Skip to content
Related Articles

Related Articles

Count of pairs from 1 to a and 1 to b whose sum is divisible by N

View Discussion
Improve Article
Save Article
  • Last Updated : 15 Jun, 2022

Given three integers a, b and N. Find the total number of distinct pairs which can be formed by selecting one integer from 1 to a and other from 1 to b, such that their sum is divisible by N.

Examples: 

Input : a = 4, b = 4, N = 4
Output : 4

Input : a = 5, b = 13, N = 3
Output : 22

Basic Approach: For a pair to be divisible by N it must contain one number from range 1 to a and other from 1 to b. 
So, for this iterate over integers from 1 to a and for each integer (i), b/N numbers are there whose sum with i will be divisible by N. Also if (i%N + b%N) >= N then 1 more pair exists whose sum is divisible by N.

For example took a = 7, b = 6 and N = 4: 

Let's check for i = 3:
  • b/N = 6/4 = 1 => there is one integer from 1 to b,
    whose sum with 3 is divisible by 4 i.e.(3, 1).
  • Also i%N + b%N = 3%4 + 6%4 = 3+2 = 5 > 4, 
    means one more integer exists from 1 to b 
    whose sum with 3 is divisible by 4 i.e.(3, 5).

Below is the implementation of the above approach: 

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
int findCountOfPairs(int a, int b, int n)
{
    int ans = 0;
 
    // Iterate over 1 to a to find distinct pairs
    for (int i = 1; i <= a; i++) {
        // For each integer from 1 to a
        // b/n integers exists such that pair
        // sum is divisible by n
        ans += b / n;
 
        // If (i%n +b%n ) >= n one more pair is possible
        ans += (i % n + b % n) >= n ? 1 : 0;
    }
 
    // Return answer
    return ans;
}
 
// Driver code
int main()
{
    int a = 5, b = 13, n = 3;
    cout << findCountOfPairs(a, b, n);
 
    return 0;
}

Java




// Java program for the above approach
class GFG
{
 
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
static int findCountOfPairs(int a, int b, int n)
{
    int ans = 0;
 
    // Iterate over 1 to a to find distinct pairs
    for (int i = 1; i <= a; i++)
    {
        // For each integer from 1 to a
        // b/n integers exists such that pair
        // sum is divisible by n
        ans += b / n;
 
        // If (i%n +b%n ) >= n one more pair is possible
        ans += (i % n + b % n) >= n ? 1 : 0;
    }
 
    // Return answer
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int a = 5, b = 13, n = 3;
    System.out.println(findCountOfPairs(a, b, n));
}
}
 
// This code has been contributed by 29AjayKumar

Python3




# Python implementation of above approach
 
# Function to find the distinct pairs from
# 1-a & 1-b such that their sum is divisible by n.
def findCountOfPairs(a, b, n):
    ans = 0
    for i in range(1, a + 1):
 
        # For each integer from 1 to a
        # b/n integers exists such that pair
        # / sum is divisible by n
        ans += b//n
 
        # If (i%n +b%n ) >= n one more pair is possible
        ans += 1 if (i % n + b % n) >= n else 0
 
    return ans
 
# Driver code
a = 5; b = 13; n = 3
print(findCountOfPairs(a, b, n))
 
# This code is contributed by Shrikant13

C#




// C# program for the above approach
using System;
 
class GFG
{
     
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
static int findCountOfPairs(int a, int b, int n)
{
    int ans = 0;
 
    // Iterate over 1 to a to find distinct pairs
    for (int i = 1; i <= a; i++)
    {
        // For each integer from 1 to a
        // b/n integers exists such that pair
        // sum is divisible by n
        ans += b / n;
 
        // If (i%n +b%n ) >= n one more pair is possible
        ans += (i % n + b % n) >= n ? 1 : 0;
    }
 
    // Return answer
    return ans;
}
 
// Driver code
static public void Main ()
{
    int a = 5, b = 13, n = 3;
    Console.WriteLine(findCountOfPairs(a, b, n));
}
}
 
// This code has been contributed by ajit.

