Split N into two integers whose addition to A and B makes them equal

Given three positive integers A, B and N, the task is to split N into two parts such that they become equal i.e. find two positive integers X and Y, such that X + Y = N and A + X = B + Y. If no such pair exists, print -1.

Examples:

Input: A = 1, B = 3, N = 4 
Output: 3 1 
Explanation: If X = 3 and Y = 1, then A + X = B + Y and X + Y =4

Input: A = 1, B = 3, N = 1 
Output: -1 
 

Naive Approach: 
The simplest approach to solve this problem is to generate all possible pairs with sum N and check for each pair, if A + X = B + Y.

Time complexity: O(N2
Auxiliary Space: O(1)

Efficient Approach: 
It can be observed that, since X + Y = N and A + X = B + Y, then X can be expressed as (N + B – A) / 2. Simply check if (N + B – A) / 2 is even or not. If it is even, calculate X and corresponding Y. Otherwise, print -1.



Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the
// splitted numbers
void findPair(int A, int B, int N)
{
    int X, Y;
 
    // Calculate X
    X = N - B + A;
 
    // If X is odd
    if (X % 2 != 0) {
 
        // No pair is possible
        cout << "-1";
    }
 
    // Otherwise
    else {
 
        // Calculate X
        X = X / 2;
 
        // Calculate Y
        Y = N - X;
        cout << X << " " << Y;
    }
}
 
// Driver Code
int main()
{
    int A = 1;
    int B = 3;
    int N = 4;
    findPair(A, B, N);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
     
// Function to calculate the
// splitted numbers
static void findPair(int A, int B, int N)
{
    int X, Y;
 
    // Calculate X
    X = N - B + A;
 
    // If X is odd
    if (X % 2 != 0)
    {
 
        // No pair is possible
        System.out.print("-1");
    }
 
    // Otherwise
    else
    {
         
        // Calculate X
        X = X / 2;
 
        // Calculate Y
        Y = N - X;
        System.out.print(X + " " + Y);
    }
}
 
//Driver function
public static void main (String[] args)
{
    int A = 1;
    int B = 3;
    int N = 4;
     
    findPair(A, B, N);
}
}
 
// This code is contributed by offbeat

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to implement
# the above approach
 
# Function to calculate the
# splitted numbers
def findPair(A, B, N):
 
    # Calculate X
    X = N - B + A
 
    # If X is odd
    if (X % 2 != 0):
 
        # No pair is possible
        print("-1")
     
    # Otherwise
    else :
 
        # Calculate X
        X = X // 2
 
        # Calculate Y
        Y = N - X
        print (X , Y)
 
# Driver Code
if __name__ == "__main__":
 
    A = 1
    B = 3
    N = 4
     
    findPair(A, B, N)
 
# This code is contributed by chitranayal

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to implement 
// the above approach 
using System;
   
class GFG{
       
// Function to calculate the 
// splitted numbers 
static void findPair(int A, int B, int N) 
    int X, Y; 
   
    // Calculate X 
    X = N - B + A; 
   
    // If X is odd 
    if (X % 2 != 0) 
    
         
        // No pair is possible 
        Console.Write("-1"); 
    
   
    // Otherwise 
    else
    
           
        // Calculate X 
        X = X / 2; 
   
        // Calculate Y 
        Y = N - X; 
        Console.Write(X + " " + Y); 
    
}
   
// Driver code
public static void Main(string[] args)
{
    int A = 1; 
    int B = 3; 
    int N = 4; 
       
    findPair(A, B, N); 
}
}
 
// This code is contributed by rutvik_56

chevron_right


Output: 

1 3

 

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Recommended Posts:


Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : chitranayal, offbeat, rutvik_56