# Maximum sum of two elements whose digit sum is equal

Given an array arr[] of N positive elements, the task is to find a pair in the array such that the digit sum of elements in the pair are equal and their sum is maximum. Print -1 if it is not possible to find the pair else print the maximum sum.

Examples:

Input: arr[] = {55, 23, 32, 46, 88}
Output: 101
(23, 32) with digit sum 5 and (55, 46)
with digit sum 10 are the only pairs
with equal digit sum. The pair with maximum
sum is (55, 46) i.e. 55 + 46 = 101

Input: arr[] = {18, 19, 23, 15}
Output: -1
Since there are no two elements
whose digit sums are equal.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Traverse through the array and use a unordered_map to store the maximum number present for a particular digit sum. For every element arr[i], find the digit sum of arr[i] and check if there is a value present in the map for this digit sum, if yes then update the answer by ans = max(ans, arr[i] + map[digSum]). Also, update the map if the value of arr[i] is greater than the value present in the map.

Below is the implementation of the above approach:

## CPP

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the ` `// digit sum of n ` `long` `digitSum(``long` `n) ` `{ ` `    ``long` `sum = 0; ` `    ``while` `(n) { ` `        ``sum += (n % 10); ` `        ``n /= 10; ` `    ``} ` `    ``return` `sum; ` `} ` ` `  `// Function to return the maximum ` `// sum possible for a pair whose ` `// digit sum is equal ` `int` `findSum(vector<``long``> arr, ``int` `n) ` `{ ` ` `  `    ``unordered_map<``long``, ``long``> mp; ` `    ``long` `ans = -1; ` ` `  `    ``// Traversing through all the ` `    ``// elements of the array ` `    ``for` `(``long` `i = 0; i < n; i++) { ` ` `  `        ``// Finding the digit sum of arr[i] ` `        ``long` `dSum = digitSum(arr[i]); ` ` `  `        ``if` `(mp[dSum] != 0) { ` `            ``ans = max(ans, mp[dSum] + arr[i]); ` `        ``} ` `        ``mp[dSum] = max(mp[dSum], arr[i]); ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``vector<``long``> arr = { 55, 23, 32, 46, 88 }; ` `    ``int` `n = arr.size(); ` ` `  `    ``cout << findSum(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` `     `  `    ``// Function to return the  ` `    ``// digit sum of n  ` `    ``static` `int` `digitSum(``int` `n)  ` `    ``{  ` `        ``int` `sum = ``0``;  ` `        ``while` `(n != ``0``) ` `        ``{  ` `            ``sum += (n % ``10``);  ` `            ``n /= ``10``;  ` `        ``}  ` `        ``return` `sum;  ` `    ``}  ` `     `  `    ``// Function to return the maximum  ` `    ``// sum possible for a pair whose  ` `    ``// digit sum is equal  ` `    ``static` `int` `findSum(``int` `arr[], ``int` `n)  ` `    ``{  ` `     `  `        ``HashMap mp = ``new` `HashMap();  ` `         `  `        ``for` `(``int` `i = ``0``; i < ``1000` `; i++) ` `        ``{ ` `            ``mp.put(i, ``0``); ` `        ``} ` `         `  `        ``int` `ans = -``1``;  ` `     `  `        ``// Traversing through all the  ` `        ``// elements of the array  ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``{  ` `     `  `            ``// Finding the digit sum of arr[i]  ` `            ``int` `dSum = digitSum(arr[i]);  ` `             `  `            ``if` `((``int``)mp.get(dSum) != ``0``) ` `            ``{  ` `                ``ans = Math.max(ans, (``int``)mp.get(dSum) + arr[i]);  ` `            ``}  ` `            ``mp.put(dSum,Math.max((``int``)mp.get(dSum), arr[i])) ; ` `        ``}  ` `        ``return` `ans;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{  ` `        ``int` `arr[] = { ``55``, ``23``, ``32``, ``46``, ``88` `};  ` `        ``int` `n = arr.length;  ` `     `  `        ``System.out.println(findSum(arr, n)); ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python

 `# Python3 implementation of the approach ` ` `  `# Function to return the ` `# digit sum of n ` `def` `digitSum(n): ` `    ``sum` `=` `0` `    ``while` `(n): ` `        ``sum` `+``=` `(n ``%` `10``) ` `        ``n ``/``/``=` `10` `    ``return` `sum` ` `  `# Function to return the maximum ` `# sum possible for a pair whose ` `# digit sum is equal ` `def` `findSum(arr, n): ` ` `  `    ``mp ``=` `dict``() ` `    ``ans ``=` `-``1` ` `  `    ``# Traversing through all the ` `    ``# elements of the array ` `    ``for` `i ``in` `range``(n): ` ` `  `        ``# Finding the digit sum of arr[i] ` `        ``dSum ``=` `digitSum(arr[i]) ` ` `  `        ``if` `(dSum ``in` `mp): ` `            ``ans ``=` `max``(ans, mp[dSum] ``+` `arr[i]) ` ` `  `        ``mp[dSum] ``=` `mp.get(dSum, ``0``) ``+` `arr[i] ` ` `  `    ``return` `ans ` ` `  `# Driver code ` `arr ``=` `[``55``, ``23``, ``32``, ``46``, ``88` `] ` `n ``=` `len``(arr) ` ` `  `print``(findSum(arr, n)) ` ` `  `# This code is contributed by mohit kumar 29 `

Output:

```101
```

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Improved By : mohit kumar 29, AnkitRai01