获得N所需的等于2的幂的最小硬币数量

给定整数N ，任务是找到要更改N美分所需的2 ⁱ形式的最小硬币数。

例子：

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

天真的方法：解决此问题的最简单方法是将硬币的所有可能值存储在数组中，并使用动态编程打印N分钱所需的最少硬币计数。
时间复杂度： O(N ² )
辅助空间： O(N)

高效方法：可以利用以下事实来优化上述方法：可以以2 s的幂的形式表示任何数字。想法是对N的设置位进行计数并打印获得的计数。请按照以下步骤解决问题：

• 遍历N的二进制表示形式中的位，并检查当前位是否已设置。如果发现为真，则增加计数。
• 最后，打印获得的总数。

下面是上述方法的实现：

C++

// C++ program for above approach
#include 
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 
  
// 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);
    }
}

输出：

4

时间复杂度： O(log ₂ (N))
辅助空间： O(1)