# 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 **2 ^{i}** required to make a change for

**N**cents.

**Examples:**

Input:N = 5Output:2Explanation:

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 = 4Output: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(N^{2})**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 **2**s. 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(log_{2}(N))**Auxiliary Space:** O(1)

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