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 <= 1001 <= nums[i] <= 104
Solution
import java.util.PriorityQueue
import kotlin.collections.HashMap
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
}
}