LeetCode in Kotlin

2815. Max Pair Sum in an Array

Easy

You are given a 0-indexed integer array nums. You have to find the maximum sum of a pair of numbers from nums such that the maximum digit in both numbers are equal.

Return the maximum sum or -1 if no such pair exists.

Example 1:

Input: nums = [51,71,17,24,42]

Output: 88

Explanation:

For i = 1 and j = 2, nums[i] and nums[j] have equal maximum digits with a pair sum of 71 + 17 = 88.

For i = 3 and j = 4, nums[i] and nums[j] have equal maximum digits with a pair sum of 24 + 42 = 66.

It can be shown that there are no other pairs with equal maximum digits, so the answer is 88.

Example 2:

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

Output: -1

Explanation: No pair exists in nums with equal maximum digits.

Constraints:

• 2 <= nums.length <= 100
• 1 <= nums[i] <= 104

Solution

import java.util.PriorityQueue
import kotlin.math.max

@Suppress("NAME_SHADOWING")
class Solution {
    fun maxSum(nums: IntArray): Int {
        // what we'll return
        var maxSum = -1
        val maximumDigitToNumber: MutableMap<Int, PriorityQueue<Int>> = HashMap()
        for (i in 1..9) {
            maximumDigitToNumber[i] = PriorityQueue(Comparator.reverseOrder())
        }
        for (n in nums) {
            maximumDigitToNumber[getMaximumDigit(n)]!!.add(n)
        }
        for ((_, value) in maximumDigitToNumber) {
            if (value.size <= 1) {
                continue
            }
            val sum = value.poll() + value.poll()
            maxSum = max(maxSum, sum)
        }
        return maxSum
    }

    private fun getMaximumDigit(n: Int): Int {
        var n = n
        var maxDigit = 1
        var nMod10 = n % 10
        while (n > 0) {
            maxDigit = max(maxDigit.toDouble(), nMod10.toDouble()).toInt()
            n /= 10
            nMod10 = n % 10
        }
        return maxDigit
    }
}

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