Given a positive integer N, the task is to find the minimum number of addition operations required to convert the number 0 to N such that in each operation any number can be multiplied by 2 or add the value 1 to it.
Examples:
Input: N = 6
Output: 1
Explanation:
Following are the operations performed to convert 0 to 6:
Add 1 –> 0 + 1 = 1.
Multiply 2 –> 1 * 2 = 2.
Add 1 –> 2 + 1 = 3.
Multiply 2 –> 3 * 2 = 6.
Therefore number of addition operations = 2.Input: N = 3
Output: 2
Approach: This problem can be solved by using the Bit Manipulation technique. In binary number representation of N, while operating each bit whenever N becomes odd (that means the least significant bit of N is set) then perform the addition operation. Otherwise, multiply by 2. The final logic to the given problem is to find the number of set bits in N.
Below is the implementation of the above approach:
C++
// C++ program for above approach#include <bits/stdc++.h>using namespace std;// Function to count number of// set bits in Nint minimumAdditionOperation( unsigned long long int N){ // Stores the count of set bits int count = 0; while (N) { // If N is odd, then it // a set bit if (N & 1 == 1) { count++; } N = N >> 1; } // Return the result return count;}// Driver Codeint main(){ int N = 6; cout << minimumAdditionOperation(N); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG { // Function to count number of // set bits in N static int minimumAdditionOperation(int N) { // Stores the count of set bits int count = 0; while (N > 0) { // If N is odd, then it // a set bit if (N % 2 == 1) { count++; } N = N >> 1; } // Return the result return count; } // Driver Code public static void main(String[] args) { int N = 6; System.out.println(minimumAdditionOperation(N)); }}// This code is contributed by dwivediyash |
Python3
# python program for above approach# Function to count number of# set bits in Ndef minimumAdditionOperation(N): # Stores the count of set bits count = 0 while (N): # If N is odd, then it # a set bit if (N & 1 == 1): count += 1 N = N >> 1 # Return the result return count# Driver Codeif __name__ == "__main__": N = 6 print(minimumAdditionOperation(N)) # This code is contributed by rakeshsahni. |
C#
// C# program for above approachusing System;public class GFG{ // Function to count number of // set bits in N static int minimumAdditionOperation(int N) { // Stores the count of set bits int count = 0; while (N != 0) { // If N is odd, then it // a set bit if ((N & 1) == 1) { count++; } N = N >> 1; } // Return the result return count; } // Driver Code static public void Main (){ int N = 6; Console.Write(minimumAdditionOperation(N)); }}// This code is contributed by AnkThon |
Javascript
<script>// Javascript program for above approach// Function to count number of// set bits in Nfunction minimumAdditionOperation(N){ // Stores the count of set bits let count = 0; while (N) { // If N is odd, then it // a set bit if (N & (1 == 1)) { count++; } N = N >> 1; } // Return the result return count;}// Driver Codelet N = 6;document.write(minimumAdditionOperation(N));// This code is contributed by saurabh_jaiswal.</script> |
2
Time Complexity: O(log N)
Auxiliary Space: O(1)
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