Count of pairs with sum N from first N natural numbers

Given a an integer N, the task is to count the number of pairs among the first N natural numbers, with sum equal to N.

Examples:

Input: N = 8
Output: 3
Explanation:
All possiblr pairs with sum 8 are { (1, 7), (2, 6), (3, 5)}

Input: N = 9
Output: 4

Naive Approach:
The simplest approach to solve the problem is to use Two Pointers. Follow the steps below to solve the problem:



  • Set i = 0 and j = N – 1 initially.
  • Iterate until i >= j, and for every pair of i, j, check if their sum is equal to N or not. If so, increase the count of pairs.
  • Move to the next pair by increasing and decreasing i and j by 1 respectively.
  • Finally, print the count of pairs obtained.

Below is the implementation of the above approach:

C++

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// C++ Program to implement
// the above approach
#include <iostream>
using namespace std;
  
int numberOfPairs(int n)
{
  
    // Stores the count of
    // pairs
    int count = 0;
    // Set the two pointers
    int i = 1, j = n - 1;
  
    while (i < j) {
  
        // Check if the sum of
        // pairs is equal to N
        if (i + j == n) {
            // Increase the count
            // of pairs
            count++;
        }
  
        // Move to the next pair
        i++;
        j--;
    }
  
    return count;
}
  
// Driver Code
int main()
{
    int n = 8;
    cout << numberOfPairs(n);
    return 0;
}

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Java

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// Java program for the above approach
import java.io.*;
  
class GFG{
      
// Function to calculate the value of count
public static int numberOfPairs(int n)
{
  
    // Stores the count of pairs
    int count = 0;
  
    // Set the two pointers
    int i = 1, j = n - 1;
  
    while (i < j)
    {
          
        // Check if the sum of
        // pairs is equal to N
        if (i + j == n)
        {
              
            // Increase the count
            // of pairs
            count++;
        }
  
        // Move to the next pair
        i++;
        j--;
    }
    return count;
}
  
// Driver code
public static void main (String[] args)
{
    int n = 8;
      
    System.out.println(numberOfPairs(n));
}
}
  
// This code is contributed by piyush3010 

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Python3

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# Python3 program for the
# above approach
def numberOfPairs(n):
    
  # Stores the count
  # of pairs
  count = 0
    
  # Set the two pointers
  i = 1
  j = n - 1
  
  while(i < j):
      
    # Check if the sum
    # of pirs is equal to n
    if (i + j) == n:
        
      # Increase the count of pairs
      count += 1
        
      # Move to the next pair
      i += 1
      j -= 1
        
  return count
  
# Driver code
if __name__=='__main__':
    
  n = 8
  print(numberOfPairs(n))
      
# This code is contributed by virusbuddah_

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C#

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// C# program for the above approach
using System;
class GFG{
       
// Function to calculate the value of count
public static int numberOfPairs(int n)
{
   
    // Stores the count of pairs
    int count = 0;
   
    // Set the two pointers
    int i = 1, j = n - 1;
   
    while (i < j)
    {
           
        // Check if the sum of
        // pairs is equal to N
        if (i + j == n)
        {
               
            // Increase the count
            // of pairs
            count++;
        }
   
        // Move to the next pair
        i++;
        j--;
    }
    return count;
}
   
// Driver code
public static void Main (string[] args)
{
    int n = 8;
       
    Console.Write(numberOfPairs(n));
}
}
   
// This code is contributed by rock_cool

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Output

3

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

Efficient Approach:
To optimize the above approach, we just need to observe if N is even or odd. If N is even, the count of possible pairs is N/2 – 1. Otherwise, it is  N/2.

Illustration:

N = 8
All possible pairs are (1, 7), (2, 6) and (3, 5)
Hence, count of possible pairs = 3 = 8/2 – 1

N = 9
All possible pairs are (1, 8), (2, 7), (3, 6) and (4, 5)
Hence, count of possible pairs = 4 = 9/2

Below is the implementation of the above approach:

C++

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// C++ program to count the number
// of pairs among the first N
// natural numbers with sum N
#include <iostream>
using namespace std;
  
// Function to return the
// count of pairs
int numberOfPairs(int n)
{
    // If n is even
    if (n % 2 == 0)
  
        // Count of pairs
        return n / 2 - 1;
  
    // Otherwise
    else
        return n / 2;
}
  
// Driver Code
int main()
{
    int n = 8;
    cout << numberOfPairs(n);
  
    return 0;
}

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Java

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// Java program to count the number
// of pairs among the first N
// natural numbers with sum N
import java.io.*;
  
class GFG{
      
// Function to calculate the value of count
public static int numberOfPairs(int n)
{
  
    // If n is even
    if (n % 2 == 0)
      
        // Count of pairs
        return n / 2 - 1;
  
    // Otherwise
    else
        return n / 2;
}
  
// Driver code
public static void main (String[] args)
{
    int n = 8;
      
    System.out.println(numberOfPairs(n));
}
}
  
// This code is contributed by piyush3010 

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Python3

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# Python3 program to count the number
# of pairs among the first N
# natural numbers with sum N
  
# Function to calculate the value of count
def numberOfPairs(n):
  
    # If n is even
    if (n % 2 == 0):
  
        # Count of pairs
        return n // 2 - 1;
  
    # Otherwise
    else:
        return n // 2;
  
# Driver code
n = 8;
  
print(numberOfPairs(n));
  
# This code is contributed by Rajput-Ji

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C#

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// C# program to count the number
// of pairs among the first N
// natural numbers with sum N
using System;
class GFG{
       
// Function to calculate the value of count
public static int numberOfPairs(int n)
{
   
    // If n is even
    if (n % 2 == 0)
       
        // Count of pairs
        return n / 2 - 1;
   
    // Otherwise
    else
        return n / 2;
}
   
// Driver code
public static void Main (string[] args)
{
    int n = 8;
       
    Console.Write(numberOfPairs(n));
}
}
   
// This code is contributed by Ritik Bansal

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Output

3

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

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