LeetCode-in-Java
869. Reordered Power of 2
Medium
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
Solution
import java.util.Arrays;
public class Solution {
public boolean reorderedPowerOf2(int n) {
int[] originalDigits = countDigits(n);
int num = 1;
for (int i = 0; i < 31; i++) {
if (Arrays.equals(originalDigits, countDigits(num))) {
return true;
}
num <<= 1;
}
return false;
}
private int noOfDigits(int n) {
if (n == 0) {
return 1;
}
return (int) Math.log10(Math.abs(n)) + 1;
}
private boolean check(int num1, int num2) {
int[] num = new int[10];
while (num1 > 0) {
num[num1 % 10]++;
num1 /= 10;
}
while (num2 > 0) {
num[num2 % 10]--;
num2 /= 10;
}
for (int i = 0; i < 10; i++) {
if (num[i] != 0) {
return false;
}
}
return true;
}
private int[] countDigits(int num) {
int[] digitCount = new int[10];
while (num > 0) {
digitCount[num % 10]++;
num /= 10;
}
return digitCount;
}
}