How do you solve Minimum Number of Flips to Reverse Binary String on LeetCode?

Minimum Number of Flips to Reverse Binary String (LeetCode #3750) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Flipping bits can be visualized as matching pairs.

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

The optimal solution for Minimum Number of Flips to Reverse Binary String runs in O(n) time and uses O(1) space. We only traverse the string once with two pointers, leading to O(n) time complexity.

What is the brute force solution for Minimum Number of Flips to Reverse Binary String?

The brute force approach for Minimum Number of Flips to Reverse Binary String is Brute Force, with O(n²) time complexity and O(1) space complexity. We can compare the binary string with its reverse character by character. For each mismatch, we count a flip. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Flips to Reverse Binary String?

Minimum Number of Flips to Reverse Binary String uses the following concepts: Math, Two Pointers, String, Bit Manipulation. The recommended patterns to study are: Two Pointers, String Manipulation.

What are the interview tips for Minimum Number of Flips to Reverse Binary String?

Explain your thought process clearly. Discuss edge cases like single-bit numbers. Practice similar problems to build confidence.

What are common mistakes when solving Minimum Number of Flips to Reverse Binary String?

Not considering the middle bit in odd-length strings. Overlooking leading zeros in binary representation.

#3750

Minimum Number of Flips to Reverse Binary String

Easy

MathTwo PointersStringBit ManipulationTwo PointersString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Using two pointers, we can efficiently compare bits from the start and end of the string, counting mismatches.

⚙️

Algorithm

3 steps

1Step 1: Convert n to its binary string representation.
2Step 2: Initialize two pointers, one at the start and one at the end of the string.
3Step 3: Move both pointers towards the center, counting mismatches until they meet.

solution.py8 lines

1def minFlips(n):
2    s = bin(n)[2:]
3    left, right, flips = 0, len(s) - 1, 0
4    while left < right:
5        if s[left] != s[right]: flips += 1
6        left += 1
7        right -= 1
8    return flips

ℹ

Complexity note: We only traverse the string once with two pointers, leading to O(n) time complexity.

1Flipping bits can be visualized as matching pairs.
2Two pointers reduce unnecessary comparisons.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.