Check if N and M can be made equal by increasing N by A and decreasing M by B

Given four numbers M, N, A and B, the task is to check whether M and N can be made equal to each other by doing any of the below operations:

  • M can be increased by A and N can be decreased by B
  • Leave both of them as it is.

Examples:

Input: M = 2, N = 8, A = 3, B = 3
Output: Yes
Explanation:
After first Operation:
M can be increased by A. Therefore, M = 2 + 3 = 5
N can be decreased by B. Therefore, N = 8 – 3 = 5
Finally, M = N = 5.

Input: M = 6, N = 4, A = 2, B = 1
Output: No

Approach: On careful observation, it can be observed that since we are increasing M and decreasing N, they can be made equal only when M is less than N. Therefore when M is less than N, there are two cases at each step:



  1. M can be increased by A and N can be decreased by B.
  2. Leave both of them as it is.

Another observation which can be made is that when M is increased and N is decreased, the absolute distance between M and N is reduced by the factor of A + B. For example:

Let M = 2, N = 14, A = 3 and B = 3.

  • In step 1, M = 5 and N = 11. The absolute distance between M and N got reduced by 6. That is, initially, the absolute distance was 12(14 – 2). After performing the given step, the absolute distance became 6(11 – 5).
  • In step 2, M = 8 and N = 8. The absolute distance between M and N again got reduced by 6 thereby making M and N equal.

From the above example, we can come to the conclusion that this problem can be solved in a constant time only by checking if the absolute distance between M and N is a multiple of (A + B) or not.

  • If it is a multiple, then M and N can be made equal.
  • Else, they cannot be made equal.

Below is the implementation of the above approach:

C++

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// C++ program to check if two numbers
// can be made equal by increasing
// the first by a and decreasing
// the second by b
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to whether the numbers
// can be made equal or not
bool checkEqualNo(int m, int n, int a, int b)
{
    if (m <= n) {
  
        // Check whether the numbers can reach
        // an equal point or not
        if ((n - m) % (a + b) == 0) {
            return true;
        }
        else {
            return false;
        }
    }
    else {
  
        // M and N cannot be made equal by
        // increasing M and decreasing N when
        // M is already greater than N
        return false;
    }
}
  
// Driver code
int main()
{
    int M = 2, N = 8;
    int A = 3, B = 3;
  
    if (checkEqualNo(M, N, A, B)) 
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
  
return 0;
}

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Java

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// Java program to check if two numbers
// can be made equal by increasing
// the first by a and decreasing
// the second by b
class GFG 
{
      
    // Function to whether the numbers
    // can be made equal or not
    static boolean checkEqualNo(int m, int n, int a, int b)
    {
        if (m <= n) {
      
            // Check whether the numbers can reach
            // an equal point or not
            if ((n - m) % (a + b) == 0) {
                return true;
            }
            else {
                return false;
            }
        }
        else {
      
            // M and N cannot be made equal by
            // increasing M and decreasing N when
            // M is already greater than N
            return false;
        }
    }
      
    // Driver code
    public static void main (String[] args)
    {
        int M = 2, N = 8;
        int A = 3, B = 3;
      
        if (checkEqualNo(M, N, A, B) == true
            System.out.println("Yes");
        else
            System.out.println("No");
      
    }
}
  
// This code is contributed by Yash_R

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Python3

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# Python3 program to check if two numbers
# can be made equal by increasing
# the first by a and decreasing
# the second by b
  
# Function to whether the numbers
# can be made equal or not
def checkEqualNo(m, n, a, b) :
    if (m <= n) :
  
        # Check whether the numbers can reach
        # an equal point or not
        if ((n - m) % (a + b) == 0) :
            return True;
      
        else :
            return False;
          
    else :
  
        # M and N cannot be made equal by
        # increasing M and decreasing N when
        # M is already greater than N
        return False;
      
# Driver code
if __name__ == "__main__" :
  
    M = 2; N = 8;
    A = 3; B = 3;
  
    if (checkEqualNo(M, N, A, B)) :
        print("Yes");
    else :
        print("No");
  
# This code is contributed by Yash_R

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C#

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// C# program to check if two numbers
// can be made equal by increasing
// the first by a and decreasing
// the second by b
using System;
  
class GFG 
{
      
    // Function to whether the numbers
    // can be made equal or not
    static bool checkEqualNo(int m, int n, int a, int b)
    {
        if (m <= n) {
      
            // Check whether the numbers can reach
            // an equal point or not
            if ((n - m) % (a + b) == 0) {
                return true;
            }
            else {
                return false;
            }
        }
        else {
      
            // M and N cannot be made equal by
            // increasing M and decreasing N when
            // M is already greater than N
            return false;
        }
    }
      
    // Driver code
    public static void Main (String[] args)
    {
        int M = 2;
        int N = 8;
        int A = 3;
        int B = 3;
      
        if (checkEqualNo(M, N, A, B) == true
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
  
// This code is contributed by Yash_R

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Output:

Yes

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