869. Reordered Power of 2 LeetCode Solution

In this guide, you will get 869. Reordered Power of 2 LeetCode Solution with the best time and space complexity. The solution to Reordered Power of problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Reordered Power of solution in C++
Reordered Power of solution in Java
Reordered Power of solution in Python
Additional Resources

869. Reordered Power of 2 LeetCode Solution image

Problem Statement of Reordered Power of

You are given an integer n. We reorder the digits in any order (including the original order) such that the leading digit is not zero.
Return true if and only if we can do this so that the resulting number is a power of two.

Example 1:

Input: n = 1
Output: true

Example 2:

Input: n = 10
Output: false

Constraints:

1 <= n <= 109

Complexity Analysis

Time Complexity:
Space Complexity:

869. Reordered Power of 2 LeetCode Solution in C++

class Solution {
 public:
  bool reorderedPowerOf2(int n) {
    int count = counter(n);

    for (int i = 0; i < 30; ++i)
      if (counter(1 << i) == count)
        return true;

    return false;
  }

 private:
  int counter(int n) {
    int count = 0;

    for (; n > 0; n /= 10)
      count += pow(10, n % 10);

    return count;
  }
};
/* code provided by PROGIEZ */

869. Reordered Power of 2 LeetCode Solution in Java

class Solution {
  public boolean reorderedPowerOf2(int n) {
    int count = counter(n);

    for (int i = 0; i < 30; ++i)
      if (counter(1 << i) == count)
        return true;

    return false;
  }

  private int counter(int n) {
    int count = 0;

    for (; n > 0; n /= 10)
      count += Math.pow(10, n % 10);

    return count;
  }
}
// code provided by PROGIEZ

869. Reordered Power of 2 LeetCode Solution in Python

class Solution:
  def reorderedPowerOf2(self, n: int) -> bool:
    count = collections.Counter(str(n))
    return any(Counter(str(1 << i)) == count for i in range(30))
 # code by PROGIEZ