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Minimum number of coins having value equal to powers of 2 required to obtain N
  • Last Updated : 04 Jan, 2021
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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);
    }
}
Output:
4

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

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