Problem objective: given an integer n, determine if its digits can be rearranged to form a power of 2 without generating a number with leading zeros.
Practical examples: n = 1 is valid because 1 = 2^0 and n = 10 is not valid because the digits 1 and 0 cannot form a power of 2 without leading zeros.
Key idea: powers of 2 show specific digit combinations. Instead of generating all permutations of n, compare the digit frequency of n with the digit frequency of each power of 2. Two numbers are permutations of each other if and only if they have the same count of each digit.
Frequency representation: build a vector of ten positions for digits 0 to 9 and fill it with the occurrences of each digit. As a unique key, that vector can be transformed into a string or a tuple. For example, the number 128 produces a counter with 1 in position 1, 1 in position 2, and 1 in position 8.
Practical strategy: precompute the frequency representations of all powers of 2 from 2^0 to 2^30 (2^30 covers the typical 32-bit integer range). Calculate the representation of n and compare it with those of the powers of 2. If any match, the rearrangement is possible.
Step-by-step algorithm: 1) implement a counter function that traverses the digits of a number and updates an array of 10 counters; 2) convert that array into a comparable key (string or tuple); 3) calculate the key of n; 4) for i from 0 to 30 calculate the key of 1 << i and compare it with the key of n; 5) return true if there is a match, false otherwise.
Descriptive version in C++: implement counter(int x) that returns an encoded integer or a string representing the frequencies; in reorderedPowerOf2 calculate count = counter(n) and for i = 0 to 30 check if counter(1 << i) == count and return true in that case, otherwise return false.
Descriptive version in JavaScript: define counter(num) as an array of size 10 initialized to zeros, convert num to a string and accumulate each digit by incrementing the array, return the array joined as a key; then compare against counter(1 << i) for i in 0..30.
Descriptive version in Python: define counter(x) that builds a list of 10 zeros and while x > 0 increment count[x % 10] and do x //= 10, return tuple(count); compare the tuple of n with the tuple of 1 << i for i in range(30).
Time and space complexity: O(1) time in practice because only 31 powers of 2 are tested and each counter processes at most the number of digits of n (bounded constant), O(1) space for the vector of 10 counters.
Final observations: the problem is about pattern recognition through digit counting, not permutation generation. The compact encoding of frequencies avoids combinatorial explosion and is efficient and robust for large integers within common ranges.
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