Minimum Number of Flips to Reverse Binary String Python/Java solution

Most implementations iterate with two indices from both ends of the string. If the characters differ, increment a counter and move both pointers inward. The same logic works in Python, Java, C++, Go, and TypeScript with O(n) time complexity.

How to solve Minimum Number of Flips to Reverse Binary String in O(n)?

Use two pointers. Initialize one pointer at the start and one at the end of the string. Move both toward the center while checking if the characters match. Whenever s[left] and s[right] differ, increment the flip counter. Each pair requires at most one flip, giving an O(n) time and O(1) space solution.

What is the best approach for Minimum Number of Flips to Reverse Binary String?

The two pointers approach is the most efficient. Start one pointer at the beginning of the string and another at the end, comparing mirrored characters. Every time the two characters differ, one flip is required to make them equal. This solution runs in O(n) time and O(1) extra space.

Is Minimum Number of Flips to Reverse Binary String asked at Google/Amazon/Meta?

Problems involving symmetric comparisons, binary strings, and two‑pointer techniques appear frequently in coding interviews at companies like Amazon, Google, and Meta. Variants of palindrome or mirrored string correction problems are common in interview question banks.

What data structure is used in Minimum Number of Flips to Reverse Binary String?

The problem mainly relies on string traversal with the two pointers technique. No advanced data structures are required. The algorithm simply compares characters at mirrored indices and counts mismatches.

What is the time complexity of Minimum Number of Flips to Reverse Binary String?

The optimal algorithm runs in O(n) time because the string is scanned once while comparing symmetric positions. Space complexity can be O(1) when using two pointers, or O(n) if you explicitly build the reversed string for comparison.

Description Hints 1 Solution Video 3:24

3750. Minimum Number of Flips to Reverse Binary String

Q: Is Minimum Number of Flips to Reverse Binary String easy or hard?

Minimum Number of Flips to Reverse Binary String is considered an Easy problem. The key idea is recognizing that mirrored positions must match. Once that symmetry is identified, a simple two‑pointer scan produces the answer in linear time.

Easy75.5% acceptance

You are given a positive integer n.

Let s be the binary representation of n without leading zeros.

The reverse of a binary string s is obtained by writing the characters of s in the opposite order.

You may flip any bit in s (change 0 → 1 or 1 → 0). Each flip affects exactly one bit.

Return the minimum number of flips required to make s equal to the reverse of its original form.

Example 1:

Input: n = 7

Output: 0

Explanation:

The binary representation of 7 is "111". Its reverse is also "111", which is the same. Hence, no flips are needed.

Example 2:

Input: n = 10

Output: 4

Explanation:

The binary representation of 10 is "1010". Its reverse is "0101". All four bits must be flipped to make them equal. Thus, the minimum number of flips required is 4.

Constraints:

1 <= n <= 10⁹

Topics

Hints open one at a time

Hint 1

Generate the reverse of the binary string and use two pointers from the ends to determine where flips are needed.

3:24

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1class Solution:
2    def minimumFlips(self, n: int) -> int:
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