Given values of A and B, find the minimum positive integer value of X that can be achieved in the equation X = P*A + P*B, Here P and Q can be zero or any positive or negative integer.
Examples:
Input: A = 3
B = 2
Output: 1
Input: A = 2
B = 4
Output: 2
Basically we need to find P and Q such that P*A > P*B and P*A – P*B is minimum positive integer. This problem can be easily solved by calculating GCD of both numbers.
For example:
For A = 2 And B = 4 Let P = 1 And Q = 0 X = P*A + Q*B = 1*2 + 0*4 = 2 + 0 = 2 (i. e GCD of 2 and 4) For A = 3 and B = 2 let P = -1 And Q = 2 X = P*A + Q*B = -1*3 + 2*2 = -3 + 4 = 1 ( i.e GCD of 2 and 3 )
Below is the implementation of above idea:
CPP
// CPP Program to find// minimum value of X// in equation X = P*A + Q*B#include <bits/stdc++.h>using namespace std;// Utility function to calculate GCDint gcd(int a, int b){ if (a == 0) return b; return gcd(b % a, a);}// Driver Codeint main(){ int a = 2; int b = 4; cout << gcd(a, b); return 0;} |
Java
// Java Program to find// minimum value of X// in equation X = P*A + Q*Bimport java.util.*;import java.lang.*;class GFG { // utility function to calculate gcd public static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Driver Program public static void main(String[] args) { int a = 2; int b = 4; System.out.println(gcd(a, b)); }} |
Python3
# Python3 Program to find# minimum value of X# in equation X = P * A + Q * B# Function to return gcd of a and bdef gcd(a, b): if a == 0: return b return gcd(b % a, a)a = 2b = 4print(gcd(a, b)) |
C#
// CSHARP Program to find// minimum value of X// in equation X = P*A + Q*Busing System;class GFG { // function to get gcd of a and b public static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Driver Code static public void Main() { int a = 2; int b = 4; Console.WriteLine(gcd(a, b)); }} |
PHP
// PHP Program to find // minimum value of X// in equation X = P*A + Q*B <?php// Function to return// gcd of a and bfunction gcd($a, $b){ if ($a == 0) return $b; return gcd($b % $a, $a);} // Driver Code$a = 2;$b = 4;echo gcd($a, $b);?> |
Javascript
<script>// javascript Program to find// minimum value of X// in equation X = P*A + Q*B // utility function to calculate gcd function gcd(a , b) { if (a == 0) return b; return gcd(b % a, a); } // Driver Program var a = 2; var b = 4; document.write(gcd(a, b));// This code contributed by umadevi9616</script> |
Output
2
Time Complexity: O(log(min(a, b)))
Auxiliary Space: O(log(min(a, b)))
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
