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Number of Groups of Sizes Two Or Three Divisible By 3
  • Difficulty Level : Medium
  • Last Updated : 16 Apr, 2021

You are given N distinct numbers. You are tasked with finding the number of groups of 2 or 3 that can be formed whose sum is divisible by three.
Examples : 
 

Input  : 1 5 7 2 9 14
Output : 13
The groups of two that can be 
formed are:
(1, 5)
(5, 7)
(1, 2)
(2, 7)
(1, 14)
(7, 14)
The groups of three are:
(1, 5, 9)
(5, 7, 9)
(1, 2, 9)
(2, 7, 9)
(2, 5, 14)
(1, 9, 14)
(7, 9, 14)

Input  : 3 6 9 12
Output : 10
All groups of 2 and 3 are valid.

 

Naive Approach : For each number, we can add it up with every other number and see if the sum is divisible by 3. We then store these sums, so that we can add each number again to check for groups of three. 
Time Complexity: O(N^2) for groups of 2, O(N^3) for groups of 3 
Auxiliary Space: O(N^2)
Optimum Approach 
If we carefully look at every number, we realize that 3 options exist: 
 

  1. The number is divisible by 3
  2. The number leaves a remainder of 1, when divided by 3
  3. The number leaves a remainder of 2, when divided by 3

Now, for groups of two being divisible by 3, either both number have to belong to category 1 (both are divisible by 3), or one number should leave a remainder 1, and the other a remainder 2. This way the remainders add up to 3, making the sum divisible by 3. 
To form a group of three, either all three numbers should give the same remainder, or one should give remainder 0, another should give 1, and the last should give 2.
In this way, we do not care about the numbers themselves, but their respective remainders. Thus by grouping them into three categories, we can find the total possible groups using a simple formula. 
Let C1 be number of elements divisible by 3. 
Let C2 be number of elements leaving remainder 1. 
Let C3 be number of elements leaving remainder 2. 
 

Answer = 
C2 * C3 + C1 * (C1 - 1) / 2    --> Groups of 2
+ C1 * (C1 - 1) * (C1 - 2) / 6 
+ C2 * (C2 - 1) * (C2 - 2) / 6 
+ C3 * (C3 - 1) * (C3 - 2) / 6 --> Groups of 3 
                   with elements of same remainder
+ C1 * C2 * C3 --> Groups of three with all
                         distinct remainders

 



CPP




// Program to find groups of 2 or 3
// whose sum is divisible by 3
#include <iostream>
using namespace std;
 
int numOfCombinations(int arr[], int N)
{
    // Initialize groups to 0
    int C[3] = { 0, 0, 0 };
 
    // Increment group with specified remainder
    for (int i = 0; i < N; ++i)
        ++C[arr[i] % 3];
 
    // Return groups using the formula
    return C[1] * C[2] + C[0] * (C[0] - 1) / 2 + C[0] * (C[0] - 1) * (C[0] - 2) / 6 + C[1] * (C[1] - 1) * (C[1] - 2) / 6 + C[2] * (C[2] - 1) * (C[2] - 2) / 6 + C[0] * C[1] * C[2];
}
 
// Driver Function
int main()
{
    int arr1[6] = { 1, 5, 7, 2, 9, 14 };
    cout << numOfCombinations(arr1, 6) << "\n";
    int arr2[4] = { 3, 6, 9, 12 };
    cout << numOfCombinations(arr2, 4) << "\n";
    return 0;
}

Java




// Program to find groups of 2 or 3
// whose sum is divisible by 3
 
class GFG {
 
    static int numOfCombinations(int arr[], int N)
    {
        // Initialize groups to 0
        int C[] = { 0, 0, 0 };
 
        // Increment group with specified remainder
        for (int i = 0; i < N; ++i)
            ++C[arr[i] % 3];
 
        // Return groups using the formula
        return C[1] * C[2] + C[0] * (C[0] - 1) / 2 + C[0] * (C[0] - 1) * (C[0] - 2) / 6 + C[1] * (C[1] - 1) * (C[1] - 2) / 6 + C[2] * (C[2] - 1) * (C[2] - 2) / 6 + C[0] * C[1] * C[2];
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr1[] = { 1, 5, 7, 2, 9, 14 };
        System.out.print(numOfCombinations(arr1, 6) + "\n");
        int arr2[] = { 3, 6, 9, 12 };
        System.out.print(numOfCombinations(arr2, 4) + "\n");
    }
}
 
// This code is contributed by Anant Agarwal.

