# Count pairs in an array having sum of elements with their respective sum of digits equal

• Last Updated : 16 Jun, 2022

Given an array arr[] consisting of N positive integers, the task is to count the number of pairs in the array, say (a, b) such that sum of a with its sum of digits is equal to sum of b with its sum of digits.

Examples:

Input: arr[] = {1, 1, 2, 2}
Output: 2
Explanation:
Following are the pairs that satisfy the given criteria:

1. (1, 1): The difference between 1+ sumOfDigits(1) and 1+sumOfDigits(1) is 0, thus they are equal.
2. (2, 2): The difference between 2+sumOfDigits(2) and 2+sumOfDigits(2) is 0 , thus they are equal.

Therefore, the total number of pairs are 2.

Input: arr[] = {105, 96, 20, 2, 87, 96}
Output: 3
Following are the pairs that satisfy the given criteria:

1. (105, 96): The difference between 105+ sumOfDigits(105) and 96+sumOfDigits(96) is 0, thus they are equal.
2. (105, 96): The difference between 105+ sumOfDigits(105) and 96+sumOfDigits(96) is 0, thus they are equal.
3. (96, 96): The difference between 96+ sumOfDigits(96) and 96+sumOfDigits(96) is 0, thus they are equal.

Input: arr[] = {1, 2, 3, 4}
Output: 0

Naive Approach: The simplest approach to solve the problem is to generate all possible pairs of the given array and count those pairs that satisfy the given criteria. After checking for all the pairs print the total count of pairs obtained.

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized by storing the sum of elements with its sum of digits in a HashMap and then count the total number of pairs formed accordingly. Follow the steps given below to solve the problem:

• Initialize an unordered_map, M that stores the frequency of the sum of elements with its sum of digits for each array element.
• Traverse the given array and increment the frequency of (arr[i] + sumOfDigits(arr[i])) in the map M.
• Initialize a variable, say count as 0 that stores the total count of resultant pairs.
• Traverse the given map M and if the frequency of any element, say F is greater than or equal to 2, then increment the value of count by (F*(F – 1))/2.
• After completing the above steps, print the value of count as the resultant count of pairs.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to find the sum of digits` `// of the number N` `int` `sumOfDigits(``int` `N)` `{` `    ``// Stores the sum of digits` `    ``int` `sum = 0;`   `    ``// If the number N is greater than 0` `    ``while` `(N) {` `        ``sum += (N % 10);` `        ``N = N / 10;` `    ``}`   `    ``// Return the sum` `    ``return` `sum;` `}`   `// Function to find the count of pairs` `// such that arr[i] + sumOfDigits(arr[i])` `// is equal to (arr[j] + sumOfDigits(arr[j])` `int` `CountPair(``int` `arr[], ``int` `n)` `{` `    ``// Stores the frequency of value` `    ``// of arr[i] + sumOfDigits(arr[i])` `    ``unordered_map<``int``, ``int``> mp;`   `    ``// Traverse the given array` `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// Find the value` `        ``int` `val = arr[i] + sumOfDigits(arr[i]);`   `        ``// Increment the frequency` `        ``mp[val]++;` `    ``}`   `    ``// Stores the total count of pairs` `    ``int` `count = 0;`   `    ``// Traverse the map mp` `    ``for` `(``auto` `x : mp) {`   `        ``int` `val = x.first;` `        ``int` `times = x.second;`   `        ``// Update the count of pairs` `        ``count += ((times * (times - 1)) / 2);` `    ``}`   `    ``// Return the total count of pairs` `    ``return` `count;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `arr[] = { 105, 96, 20, 2, 87, 96 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << CountPair(arr, N);`   `    ``return` `0;` `}`

## Java

 `/*package whatever //do not write package name here */` `import` `java.io.*;` `import` `java.util.*;` `class` `GFG ` `{` `  `  `    ``// Function to find the sum of digits` `    ``// of the number N` `    ``static` `int` `sumOfDigits(``int` `N)` `    ``{` `        ``// Stores the sum of digits` `        ``int` `sum = ``0``;`   `        ``// If the number N is greater than 0` `        ``while` `(N > ``0``) {` `            ``sum += (N % ``10``);` `            ``N = N / ``10``;` `        ``}`   `        ``// Return the sum` `        ``return` `sum;` `    ``}`   `    ``// Function to find the count of pairs` `    ``// such that arr[i] + sumOfDigits(arr[i])` `    ``// is equal to (arr[j] + sumOfDigits(arr[j])` `    ``static` `int` `CountPair(``int` `arr[], ``int` `n)` `    ``{` `        ``// Stores the frequency of value` `        ``// of arr[i] + sumOfDigits(arr[i])` `        ``HashMap mp` `            ``= ``new` `HashMap();`   `        ``// Traverse the given array` `        ``for` `(``int` `i = ``0``; i < n; i++) {`   `            ``// Find the value` `            ``int` `val = arr[i] + sumOfDigits(arr[i]);`   `            ``// Increment the frequency` `            ``if` `(mp.containsKey(val)) {` `                ``mp.put(val, mp.get(val) + ``1``);` `            ``}` `            ``else` `{` `                ``mp.put(val, ``1``);` `            ``}` `        ``}`   `        ``// Stores the total count of pairs` `        ``int` `count = ``0``;`   `        ``// Traverse the map mp` `        ``for` `(Map.Entry x :` `             ``mp.entrySet()) {`   `            ``int` `val = x.getKey();` `            ``int` `times = x.getValue();`   `            ``// Update the count of pairs` `            ``count += ((times * (times - ``1``)) / ``2``);` `        ``}`   `        ``// Return the total count of pairs` `        ``return` `count;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `arr[] = { ``105``, ``96``, ``20``, ``2``, ``87``, ``96` `};` `        ``int` `N = ``6``;` `        ``System.out.println(CountPair(arr, N));` `    ``}` `}`   `// This code is contributed by maddler.`

