LeetCode in Kotlin
2342. Max Sum of a Pair With Equal Sum of Digits
Medium
You are given a 0-indexed array nums consisting of positive integers. You can choose two indices i and j, such that i != j, and the sum of digits of the number nums[i] is equal to that of nums[j].
Return the maximum value of nums[i] + nums[j] that you can obtain over all possible indices i and j that satisfy the conditions.
Example 1:
Input: nums = [18,43,36,13,7]
Output: 54
Explanation: The pairs (i, j) that satisfy the conditions are:
-
(0, 2), both numbers have a sum of digits equal to 9, and their sum is 18 + 36 = 54.
-
(1, 4), both numbers have a sum of digits equal to 7, and their sum is 43 + 7 = 50.
So the maximum sum that we can obtain is 54.
Example 2:
Input: nums = [10,12,19,14]
Output: -1
Explanation: There are no two numbers that satisfy the conditions, so we return -1.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 109
Solution
class Solution {
fun maximumSum(nums: IntArray): Int {
val map: MutableMap<Int, Int> = HashMap()
var res = -1
for (num in nums) {
var s = 0
for (digit in num.toString().toCharArray()) {
s += Integer.valueOf(digit.code - '0'.code)
}
if (!map.containsKey(s)) {
map[s] = num
} else {
res = Math.max(res, map[s]!! + num)
map[s] = Math.max(map[s]!!, num)
}
}
return res
}
}