PHP




<?php
// PHP implementation of above approach
 
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
function findCountOfPairs($a, $b, $n)
{
    $ans = 0;
 
    // Iterate over 1 to a to find
    // distinct pairs
    for ($i = 1; $i <= $a; $i++)
    {
         
        // For each integer from 1 to a
        // b/n integers exists such that pair
        // sum is divisible by n
        $ans += (int)($b / $n);
 
        // If (i%n +b%n ) >= n one more
        // pair is possible
        $ans += (($i % $n ) +
                 ($b % $n)) >= $n ? 1 : 0;
    }
 
    // Return answer
    return $ans;
}
 
// Driver code
$a = 5;
$b = 13;
$n = 3;
echo findCountOfPairs($a, $b, $n);
 
// This code is contributed by akt_mit.
?>

Javascript




<script>
 
    // Javascript program for the above approach
     
    // Function to find the distinct pairs from
    // 1-a & 1-b such that their sum is divisible by n.
    function findCountOfPairs(a, b, n)
    {
        let ans = 0;
 
        // Iterate over 1 to a to find distinct pairs
        for (let i = 1; i <= a; i++)
        {
         
            // For each integer from 1 to a
            // b/n integers exists such that pair
            // sum is divisible by n
            ans += parseInt(b / n, 10);
 
            // If (i%n +b%n ) >= n one more pair is possible
            ans += (i % n + b % n) >= n ? 1 : 0;
        }
 
        // Return answer
        return ans;
    }
     
    let a = 5, b = 13, n = 3;
    document.write(findCountOfPairs(a, b, n));
     
    // This code is contributed by mukesh07.
     
</script>

Output: 

22

 

Time complexity: O(N)

Auxiliary Space: O(1)

Second Approach:-  This Approach Is a little Tricky. Here we find how much pair for a multiple of  N.

First:- Keep (a<b), if not then make using swap.

Second:- Start a for loop from the lowest multiple of N and go through the multiple.

  • Now Smallest element ( a ) is greater than or equally current multiple then we add ((Current_Multiple) – 1) pair to the ans.
  • Now a is smaller But b is greater than or equally current multiple than we add a to the ans.
  • Now if a and b both smaller than we count the remaining pair for add a – (current_multiple – b ) + 1.
  • Break The Loop.

Below is the implementation of the above logic.

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
int findCountOfPairs(int a, int b, int n)
{
    if (a > b)
    {
        // if first element is bigger than swap
        swap(a, b);
    }
    int temp = 1, count = 0;
    // count is store the number of pair.
    for (int i = n; temp > 0; i += n)
    {
        // we use temp for breaking a loop.
        if (a >= i)
        {
            // count when a is greater.
            temp = i - 1;
        }
        else if (b >= i)
        {
            // Count when a is smaller but
            // b is greater
            temp = a;
        }
        else if (i > b)
        {
            // Count when a and b both are smaller
            temp = a - (i - b) + 1;
        }
        if (temp > 0) //breaking condition
        {
            // For storing The pair in count.
            count += temp;
        }
    }
    // return the number of pairs.
    return count;
}
 
// Driver code
int main()
{
    int a = 5, b = 13, n = 3;
    cout << findCountOfPairs(a, b, n);
 
    return 0;
}
// contribute by Vivek Javiya

Java




// Java implementation of
// the above approach
class GFG{
     
// Function to find the
// distinct pairs from
// 1-a & 1-b such that
// their sum is divisible by n.
public static int findCountOfPairs(int a,
                                   int b,
                                   int n)
{
  if (a > b)
  {
    // if first element is
    // bigger than swap
    int temp = a;
    a = b;
    b = temp;
  }
   
  int temp = 1, count = 0;
   
  // count is store the
  // number of pair.
  for (int i = n; temp > 0; i += n)
  {
    // we use temp for
    // breaking a loop.
    if (a >= i)
    {
      // count when a
      // is greater.
      temp = i - 1;
    }
    else if (b >= i)
    {
      // Count when a is
      // smaller but
      // b is greater
      temp = a;
    }
    else if (i > b)
    {
      // Count when a and b
      // both are smaller
      temp = a - (i - b) + 1;
    }
     
    //breaking condition
    if (temp > 0)
    {
      // For storing The
      // pair in count.
      count += temp;
    }
  }
   
  // return the number
  // of pairs.
  return count;
}
   
// Driver code
public static void main(String[] args)
{
  int a = 5, b = 13, n = 3;
  System.out.print(findCountOfPairs(a,
                                    b, n));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3




# Python implementation of above approach
 
# Function to find the distinct pairs from
# 1-a & 1-b such that their sum is divisible by n.
def findCountOfPairs(a, b, n):
    if(a > b):
       
        # if first element is bigger than swap
        a, b = b, a
 
    temp = 1
    count = 0
     
    # count is store the number of pair.
    i = n
    while(temp > 0):
       
        # we use temp for breaking a loop.
        if(a >= i):
           
            # count when a is greater.
            temp = i - 1
        elif(b >= i):
           
            # Count when a is smaller but
            # b is greater
            temp = a
        elif(i > b):
           
            # Count when a and b both are smaller
            temp = a - (i - b) + 1
 
        if(temp > 0):
           
