Given two integers N and M. The task is to find the minimum number of steps to reach M from N by performing given operations.
- Multiply a number x by 2. So, x becomes 2*x.
- Subtract one from the number x. So, x becomes x-1.
Examples:
Input : N = 4, M = 6 Output : 2 Explanation : Perform operation number 2 on N. So, N becomes 3 and then perform operation number 1. Then, N becomes 6. So, the minimum number of steps is 2. Input : N = 10, M = 1 Output : 9 Explanation : Perform operation number two 9 times on N. Then N becomes 1.
Approach :
The idea is to reverse the problem as follows: We should get the number N starting from M using the operations:
- Divide the number by 2 if it is even.
- Add 1 to the number.
Now, the minimum number of operations would be:
- If N > M, return the difference between them, that is, number of steps will be adding 1 to M until it becomes equal to N.
- Else if N < M.
- Keep dividing M by 2 until it becomes less than N. If M is odd, add 1 to it first and then divide by 2. Once M is less than N, add the difference between them to the count along with the count of above operations.
Below is the implementation of the above approach:
C++
// CPP program to find minimum number // of steps to reach M from N #include <bits/stdc++.h> using namespace std;
// Function to find a minimum number // of steps to reach M from N int Minsteps( int n, int m)
{ int ans = 0;
// Continue till m is greater than n
while (m > n)
{
// If m is odd
if (m&1)
{
// add one
m++;
ans++;
}
// divide m by 2
m /= 2;
ans++;
}
// Return the required answer
return ans + n - m;
} // Driver code int main()
{ int n = 4, m = 6;
cout << Minsteps(n, m);
return 0;
} |
Java
// Java program to find minimum number // of steps to reach M from N class CFG
{ // Function to find a minimum number // of steps to reach M from N static int Minsteps( int n, int m)
{ int ans = 0 ;
// Continue till m is greater than n
while (m > n)
{
// If m is odd
if (m % 2 != 0 )
{
// add one
m++;
ans++;
}
// divide m by 2
m /= 2 ;
ans++;
}
// Return the required answer
return ans + n - m;
} // Driver code public static void main(String[] args)
{ int n = 4 , m = 6 ;
System.out.println(Minsteps(n, m));
} } // This code is contributed by Code_Mech |
Python3
# Python3 program to find minimum number # of steps to reach M from N # Function to find a minimum number # of steps to reach M from N def Minsteps(n, m):
ans = 0
# Continue till m is greater than n
while (m > n):
# If m is odd
if (m & 1 ):
# add one
m + = 1
ans + = 1
# divide m by 2
m / / = 2
ans + = 1
# Return the required answer
return ans + n - m
# Driver code n = 4
m = 6
print (Minsteps(n, m))
# This code is contributed by mohit kumar |
C#
// C# program to find minimum number // of steps to reach M from N using System;
class GFG
{ // Function to find a minimum number // of steps to reach M from N static int Minsteps( int n, int m)
{ int ans = 0;
// Continue till m is greater than n
while (m > n)
{
// If m is odd
if (m % 2 != 0)
{
// add one
m++;
ans++;
}
// divide m by 2
m /= 2;
ans++;
}
// Return the required answer
return ans + n - m;
} // Driver code public static void Main()
{ int n = 4, m = 6;
Console.WriteLine(Minsteps(n, m));
} } // This code is contributed // by Akanksha Rai |
PHP
<?php // PHP program to find minimum number // of steps to reach M from N // Function to find a minimum number // of steps to reach M from N function Minsteps( $n , $m )
{ $ans = 0;
// Continue till m is greater than n
while ( $m > $n )
{
// If m is odd
if ( $m % 2 != 0)
{
// add one
$m ++;
$ans ++;
}
// divide m by 2
$m /= 2;
$ans ++;
}
// Return the required answer
return $ans + $n - $m ;
} // Driver code $n = 4; $m = 6;
echo (Minsteps( $n , $m ));
// This code is contributed by Code_Mech ?> |
Javascript
<script> // JavaScript program to find minimum number // of steps to reach M from N // Function to find a minimum number // of steps to reach M from N function Minsteps(n, m)
{ let ans = 0;
// Continue till m is greater than n
while (m > n)
{
// If m is odd
if (m&1)
{
// add one
m++;
ans++;
}
// divide m by 2
m = Math.floor(m / 2);
ans++;
}
// Return the required answer
return ans + n - m;
} // Driver code let n = 4, m = 6;
document.write(Minsteps(n, m));
// This code is contributed by Surbhi Tyagi. </script> |
Output:
2
Time Complexity: O(log2m)
Auxiliary Space: O(1)