README.md

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

Solutions (Rust)

1. Solution

use std::collections::HashMap;

impl Solution {
    pub fn maximum_sum(nums: Vec<i32>) -> i32 {
        let mut max_pairs = HashMap::new();

        for &num in &nums {
            let digits_sum = num
                .to_string()
                .bytes()
                .map(|d| (d - b'0') as i32)
                .sum::<i32>();
            let pair = max_pairs.entry(digits_sum).or_insert((0, 0));

            if num >= pair.1 {
                *pair = (pair.1, num);
            } else if num > pair.0 {
                *pair = (num, pair.1);
            }
        }

        max_pairs
            .values()
            .filter(|&&(x, y)| x > 0)
            .map(|(x, y)| x + y)
            .max()
            .unwrap_or(-1)
    }
}