Given a positive integer K, the task is to find the minimum number of operations of the following two types, required to change 0 to K:
- Add one to the operand
- Multiply the operand by 2.
Examples:
Input: K = 1
Output: 1
Explanation:
Step 1: 0 + 1 = 1 = KInput: K = 4
Output: 3
Explanation:
Step 1: 0 + 1 = 1,
Step 2: 1 * 2 = 2,
Step 3: 2 * 2 = 4 = K
Approach:
- If K is an odd number, the last step must be adding 1 to it.
- If K is an even number, the last step is to multiply by 2 to minimise the number of steps.
- Create a dp[] table that stores in every dp[i], the minimum steps required to reach i.
![dp[i] =\begin{cases} dp[i - 1] + 1 & \text{; if i is odd} \\ dp[\frac{i}{2}] + 1 & \text{; if i is even} \end{cases}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-703f62962ea08ed468cae8d027b51118_l3.png "Rendered by QuickLaTeX.com")
Below is the implementation of the above approach:
C++
// C++ program to implement the above approach #include <bits/stdc++.h> using namespace std; // Function to find minimum operations int minOperation(int k) { // vector dp is initialised // to store the steps vector<int> dp(k + 1, 0); for (int i = 1; i <= k; i++) { dp[i] = dp[i - 1] + 1; // For all even numbers if (i % 2 == 0) { dp[i] = min(dp[i], dp[i / 2] + 1); } } return dp[k]; } // Driver Code int main() { int K = 12; cout << minOperation(k); } |
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Java
// Java program to implement the above approach class GFG{ // Function to find minimum operations static int minOperation(int k) { // dp is initialised // to store the steps int dp[] = new int[k + 1]; for(int i = 1; i <= k; i++) { dp[i] = dp[i - 1] + 1; // For all even numbers if (i % 2 == 0) { dp[i] = Math.min(dp[i], dp[i / 2] + 1); } } return dp[k]; } // Driver Code public static void main (String []args) { int K = 12; System.out.print( minOperation(K)); } } // This code is contributed by chitranayal |
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Python3
# Python3 program to implement the above approach # Function to find minimum operations def minOperation(k): # dp is initialised # to store the steps dp = [0] * (k + 1) for i in range(1, k + 1): dp[i] = dp[i - 1] + 1 # For all even numbers if (i % 2 == 0): dp[i]= min(dp[i], dp[i // 2] + 1) return dp[k] # Driver Code if __name__ == '__main__': k = 12 print(minOperation(k)) # This code is contributed by mohit kumar 29 |
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C#
// C# program to implement the above approach using System; class GFG{ // Function to find minimum operations static int minOperation(int k) { // dp is initialised // to store the steps int []dp = new int[k + 1]; for(int i = 1; i <= k; i++) { dp[i] = dp[i - 1] + 1; // For all even numbers if (i % 2 == 0) { dp[i] = Math.Min(dp[i], dp[i / 2] + 1); } } return dp[k]; } // Driver Code public static void Main() { int K = 12; Console.Write(minOperation(K)); } } // This code is contributed by Nidhi_Biet |
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Output:
5
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