# Convert 0 to N by adding 1 or multiplying by 2 in minimum steps

• Difficulty Level : Easy
• Last Updated : 03 Dec, 2021

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 ``using` `namespace` `std;` `// Function to count number of``// set bits in N``int` `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 Code``int` `main()``{``    ``int` `N = 6;``    ``cout << minimumAdditionOperation(N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `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 N``def` `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 Code``if` `__name__ ``=``=` `"__main__"``:` `    ``N ``=` `6``    ``print``(minimumAdditionOperation(N))` `    ``# This code is contributed by rakeshsahni.`

## C#

 `// C# program for above approach``using` `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

 ``

Output:

`2`

Time Complexity: O(log N)
Auxiliary Space: O(1)

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