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Formatted question description: https://leetcode.ca/all/2342.html

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

• Difficulty: Medium.
• Related Topics: Array, Hash Table, Sorting, Heap (Priority Queue).
• Similar Questions: .

## 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

## Solution (Java, C++, Python)

• 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;
}
}

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

############

# 2342. Max Sum of a Pair With Equal Sum of Digits
# https://leetcode.com/problems/max-sum-of-a-pair-with-equal-sum-of-digits/

class Solution:
def maximumSum(self, nums: List[int]) -> int:
n = len(nums)
cnt = defaultdict(list)

for x in nums:
cnt[sum(int(k) for k in str(x))].append(x)

res = -1

for values in cnt.values():
if len(values) >= 2:
values.sort()
res = max(res, values[-1] + values[-2])

return res


• 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;
}
};

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

func max(a, b int) int {
if a > b {
return a
}
return b
}

• 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
}
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).