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Minimize operations to reduce A or B to 0 by reducing A from B or B from A

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  • Last Updated : 22 Jun, 2022
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Given two numbers A and B, the task is to find the minimum number of operations required to reduce either A or B to 0, wherein each operation A can be reduced by B if A is greater than equal to B, or vice versa.

Examples:

Input: A = 5, B = 4
Output: 5
Explanation:
Reduce B from A: A=1, B=4
Reduce A from B: A=1, B=3
Reduce A from B: A=1, B=2
Reduce A from B: A=1, B=1
Reduce B from A: A=0, B=1

Input: A=1, B=1
Output: 1
Explanation:
Reduce A from B: A=0, B=0

 

Approach: The approach to solving this problem is simply to check for bigger numbers and reduce the small number from it.

  • Repeat following operations till at least one of the two numbers become 0
    • If A is greater than equal to B, reduce B from A
    • If A is smaller than A, reduce A from B
    • For each loop iteration, store the count.
  • Return the count of loop iterations at the end.

Below is the implementation of the above approach: 

C++




// C++ program to find Minimum number
// Of operations required to reduce
// Either A or B to Zero
 
#include <iostream>
using namespace std;
 
int countOperations(int num1, int num2)
{
    int cnt = 0;
    while (num1 > 0 && num2 > 0) {
        if (num1 >= num2)
            num1 -= num2;
        else
            num2 -= num1;
        cnt++;
    }
    return cnt;
}
 
// Driver Code
int main()
{
 
    int A = 5, B = 4;
    cout << countOperations(A, B);
    return 0;
}


Java




// Java program to find Minimum number
import java.io.*;
 
class GFG {
 
  // Of operations required to reduce
  // Either A or B to Zero
 
  static int countOperations(int num1, int num2)
  {
    int cnt = 0;
    while (num1 > 0 && num2 > 0) {
      if (num1 >= num2)
        num1 -= num2;
      else
        num2 -= num1;
      cnt++;
    }
    return cnt;
  }
 
  // Driver Code
  public static void main (String[] args) {
    int A = 5, B = 4;
    System.out.println(countOperations(A, B));
  }
}
 
// This code is contributed by hrithikgarg03188.


Python3




# Python code for the above approach
def countOperations(num1, num2):
    cnt = 0
    while (num1 > 0 and num2 > 0):
        if (num1 >= num2):
            num1 -= num2
        else:
            num2 -= num1
        cnt += 1
 
    return cnt
 
# Driver Code
A,B = 5,4
print(countOperations(A, B))
 
# This code is contributed by shinjanpatra


C#




// C# program to find Minimum number
// Of operations required to reduce
// Either A or B to Zero
using System;
class GFG {
 
  static int countOperations(int num1, int num2)
  {
    int cnt = 0;
    while (num1 > 0 && num2 > 0) {
      if (num1 >= num2)
        num1 -= num2;
      else
        num2 -= num1;
      cnt++;
    }
    return cnt;
  }
 
  // Driver Code
  public static void Main()
  {
 
    int A = 5, B = 4;
    Console.Write(countOperations(A, B));
  }
}
 
// This code is contributed by Samim Hossain Mondal.


Javascript




<script>
       // JavaScript code for the above approach
       function countOperations(num1, num2) {
           let cnt = 0;
           while (num1 > 0 && num2 > 0) {
               if (num1 >= num2)
                   num1 -= num2;
               else
                   num2 -= num1;
               cnt++;
           }
           return cnt;
       }
 
       // Driver Code
       let A = 5, B = 4;
       document.write(countOperations(A, B));
 
      // This code is contributed by Potta Lokesh
   </script>


 
 

Output

5

 

Time Complexity: 0(MAX(A, B)), where A and B are the two numbers given.
Auxiliary Space: 0(1)

 


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