# Count pairs in array whose sum is divisible by 4

• Difficulty Level : Easy
• Last Updated : 13 May, 2022

Given a array if ‘n’ positive integers. Count number of pairs of integers in the array that have the sum divisible by 4.
Examples :

Input: {2, 2, 1, 7, 5}
Output: 3

Explanation
Only three pairs are possible whose sum
is divisible by '4' i.e., (2, 2),
(1, 7) and (7, 5)

Input: {2, 2, 3, 5, 6}
Output: 4


Naive approach is to iterate through every pair of array but using two nested for loops and count those pairs whose sum is divisible by ‘4’. Time complexity of this approach is O(n2).

Efficient approach is to use Hashing technique. There are only three condition that can arise whose sum is divisible by ‘4’ i.e,

1. If both are divisible by 4.
2. If one of them is equal to 1 modulo 4 and other is 3 modulo 4. For instance, (1, 3), (5, 7), (5, 11).
3. If both of them is equal to 2 modulo 4 i.e., (2, 2), (2, 6), (6, 10)
Store all modulo in freq[] array such that freq[i] = number of array elements that are equal to i modulo 4.

Thus answer =>  ## C++

 // C++ Program to count pairs // whose sum divisible by '4'#include using namespace std;  // Program to count pairs whose sum divisible// by '4'int count4Divisibiles(int arr[], int n){    // Create a frequency array to count     // occurrences of all remainders when     // divided by 4    int freq = {0, 0, 0, 0};      // Count occurrences of all remainders    for (int i = 0; i < n; i++)        ++freq[arr[i] % 4];      // If both pairs are divisible by '4'    int ans = freq * (freq - 1) / 2;      // If both pairs are 2 modulo 4    ans += freq * (freq - 1) / 2;      // If one of them is equal    // to 1 modulo 4 and the    // other is equal to 3     // modulo 4    ans += freq * freq;      return ans;}  // Driver codeint main(){      int arr[] = { 2, 2, 1, 7, 5 };    int n = sizeof(arr) / sizeof(arr);      cout << count4Divisibiles(arr, n);      return 0;}

## Java

 // Java program to count pairs // whose sum divisible by '4'import java.util.*;  class Count{    public static int count4Divisibiles(int arr[] ,                                              int n )    {        // Create a frequency array to count         // occurrences of all remainders when         // divided by 4        int freq[] = {0, 0, 0, 0};        int i = 0;        int ans;                  // Count occurrences of all remainders        for (i = 0; i < n; i++)                ++freq[arr[i] % 4];                  //If both pairs are divisible by '4'        ans = freq * (freq - 1) / 2;              // If both pairs are 2 modulo 4        ans += freq * (freq - 1) / 2;              // If one of them is equal        // to 1 modulo 4 and the        // other is equal to 3         // modulo 4        ans += freq * freq;              return (ans);    }    public static void main(String[] args)    {        int arr[] = {2, 2, 1, 7, 5};        int n = 5;        System.out.print(count4Divisibiles(arr, n));    }}  // This code is contributed by rishabh_jain

## Python3

 # Python3 code to count pairs whose # sum is divisible by '4'  # Function to count pairs whose # sum is divisible by '4'def count4Divisibiles( arr , n ):          # Create a frequency array to count     # occurrences of all remainders when     # divided by 4    freq = [0, 0, 0, 0]          # Count occurrences of all remainders    for i in range(n):        freq[arr[i] % 4]+=1              #If both pairs are divisible by '4'    ans = freq * (freq - 1) / 2          # If both pairs are 2 modulo 4    ans += freq * (freq - 1) / 2          # If one of them is equal    # to 1 modulo 4 and the    # other is equal to 3     # modulo 4    ans += freq * freq          return int(ans)  # Driver codearr = [2, 2, 1, 7, 5]n = len(arr)print(count4Divisibiles(arr, n))  # This code is contributed by "Sharad_Bhardwaj".

## C#

 // C# program to count pairs // whose sum divisible by '4'using System;  class Count{    public static int count4Divisibiles(int []arr ,                                             int n )    {        // Create a frequency array to count         // occurrences of all remainders when         // divided by 4        int []freq = {0, 0, 0, 0};        int i = 0;        int ans;                  // Count occurrences of all remainders        for (i = 0; i < n; i++)            ++freq[arr[i] % 4];                  //If both pairs are divisible by '4'        ans = freq * (freq - 1) / 2;              // If both pairs are 2 modulo 4        ans += freq * (freq - 1) / 2;              // If one of them is equal        // to 1 modulo 4 and the        // other is equal to 3         // modulo 4        ans += freq * freq;              return (ans);    }          // Driver code    public static void Main()    {        int []arr = {2, 2, 1, 7, 5};        int n = 5;        Console.WriteLine(count4Divisibiles(arr, n));    }}  // This code is contributed by vt_m

## PHP

 

Output :

 3


Time complexity: O(n)
Auxiliary space: O(1)

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