          # breaking condition
            # For storing The pair in count.
            count += temp
        i += n
 
    # return the number of pairs.
    return count
 
# Driver code
a = 5
b = 13
n = 3
 
print(findCountOfPairs(a, b, n))
 
# This code is contributed by avanitrachhadiya2155

C#




// C# implementation of
// the above approach
using System;
 
class GFG
{
     
    // Function to find the
    // distinct pairs from
    // 1-a & 1-b such that
    // their sum is divisible by n.
    static int findCountOfPairs(int a, int b, int n)
    {
      if (a > b)
      {
         
        // if first element is
        // bigger than swap
        int temp1 = a;
        a = b;
        b = temp1;
      }
        
      int temp = 1, count = 0;
        
      // count is store the
      // number of pair.
      for (int i = n; temp > 0; i += n)
      {
         
        // we use temp for
        // breaking a loop.
        if (a >= i)
        {
           
          // count when a
          // is greater.
          temp = i - 1;
        }
        else if (b >= i)
        {
           
          // Count when a is
          // smaller but
          // b is greater
          temp = a;
        }
        else if (i > b)
        {
           
          // Count when a and b
          // both are smaller
          temp = a - (i - b) + 1;
        }
          
        // breaking condition
        if (temp > 0)
        {
           
          // For storing The
          // pair in count.
          count += temp;
        }
      }
        
      // return the number
      // of pairs.
      return count;
    }
   
  // Driver code
  static void Main()
  {
      int a = 5, b = 13, n = 3;
      Console.WriteLine(findCountOfPairs(a, b, n));
  }
}
 
// This code is contributed by divyesh072019

Javascript




<script>
 
    // Javascript implementation of
    // the above approach
     
    // Function to find the
    // distinct pairs from
    // 1-a & 1-b such that
    // their sum is divisible by n.
    function findCountOfPairs(a, b, n)
    {
      if (a > b)
      {
          
        // if first element is
        // bigger than swap
        let temp1 = a;
        a = b;
        b = temp1;
      }
         
      let temp = 1, count = 0;
         
      // count is store the
      // number of pair.
      for (let i = n; temp > 0; i += n)
      {
          
        // we use temp for
        // breaking a loop.
        if (a >= i)
        {
            
          // count when a
          // is greater.
          temp = i - 1;
        }
        else if (b >= i)
        {
            
          // Count when a is
          // smaller but
          // b is greater
          temp = a;
        }
        else if (i > b)
        {
            
          // Count when a and b
          // both are smaller
          temp = a - (i - b) + 1;
        }
           
        // breaking condition
        if (temp > 0)
        {
            
          // For storing The
          // pair in count.
          count += temp;
        }
      }
         
      // return the number
      // of pairs.
      return count;
    }
     
    let a = 5, b = 13, n = 3;
      document.write(findCountOfPairs(a, b, n));
     
</script>

Output :

22

Time Complexity: O(n)

Auxiliary Space: O(1)
Efficient Approach: For solving the problem efficiently break it into four-part and solve as: 

  • Each integer from range 1 to N*(a/N) will have exactly b/N integers from 1 to N*(b/N) whose sum is divisible by N.
  • There exist a/N integers from range 1 to N*(a/N) which can form pairs with b%N integer ranging from N*(b/N) to b.
  • There exist a%N integers from range N*(a/N) to a which can form pairs with b/N integer ranging from 1 to N*(b/N).
  • There exists (a%N + b%N)/N integers from range N*(a/N) to a and from N*(b/N) to b which can form pair whose sum is divisible by N.

Below is the implementation of the above approach:

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
int findCountOfPairs(int a, int b, int n)
{
    int ans = 0;
 
    // pairs from 1 to n*(a/n) and 1 to n*(b/n)
    ans += n * (a / n) * (b / n);
 
    // pairs from 1 to n*(a/n) and n*(b/n) to b
    ans += (a / n) * (b % n);
 
    // pairs from n*(a/n) to a and 1 to n*(b/n)
    ans += (a % n) * (b / n);
 
    // pairs from n*(a/n) to a and  n*(b/n) to b
    ans += ((a % n) + (b % n)) / n;
 
    // Return answer
    return ans;
}
 
// Driver code
int main()
{
    int a = 5, b = 13, n = 3;
    cout << findCountOfPairs(a, b, n);
 
    return 0;
}

Java




// Java implementation of above approach
import java.io.*;
 
class GFG
{
     
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
static int findCountOfPairs(int a, int b, int n)
{
    int ans = 0;
 
