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Count distinct sum of pairs possible from a given range
  • Difficulty Level : Expert
  • Last Updated : 26 Mar, 2021

Given two positive integers L and R ( where L ≤ R ), the task is to count the number of distinct integers that can be obtained by adding any pair of integers from the range [L, R].

Examples: 

Input: L = 3, R = 5
Output: 11
Explanation: All possible distinct sum of pairs are as follows:

  1. (3, 3). Sum = 6.
  2. (3, 4). Sum = 7.
  3. (3, 5). Sum = 8.
  4. (4, 5). Sum = 9.
  5. (5, 5). Sum = 10.

Therefore, the count of distinct sums is 5.

Input: L = 12, R = 14
Output: 5



Naive Approach: The simplest approach to solve the given problem is to find the sum of all possible pairs of numbers from the range [L, R] and print the count of all distinct sums obtained. 

Time Complexity: O((L – R)2)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following observations: 

  • Since, the given range [L, R] is continuous, the range of numbers obtained by adding will also be continuous.
  • The minimum and maximum sum of pairs from the range are given by 2 * L and 2 * R respectively.
  • Therefore, the count distinct sum of pairs is given by (2 * R – 2 * L + 1).

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count distinct sum of
// pairs possible from the range [L, R]
int distIntegers(int L, int R)
{
    // Return the count of
    // distinct sum of pairs
    return 2 * R - 2 * L + 1;
}
 
// Driver Code
int main()
{
    int L = 3, R = 8;
    cout << distIntegers(L, R);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
 
class GFG{
     
// Function to count distinct sum of
// pairs possible from the range [L, R]
static int distIntegers(int L, int R)
{
     
    // Return the count of
    // distinct sum of pairs
    return 2 * R - 2 * L + 1;
}
  
// Driver Code
public static void main (String[] args)
{
    int L = 3, R = 8;
     
    System.out.println(distIntegers(L, R));
}
}
 
// This code is contributed by rag2127

Python3




# Python3 program for the above approach
 
# Function to count distinct sum of
# pairs possible from the range [L, R]
def distIntegers(L, R):
   
    # Return the count of
    # distinct sum of pairs
    return 2 * R - 2 * L + 1
 
# Driver Code
if __name__ == '__main__':
    L, R = 3, 8
    print (distIntegers(L, R))
 
# This code is contributed by mohit kumar 29.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to count distinct sum of
// pairs possible from the range [L, R]
static int distIntegers(int L, int R)
{
     
    // Return the count of
    // distinct sum of pairs
    return 2 * R - 2 * L + 1;
}
 
// Driver Code
static public void Main()
{
    int L = 3, R = 8;
     
    Console.Write(distIntegers(L, R));
}
}
 
// This code is contributed by avijitmondal1998

Javascript




<script>
 
    // Function to count distinct sum of
    // pairs possible from the range [L, R]
    function distIntegers(L,R)
    {
        // Return the count of
        // distinct sum of pairs
        return 2 * R - 2 * L + 1;
    }
     
    // Driver Code
    let L = 3, R = 8;
    document.write(distIntegers(L, R));
     
    // This code is contributed by avanitrachhadiya2155
     
</script>
Output: 
11

 

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

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