LeetCode 3750: Minimum Number of Flips to Reverse Binary String ·

Table of Contents

Meta Data
#

Difficulty: easy First Attempt: 2025-11-22

Submissions: Accepted (first try)

Approach
#

Write n as a binary string s without leading zeros and turn it into its own reverse, one bit flip at a time.

Position i must become s[len - 1 - i], and needs one flip when they differ, so the answer is the number of positions where s and its reverse differ.

From a converging-pointer view: for each pair i and len - 1 - i, if the two characters differ, both must flip, so every mismatched pair contributes 2.

My Solution
#

class Solution:
    def minimumFlips(self, n: int) -> int:
        s = bin(n)[2:]
        return sum(a != b for a, b in zip(s, reversed(s)))

My original built the reversed string first and compared position by position in a loop, which is the same logic.

Complexity
#

Time O(log n), space O(log n).

Mistakes to Avoid
#

Forgetting to strip the leading 0b from bin(n).
Reversing into “another number” and comparing values: the comparison is bit by bit.

Interview Explanation
#

I convert n to a binary string and compare it with its reverse position by position; the number of mismatches is the number of flips, since each mismatched bit needs exactly one flip. Time is proportional to the number of bits, O(log n).