Reordered power of 2

Hash Table Math Sorting Counting Enumeration

66.0% acceptance

Feb 22, 2026

2541

499

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.

Solution

Rust

Time O(2^n)

Space O(n)

LeetCode

solution.rs

/*
 * 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
 */

fn sorted_digits(mut n: i32) -> [u8; 10] {
  let mut cnt = [0u8; 10];
  while n > 0 { cnt[(n % 10) as usize] += 1; n /= 10; }
  cnt
}

impl Solution {
  pub fn reordered_power_of2(n: i32) -> bool {
    let target = sorted_digits(n);
    let mut p = 1i32;
    for _ in 0..31 {
      if sorted_digits(p) == target { return true; }
      p = p.saturating_mul(2);
    }
    false
  }
}