Given two integers N and M and the task is to convert N to M with the following operations:
- Multiply N by 2 i.e. N = N * 2.
- Subtract 1 from N i.e. N = N – 1.
Examples:
Input: N = 4, M = 6
Output: 2
Perform operation 2: N = N – 1 = 4 – 1 = 3
Perform operation 1: N = N * 2 = 3 * 2 = 6Input: N = 10, M = 1
Output: 9
Approach: Create an array dp[] of size MAX = 105 + 5 to store the answer in order to prevent the same computation again and again and initialize all the array elements with -1.
- If N ? 0 or N ? MAX means it can not be converted to M so return MAX.
- If N = M then return 0 as N got converted to M.
- Else find the value at dp[N] if it is not -1, it means it has been calculated earlier so return dp[N].
- If it is -1 then will call the recursive function as 2 * N and N – 1 and return the minimum because if N is odd then it can be reached only by performing N – 1 operation and if N is even then 2 * N operations have to be performed so check both the possibilities and return the minimum.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 5;int n, m;int dp[N];// Function to return the minimum// number of given operations// required to convert n to mint minOperations(int k){ // If k is either 0 or out of range // then return max if (k <= 0 || k >= 2e4) { return 1e9; } // If k = m then conversion is // complete so return 0 if (k == m) { return 0; } int& ans = dp[k]; // If it has been calculated earlier if (ans != -1) { return ans; } ans = 1e9; // Call for 2*k and k-1 and return // the minimum of them. If k is even // then it can be reached by 2*k operations // and If k is odd then it can be reached // by k-1 operations so try both cases // and return the minimum of them ans = 1 + min(minOperations(2 * k), minOperations(k - 1)); return ans;}// Driver codeint main(){ n = 4, m = 6; memset(dp, -1, sizeof(dp)); cout << minOperations(n); return 0;} |
Java
// Java implementation of the approach import java.util.*;class GFG{ static final int N = 10000; static int n, m; static int[] dp = new int[N]; // Function to return the minimum // number of given operations // required to convert n to m static int minOperations(int k) { // If k is either 0 or out of range // then return max if (k <= 0 || k >= 10000) return 1000000000; // If k = m then conversion is // complete so return 0 if (k == m) return 0; dp[k] = dp[k]; // If it has been calculated earlier if (dp[k] != -1) return dp[k]; dp[k] = 1000000000; // Call for 2*k and k-1 and return // the minimum of them. If k is even // then it can be reached by 2*k operations // and If k is odd then it can be reached // by k-1 operations so try both cases // and return the minimum of them dp[k] = 1 + Math.min(minOperations(2 * k), minOperations(k - 1)); return dp[k]; } // Driver Code public static void main(String[] args) { n = 4; m = 6; Arrays.fill(dp, -1); System.out.println(minOperations(n)); }}// This code is contributed by// sanjeev2552 |
Python3
# Python3 implementation of the approachN = 1000dp = [-1] * N# Function to return the minimum# number of given operations# required to convert n to mdef minOperations(k): # If k is either 0 or out of range # then return max if (k <= 0 or k >= 1000): return 1e9 # If k = m then conversion is # complete so return 0 if (k == m): return 0 dp[k] = dp[k] # If it has been calculated earlier if (dp[k] != -1): return dp[k] dp[k] = 1e9 # Call for 2*k and k-1 and return # the minimum of them. If k is even # then it can be reached by 2*k operations # and If k is odd then it can be reached # by k-1 operations so try both cases # and return the minimum of them dp[k] = 1 + min(minOperations(2 * k), minOperations(k - 1)) return dp[k]# Driver codeif __name__ == '__main__': n = 4 m = 6 print(minOperations(n)) # This code is contributed by ashutosh450 |
C#
// C# implementation of the approach using System;using System.Linq;class GFG{ static int N = 10000; static int n, m; static int[] dp = Enumerable.Repeat(-1, N).ToArray(); // Function to return the minimum // number of given operations // required to convert n to m static int minOperations(int k) { // If k is either 0 or out of range // then return max if (k <= 0 || k >= 10000) return 1000000000; // If k = m then conversion is // complete so return 0 if (k == m) return 0; dp[k] = dp[k]; // If it has been calculated earlier if (dp[k] != -1) return dp[k]; dp[k] = 1000000000; // Call for 2*k and k-1 and return // the minimum of them. If k is even // then it can be reached by 2*k operations // and If k is odd then it can be reached // by k-1 operations so try both cases // and return the minimum of them dp[k] = 1 + Math.Min(minOperations(2 * k), minOperations(k - 1)); return dp[k]; } // Driver Code public static void Main(String[] args) { n = 4; m = 6; //Arrays.fill(dp, -1); Console.Write(minOperations(n)); }}// This code is contributed by// Mohit kumar 29 |
Javascript
<script> let N = 10000; let n, m; let dp = new Array(N); function minOperations(k) { // If k is either 0 or out of range // then return max if (k <= 0 || k >= 10000) return 1000000000; // If k = m then conversion is // complete so return 0 if (k == m) return 0; dp[k] = dp[k]; // If it has been calculated earlier if (dp[k] != -1) return dp[k]; dp[k] = 1000000000; // Call for 2*k and k-1 and return // the minimum of them. If k is even // then it can be reached by 2*k operations // and If k is odd then it can be reached // by k-1 operations so try both cases // and return the minimum of them dp[k] = 1 + Math.min(minOperations(2 * k), minOperations(k - 1)); return dp[k]; } // Driver Code n = 4; m = 6; for(let i = 0; i < dp.length; i++) { dp[i] = -1; } document.write(minOperations(n));// This code is contributed by unknown2108</script> |
Output:
2
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