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Find all powers of 2 less than or equal to a given number
• Difficulty Level : Easy
• Last Updated : 23 Jun, 2020

Given a positive number N, the task is to find out all the perfect powers of two which are less than or equal to the given number N.

Examples:

Input: N = 63
Output: 32 16 8 4 2 1
Explanation:
There are total of 6 powers of 2, which are less than or equal to the given number N.

Input: N = 193
Output: 128 64 32 16 8 4 2 1
Explaination:
There are total of 8 powers of 2, which are less than or equal to the given number N.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: The idea is to traverse each number from N to 1 and check if it is a perfect power of 2 or not. If yes, then print that number.

Another Approach: The idea is to find all powers of 2 and simply print the powers that are lesser than or equal to N.

Another Approach: The idea is based on the concept that all powers of 2 has all bits set, in its binary form. Bitset function is used in this approach solve the above problem. Below are the steps:

1. Find the largest power of 2(say temp) which is used to evaluate the number less than or equal to N.
2. Initialise an bitset array arr[] of maximum size 64, to store the binary representation of the given number N.
3. Reset all the bits in the bitset array using reset() function.
4. Iterate a loop from total to 0, and sequentially make each bit 1, and find the value of that binary expression and then reset the bit.

Below is the implementation of the above approach:

 `// C++ program for the above approach`` ` `#include ``using` `namespace` `std;``const` `int` `MAX = 64;`` ` `// Function to return max exponent of``// 2 which evaluates a number less``// than or equal to N``int` `max_exponent(``int` `n)``{``    ``return` `(``int``)(log2(n));``}`` ` `// Function to print all the powers``// of 2 less than or equal to N``void` `all_powers(``int` `N)``{``    ``bitset<64> arr(N);`` ` `    ``// Reset all the bits``    ``arr.reset();`` ` `    ``int` `total = max_exponent(N);`` ` `    ``// Iterate from total to 0``    ``for` `(``int` `i = total; i >= 0; i--) {`` ` `        ``// Reset the next bit``        ``arr.reset(i + 1);`` ` `        ``// Set the current bit``        ``arr.set(i);`` ` `        ``// Value of the binary expression``        ``cout << arr.to_ulong() << ``" "``;``    ``}``}`` ` `// Driver Code``int` `main()``{``    ``// Given Number``    ``int` `N = 63;`` ` `    ``// Function Call``    ``all_powers(N);``    ``return` `0;``}`
Output:
```32 16 8 4 2 1
```

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

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