LeetCode in Kotlin

3852. Smallest Pair With Different Frequencies

Easy

You are given an integer array nums.

Consider all pairs of distinct values x and y from nums such that:

• x < y
• x and y have different frequencies in nums.

Among all such pairs:

• Choose the pair with the smallest possible value of x.
• If multiple pairs have the same x, choose the one with the smallest possible value of y.

Return an integer array [x, y]. If no valid pair exists, return [-1, -1].

Example 1:

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

Output: [1,3]

Explanation:

The smallest value is 1 with a frequency of 2, and the smallest value greater than 1 that has a different frequency from 1 is 3 with a frequency of 1. Thus, the answer is [1, 3].

Example 2:

Input: nums = [1,5]

Output: [-1,-1]

Explanation:

Both values have the same frequency, so no valid pair exists. Return [-1, -1].

Example 3:

Input: nums = [7]

Output: [-1,-1]

Explanation:

There is only one value in the array, so no valid pair exists. Return [-1, -1].

Constraints:

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

Solution

class Solution {
    fun minDistinctFreqPair(nums: IntArray): IntArray? {
        val freq = IntArray(101)
        for (num in nums) {
            freq[num]++
        }
        var first = -1
        var second = -1
        for (i in 1..100) {
            if (freq[i] != 0) {
                if (first == -1) {
                    first = i
                } else if (freq[i] != freq[first]) {
                    second = i
                    break
                }
            }
        }
        return if (second == -1) intArrayOf(-1, -1) else intArrayOf(first, second)
    }
}

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