Algorithm
Problem Name: 869. Reordered Power of 2
Problem Link: https://leetcode.com/problems/reordered-power-of-2/
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
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
start codi
class Solution {
public boolean reorderedPowerOf2(int n) {
int[] digitCount = count(n);
for (int i = 0; i < 31; i++) {
if (Arrays.equals(digitCount, count(1 << i))) {
return true;
}
}
return false;
}
private int[] count(int n) {
int[] result = new int[10];
while (n > 0) {
result[n % 10]++;
n /= 10;
}
return result;
}
}
Input
n = 1
Output
true
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const reorderedPowerOf2 = function(N) {
const A = count(N);
for (let i = 0; i < 31; i++) {
if (arrayEqual(A, count(1 << i))) return true;
}
return false;
};
function count(num) {
const res = [];
while (num > 0) {
addOne(res, num % 10);
num = parseInt(num / 10);
}
return res;
}
function addOne(arr, idx) {
if (arr[idx]) {
arr[idx] += 1;
return;
}
arr[idx] = 1;
}
function arrayEqual(a1, a2) {
return JSON.stringify(a1) === JSON.stringify(a2);
}
console.log(reorderedPowerOf2(1));
console.log(reorderedPowerOf2(10));
console.log(reorderedPowerOf2(16));
console.log(reorderedPowerOf2(24));
console.log(reorderedPowerOf2(46));
Input
n = 1
Output
true
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def reorderedPowerOf2(self, N):
cnt = collections.Counter(str(N))
return any(cnt == collections.Counter(str(1 << c)) for c in range(32))