Given two integers A and B. the task is to find the minimum number of operations required to make A and B equal. In each operation, either of the below steps can be performed:
- Increment either A or B with its initial value.
- Increment both A and B with their initial value
Examples:
Input: A = 4, B = 10
Output: 4
Explanation:
Initially A = 4, B = 10
Operation 1: Increment A only: A = A + 4 = 8
Operation 2: Increment A only: A = A + 4 = 12
Operation 3: Increment A only: A = A + 4 = 16
Operation 4: Increment A and B: A = A + 4 = 20 and B = B + 10 = 20
They are equal now.Input: A = 7, B = 23
Output: 22
Explanation:
Initially A = 7, B = 23
Operation 1 – 7: Increment A and B: A = 56 and B = 161
Operation 8 – 22: Increment A: A = 161 and B = 161
They are equal now.
Approach: This problem can be solved using GCD.
- If A is greater than B, then swap A and B.
- Now reduce B, such that gcd of A and B becomes 1.
- Hence the minimum operations required to reach equal value is (B – 1).
For example: If A = 4, B = 10:
- Step 1: Compare 4 and 10, as we always need B as the greater value. Here already B is greater than A. So, now no swap is required.
- Step 2: GCD(4, 10) = 2. So, we reduce B to B/2. Now A = 4 and B = 5.
GCD(4, 5) = 1, which was the target. - Step 3: (Current value of B – 1) will be the required count. Here, Current B = 5. So (5 – 1 = 4), i.e. total 4 operations are required.
Below is the implementation of the above approach.
C++
// C++ program to find minimum// operations required to// make two numbers equal #include <bits/stdc++.h>using namespace std; // Function to return the// minimum operations requiredlong long int minOperations( long long int A, long long int B){ // Keeping B always greater if (A > B) swap(A, B); // Reduce B such that // gcd(A, B) becomes 1. B = B / __gcd(A, B); return B - 1;} // Driver codeint main(){ long long int A = 7, B = 15; cout << minOperations(A, B) << endl; return 0;} |
Java
// Java program to find minimum// operations required to// make two numbers equalclass GFG{ // Function to return the// minimum operations requiredstatic int minOperations( int A, int B){ // Keeping B always greater if (A > B) { A = A+B; B = A-B; A = A-B; } // Reduce B such that // gcd(A, B) becomes 1. B = B / __gcd(A, B); return B - 1;}static int __gcd(int a, int b) { return b == 0? a:__gcd(b, a % b); } // Driver codepublic static void main(String[] args){ int A = 7, B = 15; System.out.print(minOperations(A, B) +"\n"); }} // This code contributed by sapnasingh4991 |
Python3
# Python program to find minimum # operations required to # make two numbers equal import math # Function to return the # minimum operations required def minOperations(A, B): # Keeping B always greater if (A > B): swap(A, B) # Reduce B such that # gcd(A, B) becomes 1. B = B // math.gcd(A, B); return B - 1 # Driver code A = 7B = 15 print(minOperations(A, B)) # This code is contributed by Sanjit_Prasad |
C#
// C# program to find minimum// operations required to// make two numbers equalusing System; class GFG{ // Function to return the// minimum operations requiredstatic int minOperations( int A, int B){ // Keeping B always greater if (A > B) { A = A+B; B = A-B; A = A-B; } // Reduce B such that // gcd(A, B) becomes 1. B = B / __gcd(A, B); return B - 1;}static int __gcd(int a, int b) { return b == 0? a:__gcd(b, a % b); } // Driver codepublic static void Main(String[] args){ int A = 7, B = 15; Console.Write(minOperations(A, B) +"\n");}} // This code is contributed by sapnasingh4991 |
14
Time Complexity: O(log(max(A, B))
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