Starting with a positive 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 in a way such that the resulting number is a power of 2.
Example 1:
Input: 1 Output: true
Example 2:
Input: 10 Output: false
Example 3:
Input: 16 Output: true
Example 4:
Input: 24 Output: false
Example 5:
Input: 46 Output: true
Note:
1 <= N <= 10^9struct Solution;
use std::collections::HashSet;
impl Solution {
fn reordered_power_of2(n: i32) -> bool {
let mut hs: HashSet<Vec<char>> = HashSet::new();
for i in 0..32 {
let mut v: Vec<char> = (1 << i).to_string().chars().collect();
v.sort_unstable();
hs.insert(v);
}
let mut v: Vec<char> = n.to_string().chars().collect();
v.sort_unstable();
hs.contains(&v)
}
}
#[test]
fn test() {
let n = 1;
let res = true;
assert_eq!(Solution::reordered_power_of2(n), res);
let n = 10;
let res = false;
assert_eq!(Solution::reordered_power_of2(n), res);
let n = 16;
let res = true;
assert_eq!(Solution::reordered_power_of2(n), res);
let n = 24;
let res = false;
assert_eq!(Solution::reordered_power_of2(n), res);
let n = 46;
let res = true;
assert_eq!(Solution::reordered_power_of2(n), res);
}