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

## Description

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

Solution 1: Hash Table

We can use a hash table $d$ to record the maximum value corresponding to each digit sum, and initialize an answer variable $ans = -1$.

Next, we traverse the array $nums$. For each number $v$, we calculate its digit sum $x$. If $x$ exists in the hash table $d$, then we update the answer $ans = \max(ans, d[x] + v)$. Then update the hash table $d[x] = \max(d[x], v)$.

Finally, return the answer $ans$.

Since the maximum element in $nums$ is $10^9$, the maximum digit sum is $9 \times 9 = 81$. We can directly define an array $d$ of length $100$ to replace the hash table.

The time complexity is $O(n \times \log M)$, and the space complexity is $O(D)$. Here, $n$ is the length of the array $nums$, and $M$ and $D$ are the maximum value of the elements in the array $nums$ and the maximum value of the digit sum, respectively. In this problem, $M \leq 10^9$, $D \leq 81$.

• class Solution {
public int maximumSum(int[] nums) {
int[] d = new int[100];
int ans = -1;
for (int v : nums) {
int x = 0;
for (int y = v; y > 0; y /= 10) {
x += y % 10;
}
if (d[x] > 0) {
ans = Math.max(ans, d[x] + v);
}
d[x] = Math.max(d[x], v);
}
return ans;
}
}

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

• class Solution:
def maximumSum(self, nums: List[int]) -> int:
d = defaultdict(int)
ans = -1
for v in nums:
x, y = 0, v
while y:
x += y % 10
y //= 10
if x in d:
ans = max(ans, d[x] + v)
d[x] = max(d[x], v)
return ans


• func maximumSum(nums []int) int {
d := [100]int{}
ans := -1
for _, v := range nums {
x := 0
for y := v; y > 0; y /= 10 {
x += y % 10
}
if d[x] > 0 {
ans = max(ans, d[x]+v)
}
d[x] = max(d[x], v)
}
return ans
}

• function maximumSum(nums: number[]): number {
const d: number[] = Array(100).fill(0);
let ans = -1;
for (const v of nums) {
let x = 0;
for (let y = v; y; y = (y / 10) | 0) {
x += y % 10;
}
if (d[x]) {
ans = Math.max(ans, d[x] + v);
}
d[x] = Math.max(d[x], v);
}
return ans;
}


• impl Solution {
pub fn maximum_sum(nums: Vec<i32>) -> i32 {
let mut d = vec![0; 100];
let mut ans = -1;

for &v in nums.iter() {
let mut x: usize = 0;
let mut y = v;
while y > 0 {
x += (y % 10) as usize;
y /= 10;
}
if d[x] > 0 {
ans = ans.max(d[x] + v);
}
d[x] = d[x].max(v);
}

ans
}
}