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

(105, 96): The difference between 105+ sumOfDigits(105) and 96+sumOfDigits(96) is 0, thus they are equal.
(105, 96): The difference between 105+ sumOfDigits(105) and 96+sumOfDigits(96) is 0, thus they are equal.
(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 <bits/stdc++.h> 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[0]);     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<Integer, Integer> mp             = new HashMap<Integer, Integer>();          // 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<Integer, Integer> 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# 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

<script>         // JavaScript Program to implement         // the above approach         // Function to find the sum of digits         // of the number N         function sumOfDigits(N) {             // Stores the sum of digits             let sum = 0;              // If the number N is greater than 0             while (N != 0) {                 sum = sum + (N % 10);                 N = Math.floor(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])         function CountPair(arr, n) {             // Stores the frequency of value             // of arr[i] + sumOfDigits(arr[i])             let mp = new Map();              // Traverse the given array             for (let i = 0; i < n; i++) {                  // Find the value                   let val = arr[i] + sumOfDigits(arr[i]);                 // Increment the frequency                 if (mp.has(val)) {                     mp.set(val, mp.get(val) + 1);                 }                 else {                     mp.set(val, 1);                 }             }              // Stores the total count of pairs             let count = 0;              // Traverse the map mp             for (let [key, value] of mp) {                  // Update the count of pairs                   count = count + (value * (value - 1)) / 2;             }              // Return the total count of pairs             return count;         }          // Driver Code          let arr = [105, 96, 20, 2, 87, 96];         let N = arr.length;         document.write(CountPair(arr, N))   // This code is contributed by Potta Lokesh     </script>

Output

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

Count pairs in an array such that both elements has equal set bits

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

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