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One approach to solve the problem is to leverage the fact that two numbers with the same digits in different orders will have the same digit frequency (or count of each digit). For the given number n, we aim to check if its digits can be rearranged to form any power of two. To achieve this:
10^9 and store their digit frequency in a set.n, calculate its digit frequency and check if it exists in the set.Time Complexity: O(1) because the number of power of twos checked (31) and the frequency array size (10) are constant.
Space Complexity: O(1) as only small fixed-size integer arrays are used.
1#include <stdbool.h>
2#include <stdio.h>
3#include <string.h>
4
5bool reorderedPowerOf2(int n) {
6    int countDigits[10] = {0};
7    int num = n;
8    while (num > 0) {
9        countDigits[num % 10]++;
10        num /= 10;
11    }
12
13    for (int i = 0; i < 31; ++i) {
14        int powerDigits[10] = {0};
15        int powerOfTwo = 1 << i;
16        num = powerOfTwo;
17        while (num > 0) {
18            powerDigits[num % 10]++;
19            num /= 10;
20        }
21
22        bool match = true;
23        for (int j = 0; j < 10; ++j) {
24            if (countDigits[j] != powerDigits[j]) {
25                match = false;
26                break;
27            }
28        }
29
30        if (match) {
31            return true;
32        }
33    }
34
35    return false;
36}
37
38int main() {
39    int n = 1;
40    printf("%s\n", reorderedPowerOf2(n) ? "true" : "false");
41    return 0;
42}The implementation uses an array to count the frequency of each digit in n. It then iterates through all powers of two that can exist within the limits (2^0 to 2^30). For each power, it calculates the digit frequency and compares it to n's. If a match is found, it returns true. If no match is found through all powers, it returns false.
A backtracking method can be employed by generating permutations of the digits of the number n and checking if any results in a power of two. Given the constraint, this method needs optimization to efficiently discard invalid permutations early.
n's digits where the leading digit isn't zero.Not provided due to complexity constraints in inline format.
1
Python uses the itertools.permutations for generating permutations of the string representation of n. It filters permutations to exclude those starting with '0' and verifies if constructed numbers are powers of two.