## Python3

 `# Python 3 program for the above approach`   `# Function to find the sum of digits` `# of the number N` `def` `sumOfDigits(N):` `    ``# Stores the sum of digits` `    ``sum` `=` `0`   `    ``# If the number N is greater than 0` `    ``while` `(N):` `        ``sum` `+``=` `(N ``%` `10``)` `        ``N ``=` `N ``/``/` `10`   `    ``# Return the sum` `    ``return` `sum`   `# Function to find the count of pairs` `# such that arr[i] + sumOfDigits(arr[i])` `# is equal to (arr[j] + sumOfDigits(arr[j])` `def` `CountPair(arr, n):` `    ``# Stores the frequency of value` `    ``# of arr[i] + sumOfDigits(arr[i])` `    ``mp ``=` `{}`   `    ``# Traverse the given array` `    ``for` `i ``in` `range``(n):` `        ``# Find the value` `        ``val ``=` `arr[i] ``+` `sumOfDigits(arr[i])`   `        ``# Increment the frequency` `        ``if` `val ``in` `mp:` `            ``mp[val] ``+``=` `1` `        ``else``:` `            ``mp[val] ``=` `1`   `    ``# Stores the total count of pairs` `    ``count ``=` `0`   `    ``# Traverse the map mp` `    ``for` `key,value ``in` `mp.items():` `        ``val ``=` `key` `        ``times ``=` `value`   `        ``# Update the count of pairs` `        ``count ``+``=` `((times ``*` `(times ``-` `1``)) ``/``/` `2``)`   `    ``# Return the total count of pairs` `    ``return` `count`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[``105``, ``96``, ``20``, ``2``, ``87``, ``96``]` `    ``N ``=` `len``(arr)` `    ``print``(CountPair(arr, N))` `    `  `    ``# This code is contributed by SURENDRA_GANGWAR.`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG{`   `  ``// Function to find the sum of digits` `// of the number N` `static` `int` `sumOfDigits(``int` `N)` `{` `  `  `    ``// Stores the sum of digits` `    ``int` `sum = 0;`   `    ``// If the number N is greater than 0` `    ``while` `(N>0) {` `        ``sum += (N % 10);` `        ``N = N / 10;` `    ``}`   `    ``// Return the sum` `    ``return` `sum;` `}`   `// Function to find the count of pairs` `// such that arr[i] + sumOfDigits(arr[i])` `// is equal to (arr[j] + sumOfDigits(arr[j])` `static` `int` `CountPair(``int` `[]arr, ``int` `n)` `{` `    ``// Stores the frequency of value` `    ``// of arr[i] + sumOfDigits(arr[i])` `    ``Dictionary<``int``, ``int``> mp = ``new` `Dictionary<``int``,``int``>();`   `    ``// Traverse the given array` `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// Find the value` `        ``int` `val = arr[i] + sumOfDigits(arr[i]);`   `        ``// Increment the frequency` `        ``if``(mp.ContainsKey(val))` `         ``mp[val]++;` `        ``else` `         ``mp.Add(val,1);` `    ``}`   `    ``// Stores the total count of pairs` `    ``int` `count = 0;`   `    ``// Traverse the map mp` `    ``foreach``(KeyValuePair<``int``, ``int``> entry ``in` `mp) {`   `        ``int` `val = entry.Key;` `        ``int` `times = entry.Value;`   `        ``// Update the count of pairs` `        ``count += ((times * (times - 1)) / 2);` `    ``}`   `    ``// Return the total count of pairs` `    ``return` `count;` `}`   `// Driver Code` `public` `static` `void` `Main()` `{` `    ``int` `[]arr = { 105, 96, 20, 2, 87, 96 };` `    ``int` `N = arr.Length;` `    ``Console.Write(CountPair(arr, N));` `}` `}`   `// This code is contributed by ipg2016107.`

## Javascript

 ``

Output

`3`

Time Complexity: O(N)
Auxiliary Space: O(N)

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