    // pairs from 1 to n*(a/n) and 1 to n*(b/n)
    ans += n * (a / n) * (b / n);
 
    // pairs from 1 to n*(a/n) and n*(b/n) to b
    ans += (a / n) * (b % n);
 
    // pairs from n*(a/n) to a and 1 to n*(b/n)
    ans += (a % n) * (b / n);
 
    // pairs from n*(a/n) to a and n*(b/n) to b
    ans += ((a % n) + (b % n)) / n;
 
    // Return answer
    return ans;
}
 
// Driver code
public static void main (String[] args)
{
    int a = 5, b = 13, n = 3;
    System.out.println (findCountOfPairs(a, b, n));
}
}
 
// This code is contributed by ajit..

Python3




# Python 3 implementation of above approach
 
# Function to find the distinct pairs from
# 1-a & 1-b such that their sum is divisible by n.
def findCountOfPairs(a, b, n):
    ans = 0
 
    # pairs from 1 to n*(a/n) and 1 to n*(b/n)
    ans += n * int(a / n) * int(b / n)
 
    # pairs from 1 to n*(a/n) and n*(b/n) to b
    ans += int(a / n) * (b % n)
 
    # pairs from n*(a/n) to a and 1 to n*(b/n)
    ans += (a % n) * int(b / n)
 
    # pairs from n*(a/n) to a and n*(b/n) to b
    ans += int(((a % n) + (b % n)) / n);
 
    # Return answer
    return ans
 
# Driver code
if __name__ == '__main__':
    a = 5
    b = 13
    n = 3
    print(findCountOfPairs(a, b, n))
 
# This code is contributed by
# Surendra_Gangwar

C#




// C# implementation of above approach
using System;
 
class GFG
{
         
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
 
static int findCountOfPairs(int a, int b, int n)
{
    int ans = 0;
 
    // pairs from 1 to n*(a/n) and 1 to n*(b/n)
    ans += n * (a / n) * (b / n);
 
    // pairs from 1 to n*(a/n) and n*(b/n) to b
    ans += (a / n) * (b % n);
 
    // pairs from n*(a/n) to a and 1 to n*(b/n)
    ans += (a % n) * (b / n);
 
    // pairs from n*(a/n) to a and n*(b/n) to b
    ans += ((a % n) + (b % n)) / n;
 
    // Return answer
    return ans;
}
 
// Driver code
    static public void Main (){
    int a = 5, b = 13, n = 3;
    Console.WriteLine(findCountOfPairs(a, b, n));
}
}
 
// This code is contributed by @Tushil

PHP




<?php
// PHP implementation of above approach
 
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
function findCountOfPairs($a, $b, $n)
{
    $ans = 0;
 
    // pairs from 1 to n*(a/n) and 1 to n*(b/n)
    $ans += $n * (int)($a / $n) * (int)($b / $n);
 
    // pairs from 1 to n*(a/n) and n*(b/n) to b
    $ans += (int)($a / $n) * ($b % $n);
 
    // pairs from n*(a/n) to a and 1 to n*(b/n)
    $ans += ($a % $n) * (int)($b / $n);
 
    // pairs from n*(a/n) to a and n*(b/n) to b
    $ans += (($a % $n) + (int)($b % $n)) / $n;
 
    // Return answer
    return $ans;
}
 
// Driver code
$a = 5;
$b = 13;
$n = 3;
echo findCountOfPairs($a, $b, $n);
 
// This code is contributed by ajit.
?>

Javascript




<script>
    // Javascript implementation of above approach
     
    // Function to find the distinct pairs from
    // 1-a & 1-b such that their sum is divisible by n.
    function findCountOfPairs(a, b, n)
    {
        let ans = 0;
 
        // pairs from 1 to n*(a/n) and 1 to n*(b/n)
        ans += n * parseInt(a / n, 10) * parseInt(b / n, 10)
 
        // pairs from 1 to n*(a/n) and n*(b/n) to b
        ans += parseInt(a / n, 10) * parseInt(b % n, 10);
 
        // pairs from n*(a/n) to a and 1 to n*(b/n)
        ans += parseInt(a % n, 10) * parseInt(b / n, 10);
 
        // pairs from n*(a/n) to a and  n*(b/n) to b
        ans += parseInt(((a % n) + (b % n)) / n, 10);
 
        // Return answer
        return ans;
    }
     
    let a = 5, b = 13, n = 3;
    document.write(findCountOfPairs(a, b, n));
     
</script>

Output: 

22

 

Time Complexity: O(1)

Auxiliary Space: O(1)


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!