Reordered Power of 2 - Solution & Explanation

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:

• Precompute all powers of two up to 10^9 and store their digit frequency in a set.
• For the input number n, calculate its digit frequency and check if it exists in the set.

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.

• Generate all valid permutations of n's digits where the leading digit isn't zero.
• Track numbers that form powers of two during permutations.

Permutations in C would require explicit handling of stateful recursion and result collections, a task not directly suited for this predefined response format.