Leetcode – Minimum Changes To Make Alternating Binary String Solution

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Leetcode is a platform that provides a collection of coding challenges to help you prepare for coding interviews. One such interesting problem is “Minimum Changes To Make Alternating Binary String.” This article aims to dissect the problem and provide a comprehensive guide to solving it in Python.

Table of Contents

  1. Problem Statement
  2. Understanding the Problem
  3. Brute-force Approach
  4. Optimized Approach
  5. Code Walkthrough
  6. Testing the Solution
  7. Time and Space Complexity Analysis
  8. Tips and Tricks
  9. Conclusion

1. Problem Statement

Given a string s of length n, you need to find the minimum number of changes needed to make the string alternating. An alternating string is a string where no two adjacent characters are the same. Your task is to return the minimum number of changes required.

Example:

Input: s = "01010"
Output: 0

Input: s = "1111"
Output: 2

2. Understanding the Problem

Before diving into the coding part, it’s crucial to understand what the problem is asking. In this problem, we have a binary string consisting of ‘0’s and ‘1’s. We want to make the string “alternating” by changing the minimum number of characters.

3. Brute-force Approach

A naive approach to this problem would be to iterate through the string and for every character that doesn’t alternate, flip it and count the flips. While this would work, it is not efficient, especially for long strings.

4. Optimized Approach

An optimized approach would involve traversing the string only once and keeping a count of characters that should be flipped to make the string alternating. Essentially, you can choose one of two possible alternating strings: “010101…” or “101010…” and then find out which one requires fewer changes.

5. Code Walkthrough

Here is the Python code using the optimized approach:

def minOperations(s):
    changes1, changes2 = 0, 0
    
    # Count changes for "0101..." pattern
    for i in range(len(s)):
        if i % 2 == 0:
            if s[i] != '0':
                changes1 += 1
        else:
            if s[i] != '1':
                changes1 += 1
                
    # Count changes for "1010..." pattern
    for i in range(len(s)):
        if i % 2 == 0:
            if s[i] != '1':
                changes2 += 1
        else:
            if s[i] != '0':
                changes2 += 1
                
    return min(changes1, changes2)

# Test the function
print(minOperations("01010"))  # Output should be 0
print(minOperations("1111"))  # Output should be 2

6. Testing the Solution

It’s important to test your solution on multiple test cases including edge cases:

  • Strings with all ‘0’s or all ‘1’s
  • Strings with alternate patterns already
  • Strings with length 1 or 2

7. Time and Space Complexity Analysis

The time complexity of this solution is O(n) where n is the length of the string, as we traverse the string twice. The space complexity is O(1) as we only use a constant amount of extra space.

8. Tips and Tricks

  • You can combine both for loops into one to further optimize the code.
  • Always validate the input before processing. In this case, you can check whether the input string is not empty and contains only ‘0’s and ‘1’s.

9. Conclusion

The “Minimum Changes To Make Alternating Binary String” problem on LeetCode is an interesting problem that tests your understanding of string manipulation and pattern recognition. While a brute-force approach could work, an optimized approach greatly reduces the time complexity. This article aimed to provide you with a deep understanding of the problem and an efficient way to solve it in Python.

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