Minimum number operations required to convert n to m | Set-2
Given two integers n and m and a and b, in a single operation n can be multiplied by either a or b. The task is to convert n to m with a minimum number of given operations. If it is impossible to convert n to m with the given operation then print -1.
Examples:
Input: n = 120, m = 51840, a = 2, b = 3
Output: 7
120 * 2 * 2 * 2 * 2 * 3 * 3 * 3 = 51840
Input: n = 10, m = 50, a = 5, b = 7
Output: 1
10 * 5 = 50
In the previous post, we discussed an approach using division.
In this post, we will use an approach that finds the minimum number of operations using recursion. The recursion will consist of two states, the number being multiplied by a or by b, and counting the steps. The minimum of both steps will be the answer.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
#define MAXN 10000000
int minimumSteps( int n, int m, int a, int b)
{
if (n > m)
return MAXN;
if (n == m)
return 0;
return min(1 + minimumSteps(n * a, m, a, b),
1 + minimumSteps(n * b, m, a, b));
}
int main()
{
int n = 120, m = 51840;
int a = 2, b = 3;
cout << minimumSteps(n, m, a, b);
return 0;
}
|
Java
class GFG
{
static int MAXN = 10000000 ;
static int minimumSteps( int n, int m, int a, int b)
{
if (n > m)
return MAXN;
if (n == m)
return 0 ;
return Math.min( 1 + minimumSteps(n * a, m, a, b),
1 + minimumSteps(n * b, m, a, b));
}
public static void main (String[] args)
{
int n = 120 , m = 51840 ;
int a = 2 , b = 3 ;
System.out.println(minimumSteps(n, m, a, b));
}
}
|
Python3
MAXN = 10000000
def minimumSteps(n, m, a, b):
if (n > m):
return MAXN
if (n = = m):
return 0
return min ( 1 + minimumSteps(n * a, m, a, b),
1 + minimumSteps(n * b, m, a, b))
if __name__ = = '__main__' :
n = 120
m = 51840
a = 2
b = 3
print (minimumSteps(n, m, a, b))
|
C#
using System;
class GFG
{
static int MAXN = 10000000;
static int minimumSteps( int n, int m, int a, int b)
{
if (n > m)
return MAXN;
if (n == m)
return 0;
return Math.Min(1 + minimumSteps(n * a, m, a, b),
1 + minimumSteps(n * b, m, a, b));
}
public static void Main ()
{
int n = 120, m = 51840;
int a = 2, b = 3;
Console.WriteLine(minimumSteps(n, m, a, b));
}
}
|
PHP
<?php
$MAXN = 10000000;
function minimumSteps( $n , $m , $a , $b )
{
global $MAXN ;
if ( $n > $m )
return $MAXN ;
if ( $n == $m )
return 0;
return min(1 + minimumSteps( $n * $a , $m , $a , $b ),
1 + minimumSteps( $n * $b , $m , $a , $b ));
}
$n = 120; $m = 51840;
$a = 2; $b = 3;
echo minimumSteps( $n , $m , $a , $b );
?>
|
Javascript
<script>
var MAXN = 10000000;
function minimumSteps(n , m , a , b) {
if (n > m)
return MAXN;
if (n == m)
return 0;
return Math.min(1 + minimumSteps(n * a, m, a, b),
1 + minimumSteps(n * b, m, a, b));
}
var n = 120, m = 51840;
var a = 2, b = 3;
document.write(minimumSteps(n, m, a, b));
</script>
|
Time Complexity: O(m2)
Auxiliary Space: O(m2 + MAXN)
Last Updated :
09 Jun, 2022
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