Python3




# Program to find groups of 2 or 3
# whose sum is divisible by 3
 
def numOfCombinations(arr, N):
    # Initialize groups to 0
    C = [0, 0, 0]
  
    # Increment group with
    # specified remainder
    for i in range(N):
         
        C[arr[i] % 3]= C[arr[i] % 3]+1
  
    # Return groups using the formula
    return (C[1] * C[2] + C[0] * (C[0] - 1) / 2 + C[0] * (C[0] - 1) * (C[0] - 2) / 6 + C[1] * (C[1] - 1) * (C[1] - 2) / 6 + C[2] * (C[2] - 1) * (C[2] - 2) / 6 + C[0] * C[1] * C[2])
 
# Driver code
 
arr1 = [1, 5, 7, 2, 9, 14]
print(int(numOfCombinations(arr1, 6)))
arr2 = [3, 6, 9, 12]
print(int(numOfCombinations(arr2, 4)))
 
 
# This code is contributed
# by Anant Agarwal.

C#




// C# Program to find groups of 2 or
// 3 whose sum is divisible by 3
using System;
 
class GFG {
 
    // Function to find number of combinations
    static int numOfCombinations(int[] arr,
                                 int N)
    {
 
        // Initialize groups to 0
        int[] C = { 0, 0, 0 };
 
        // Increment group with specified remainder
        for (int i = 0; i < N; ++i)
            ++C[arr[i] % 3];
 
        // Return groups using the formula
        return C[1] * C[2] + C[0] * (C[0] - 1) / 2 + C[0] * (C[0] - 1) * (C[0] - 2) / 6 + C[1] * (C[1] - 1) * (C[1] - 2) / 6 + C[2] * (C[2] - 1) * (C[2] - 2) / 6 + C[0] * C[1] * C[2];
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr1 = { 1, 5, 7, 2, 9, 14 };
        Console.WriteLine(numOfCombinations(arr1, 6));
        int[] arr2 = { 3, 6, 9, 12 };
        Console.WriteLine(numOfCombinations(arr2, 4));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
 
// PHP Program to find
// groups of 2 or 3
// whose sum is divisible by 3
 
 
function numOfCombinations($arr, $N)
{
    // Initialize groups to 0
    $C = array(0, 0, 0);
 
    // Increment group with
    // specified remainder
    for ($i = 0; $i < $N; ++$i)
        ++$C[$arr[$i] % 3];
 
    // Return groups using the formula
    return $C[1] * $C[2] +
           $C[0] * ($C[0] - 1) / 2 +
           $C[0] * ($C[0] - 1) *
                   ($C[0] - 2) / 6 +
           $C[1] * ($C[1] - 1) *
                   ($C[1] - 2) / 6 +
           $C[2] * ($C[2] - 1) *
                   ($C[2] - 2) / 6 +
           $C[0] * $C[1] * $C[2];
}
 
// Driver Code
$arr1 = array(1, 5, 7, 2, 9, 14);
echo numOfCombinations($arr1, 6), "\n";
 
$arr2 = array(3, 6, 9, 12);
    echo numOfCombinations($arr2, 4), "\n";
 
// This code is contributed by ajit
?>

Javascript




<script>
// Javascript Program to find
// groups of 2 or 3
// whose sum is divisible by 3
function numOfCombinations(arr, N)
{
 
    // Initialize groups to 0
    let C = [0, 0, 0];
 
    // Increment group with
    // specified remainder
    for (let i = 0; i < N; ++i)
        ++C[arr[i] % 3];
 
    // Return groups using the formula
    return C[1] * C[2] +
        C[0] * (C[0] - 1) / 2 +
        C[0] * (C[0] - 1) *
                (C[0] - 2) / 6 +
        C[1] * (C[1] - 1) *
                (C[1] - 2) / 6 +
        C[2] * (C[2] - 1) *
                (C[2] - 2) / 6 +
        C[0] * C[1] * C[2];
}
 
// Driver Code
let arr1 = [1, 5, 7, 2, 9, 14];
document.write(numOfCombinations(arr1, 6) + "<br>");
 
let arr2 = [3, 6, 9, 12];
document.write(numOfCombinations(arr2, 4) + "<br>");
 
// This code is contributed by gfgking.
</script>

Output : 
 

13
10

Asked in Amazon
This article is contributed by Aditya Kamath. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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