Minimum operations required to make two numbers equal
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++
#include <bits/stdc++.h>
using namespace std;
long long int minOperations(
long long int A,
long long int B)
{
if (A > B)
swap(A, B);
B = B / __gcd(A, B);
return B - 1;
}
int main()
{
long long int A = 7, B = 15;
cout << minOperations(A, B)
<< endl;
return 0;
}
|
Java
class GFG{
static int minOperations(
int A,
int B)
{
if (A > B) {
A = A+B;
B = A-B;
A = A-B;
}
B = B / __gcd(A, B);
return B - 1 ;
}
static int __gcd( int a, int b)
{
return b == 0 ? a:__gcd(b, a % b);
}
public static void main(String[] args)
{
int A = 7 , B = 15 ;
System.out.print(minOperations(A, B)
+ "\n" );
}
}
|
Python3
import math
def minOperations(A, B):
if (A > B):
swap(A, B)
B = B / / math.gcd(A, B);
return B - 1
A = 7
B = 15
print (minOperations(A, B))
|
C#
using System;
class GFG{
static int minOperations(
int A,
int B)
{
if (A > B) {
A = A+B;
B = A-B;
A = A-B;
}
B = B / __gcd(A, B);
return B - 1;
}
static int __gcd( int a, int b)
{
return b == 0? a:__gcd(b, a % b);
}
public static void Main(String[] args)
{
int A = 7, B = 15;
Console.Write(minOperations(A, B)
+ "\n" );
}
}
|
Javascript
<script>
function minOperations(A, B)
{
if (A > B) {
A = A + B;
B = A - B;
A = A - B;
}
B = B / __gcd(A, B);
return B - 1;
}
function __gcd(a , b) {
return b == 0 ? a : __gcd(b, a % b);
}
var A = 7, B = 15;
document.write(minOperations(A, B) + "\n" );
</script>
|
Time Complexity: O(log(max(A, B))
Auxiliary Space: O(log(max(A, B))
Last Updated :
24 Nov, 2021
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