2342. Max Sum of a Pair With Equal Sum of Digits

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

JAVA

• class Solution {
      public int maximumSum(int[] nums) {
          Map<Integer, Integer> map = new HashMap<>();
          int res = -1;
          for (int num : nums) {
              int s = 0;
              for (char digit : String.valueOf(num).toCharArray()) {
                  s += Integer.valueOf(digit - '0');
              }
              if (!map.containsKey(s)) {
                  map.put(s, num);
              } else {
                  res = Math.max(res, map.get(s) + num);
                  map.put(s, Math.max(map.get(s), num));
              }
          }
          return res;
      }
  }
  
  ############
  
  class Solution {
      public int maximumSum(int[] nums) {
          int ans = -1;
          int[] d = new int[100];
          for (int v : nums) {
              int y = 0;
              for (int x = v; x > 0; x /= 10) {
                  y += x % 10;
              }
              if (d[y] > 0) {
                  ans = Math.max(ans, d[y] + v);
              }
              d[y] = Math.max(d[y], v);
          }
          return ans;
      }
  }

C++

• class Solution {
  public:
      int maximumSum(vector<int>& nums) {
          vector<vector<int>> d(100);
          for (int& v : nums) {
              int y = 0;
              for (int x = v; x > 0; x /= 10) {
                  y += x % 10;
              }
              d[y].emplace_back(v);
          }
          int ans = -1;
          for (auto& vs : d) {
              if (vs.size() > 1) {
                  sort(vs.rbegin(), vs.rend());
                  ans = max(ans, vs[0] + vs[1]);
              }
          }
          return ans;
      }
  };

PYTHON

• from collections import defaultdict
  
  class Solution:
      def maximumSum(self, nums):
          digit_sums = defaultdict(list)
  
          # Group numbers by their digit sum
          for num in nums:
              digit_sum = sum(int(d) for d in str(num))
              digit_sums[digit_sum].append(num)
  
          max_sum = -1
  
          # Find the largest sum of two numbers with the same digit sum
          for group in digit_sums.values():
              if len(group) > 1:
                  group.sort(reverse=True)  # Sort descending
                  max_sum = max(max_sum, group[0] + group[1])  # Take the two largest numbers
  
          return max_sum