Minimum number of coins having value equal to powers of 2 required to obtain N

Given an integer N, the task is to find the minimum number of coins of the form 2ⁱ required to make a change for N cents.

Examples:

Input: N = 5 
Output: 2 
Explanation:
Possible values of coins are: {1, 2, 4, 8, …} 
Possible ways to make change for N cents are as follows: 
5 = 1 + 1 + 1 + 1 + 1 
5 = 1 + 2 + 2 
5 = 1 + 4 (Minimum) 
Therefore, the required output is 2

Input: N = 4 
Output: 4

Naive Approach: The simplest approach to solve this problem is to store all possible values of the coins in an array and print the minimum count of coins required to make a change for N cents using Dynamic programming. 
Time Complexity: O(N²)
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimized using the fact that any number can be represented in the form of a power of 2s. The idea is to count the set bits of N and print the count obtained. Follow the steps below to solve the problem:

• Iterate over the bits in the binary representation of N and check if the current bit is set or not. If found to be true, then increment the count.
• Finally, print the total count obtained.

Below is the implementation of the above approach:

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count of set bit in N
void count_setbit(int N)
{

    // Stores count of set bit in N
    int result = 0;

    // Iterate over the range [0, 31]
    for (int i = 0; i < 32; i++) {

        // If current bit is set
        if ((1 << i) & N) {

            // Update result
            result++;
        }
    }
    cout << result << endl;
}

// Driver Code
int main()
{
    int N = 43;

    count_setbit(N);
    return 0;
}

C

// C program for above approach
#include <stdio.h>

// Function to count of set bit in N
void count_setbit(int N)
{

    // Stores count of set bit in N
    int result = 0;

    // Iterate over the range [0, 31]
    for (int i = 0; i < 32; i++) {

        // If current bit is set
        if ((1 << i) & N) {

            // Update result
            result++;
        }
    }
    printf("%d\n", result);
}

// Driver Code
int main()
{
    int N = 43;

    count_setbit(N);
    return 0;
}

Java

// Java program for above approach
public class Main {

    // Function to count of set bit in N
    public static void count_setbit(int N)
    {
        // Stores count of set bit in N
        int result = 0;

        // Iterate over the range [0, 31]
        for (int i = 0; i < 32; i++) {

            // If current bit is set
            if (((1 << i) & N) > 0) {

                // Update result
                result++;
            }
        }

        System.out.println(result);
    }

    // Driver Code
    public static void main(String[] args)
    {

        int N = 43;
        count_setbit(N);
    }
}

Python

# Python program for above approach

# Function to count of set bit in N
def count_setbit(N):

    # Stores count of set bit in N
    result = 0

    # Iterate over the range [0, 31]
    for i in range(32):

       # If current bit is set
       if( (1 << i) & N ):

           # Update result
           result = result + 1

    print(result)

if __name__ == '__main__':

    N = 43
    count_setbit(N)

C#

// C# program for above approach
using System;
class GFG {

    // Function to count of setbit in N
    static void count_setbit(int N)
    {

        // Stores count of setbit in N
        int result = 0;

        // Iterate over the range [0, 31]
        for (int i = 0; i < 32; i++) {

            // If current bit is set
            if (((1 << i) & N) > 0) {

                // Update result
                result++;
            }
        }

        Console.WriteLine(result);
    }

    // Driver Code
    static void Main()
    {

        int N = 43;
        count_setbit(N);
    }
}

Javascript

<script>
// Javascript program to implement
// the above approach

    // Function to count of set bit in N
    function count_setbit(N)
    {
        // Stores count of set bit in N
        let result = 0;

        // Iterate over the range [0, 31]
        for (let i = 0; i < 32; i++) {

            // If current bit is set
            if (((1 << i) & N) > 0) {

                // Update result
                result++;
            }
        }
        document.write(result);
    }

// Driver Code
        let N = 43;
        count_setbit(N);

    // This code is contributed by souravghosh0416.
</script>

Output:

4

Time Complexity: O(log₂(N))
Auxiliary Space: O(1)