Convert a number m to n with minimum operations. The operations allowed are :
- Multiply by 2, i.e., do m = 2 * m
- Subtract 1, i.e., do m = m – 1
Print -1 if it is not possible to convert.
Examples :
Input : m = 3, n = 11 Output : 3 1st operation: *2 = 3*2 = 6 2nd operation: *2 = 6*2 = 12 3rd operation: -1 = 12-1 = 11 Input : m = 15, n = 20 Output : 6 1st operation: -1 '5' times = 15 + (-1*5) = 10 2nd operation: *2 = 10*2 = 20 Input : m = 0, n = 8 Output : -1 Using the given set of operations 'm' cannot be converted to 'n' Input : m = 12, n = 8 Output : 4
The idea is based on below facts.
1) If m is less than 0 and n is greater than 0, then not possible.
2) If m is greater than n, then we can reach n using subtractions only.
3) Else (m is less than n), we must do m*2 operations. Following two cases arise.
……a) If n is odd, we must do a -1 operation to reach it.
……b) If n is even, we must do a *2 operation to reach it.
Algorithm:
int convert(m,n)
if (m == n)
return 0;
// not possible
if (m <= 0 && n > 0)
return -1;
// m is greater than n
if (m > n)
return m-n;
// n is odd
if (n % 2 == 1)
// perform '-1'
return 1 + convert(m, n+1);
// n is even
else
// perform '*2'
return 1 + convert(m, n/2);Note: The list of operations so generated should be applied in reverse order.
For example:
m = 3, n = 11
convert(3,11)
| --> n is odd: operation '-1'
convert(3,12)
| --> n is even: operation '*2'
convert(3,6)
| --> n is even: operation '*2'
convert(3,3)
| --> m == n
return
Therefore, the sequence of operations is '*2', '*2', '-1'.
C++
// C++ implementation to convert// a number m to n using minimum// number of given operations#include <bits/stdc++.h>using namespace std;// Function to find minimum// number of given operations// to convert m to nint convert(int m, int n){ if (m == n) return 0; // only way is to do // -1 (m - n) times if (m > n) return m - n; // not possible if (m <= 0 && n > 0) return -1; // n is greater and n is odd if (n % 2 == 1) // perform '-1' on m // (or +1 on n) return 1 + convert(m, n + 1); // n is even else // perform '*2' on m // (or n/2 on n) return 1 + convert(m, n / 2);}// Driver codeint main(){ int m = 3, n = 11; cout << "Minimum number of operations : " << convert(m, n); return 0;} |
Java
// Java implementation to convert// a number m to n using minimum// number of given operationsclass ConvertNum{ // function to find minimum // number of given operations // to convert m to n static int convert(int m, int n) { if (m == n) return 0; // only way is to do // -1 (m - n) times if (m > n) return m - n; // not possible if (m <= 0 && n > 0) return -1; // n is greater and n is odd if (n % 2 == 1) // perform '-1' on m // (or +1 on n) return 1 + convert(m, n + 1); // n is even else // perform '*2' on m // (or n / 2 on n) return 1 + convert(m, n / 2); } // Driver Code public static void main (String[] args) { int m = 3, n = 11; System.out.println("Minimum number of " + "operations : " + convert(m, n)); }} |
Python3
# Python implementation to convert# a number m to n using minimum# number of given operations# Function to find minimum# number of given operations# to convert m to ndef conver(m, n): if(m == n): return 0 # only way is to do # -1(m - n): times if(m > n): return m - n # not possible if(m <= 0 and n > 0): return -1 # n is greater and n is odd if(n % 2 == 1): # perform '-1' on m #(or +1 on n): return 1 + conver(m, n + 1) # n is even else: # perform '*2' on m #(or n/2 on n): return 1 + conver(m, n / 2)# Driver codem = 3n = 11print("Minimum number of operations :", conver(m, n))# This code is contributed by# Sanjit_Prasad |
C#
// C# implementation to convert// a number m to n using minimum// number of given operationsusing System;class GFG{ // function to find minimum // number of given operations // to convert m to n static int convert(int m, int n) { if (m == n) return 0; // only way is to do // -1 (m - n) times if (m > n) return m - n; // not possible if (m <= 0 && n > 0) return -1; // n is greater and n is odd if (n % 2 == 1) // perform '-1' on m // (or +1 on n) return 1 + convert(m, n + 1); // n is even else // perform '*2' on m // (or n / 2 on n) return 1 + convert(m, n / 2); } // Driver code public static void Main () { int m = 3, n = 11; Console.Write("Minimum number of " + "operations : " + convert(m, n)); }}// This code is contributed by nitin mittal |
PHP
<?php// PHP implementation to convert// a number m to n using minimum// number of given operations// Function to find minimum// number of given operations// to convert m to nfunction convert($m, $n){ if ($m == $n) return 0; // only way is to do // -1 (m - n) times if ($m > $n) return $m - $n; // not possible if ($m <= 0 && $n > 0) return -1; // n is greater and n is odd if ($n % 2 == 1) // perform '-1' on m // (or +1 on n) return 1 + convert($m, $n + 1); // n is even else // perform '*2' on m // (or n/2 on n) return 1 + convert($m, $n / 2);}// Driver code{ $m = 3; $n = 11; echo "Minimum number of ". "operations : ", convert($m, $n); return 0;} // This code is contributed// by nitin mittal.?> |
Javascript
<script>// javascript implementation to convert// a number m to n using minimum// number of given operations// function to find minimum// number of given operations// to convert m to nfunction convert(m , n){ if (m == n) return 0; // only way is to do // -1 (m - n) times if (m > n) return m - n; // not possible if (m <= 0 && n > 0) return -1; // n is greater and n is odd if (n % 2 == 1) // perform '-1' on m // (or +1 on n) return 1 + convert(m, n + 1); // n is even else // perform '*2' on m // (or n / 2 on n) return 1 + convert(m, n / 2);}// Driver Codevar m = 3, n = 11;document.write("Minimum number of " + "operations : " + convert(m, n));// This code is contributed by Princi Singh</script> |
Output :
Minimum number of operations : 3
References :
http://tech.queryhome.com/112705/convert-number-with-minimum-operations-operations-allowed
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