900. Reordered Power Of 2¶

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900. 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¶

class Solution {
    public boolean reorderedPowerOf2(int N) {
        char[] a1 = String.valueOf(N).toCharArray();
        Arrays.sort(a1);
        String s1 = new String(a1);
        for (int i = 0; i < 31; i++) {
            char[] a2 = String.valueOf((int)(1 << i)).toCharArray();
            Arrays.sort(a2);
            String s2 = new String(a2);
            if (s1.equals(s2)) return true;
        }
        return false;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here