869. Reordered Power of 2

Description

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

 

Solutions

Solution: Bit Manipulation

• Time complexity: O(1)
• Space complexity: O(1)

 

JavaScript

/**
 * @param {number} n
 * @return {boolean}
 */
const reorderedPowerOf2 = function (n) {
  const counter = num => {
    let result = 0;

    while (num) {
      result += Math.pow(10, num % 10);
      num = Math.floor(num / 10);
    }

    return result;
  };

  const count = counter(n);

  for (let index = 0; index < 30; index++) {
    const num = 1 << index;

    if (counter(num) === count) return true;
  }

  return false;
};