- Difficulty: Medium
- Tags: LeetCode, Medium, Array, Hash Table, Sorting, Heap (Priority Queue), leetcode-2342, O(nlogr), O(n), Greedy
Problem
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 <= 105
1 <= nums[i] <= 109