwise return false.

Example 1:

Input: nums1 = [1,4,7]

Output: true

Explanation:

• Set nums2[0] = nums1[0] = 1.
• Set nums2[1] = nums1[1] - nums1[0] = 4 - 1 = 3.
• Set nums2[2] = nums1[2] = 7.
• nums2 = [1, 3, 7], and all elements are odd. Thus, the answer is true.

Example 2:

Input: nums1 = [2,3]

Output: false

Explanation:

It is not possible to construct nums2 such that all elements have the same parity. Thus, the answer is false.

Example 3:

Input: nums1 = [4,6]

Output: true

Explanation:

• Set nums2[0] = nums1[0] = 4.
• Set nums2[1] = nums1[1] = 6.
• nums2 = [4, 6], and all elements are even. Thus, the answer is true.

Constraints:

• 1 <= n == nums1.length <= 105
• 1 <= nums1[i] <= 109
• nums1 consists of distinct integers.

Solution

import kotlin.math.min

class Solution {
    fun uniformArray(nums1: IntArray): Boolean {
        var min = Int.MAX_VALUE
        for (x in nums1) {
            min = min(min, x)
        }
        if (min % 2 == 1) {
            return true
        }
        for (x in nums1) {
            if (x % 2 == 1) {
                return false
            }
        }
        return true
    }
}

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