#2914

Medium

auto_awesomeString-driven solution strategy

LeetCode Problem Workspace

Minimum Number of Changes to Make Binary String Beautiful

This problem involves determining the minimum number of changes to make a binary string beautiful using string-driven strategies.

string

Problem Statement

You are given a 0-indexed binary string s with an even length. The goal is to modify the string by changing some of its characters to '0' or '1' so that it becomes beautiful. A string is considered beautiful if you can partition it into one or more substrings such that each substring contains an even number of identical characters.

For example, if you have a string s = '1001', the minimum number of changes to make it beautiful would be 2. You can change s[1] to 1 and s[3] to 0, resulting in 1100, which is beautiful as it can be split into '11|00'. The challenge is to determine the minimum number of changes required for any given input string.

Examples

Example 1

Input: s = "1001"

Output: 2

We change s[1] to 1 and s[3] to 0 to get string "1100". It can be seen that the string "1100" is beautiful because we can partition it into "11|00". It can be proven that 2 is the minimum number of changes needed to make the string beautiful.

Example 2

Input: s = "10"

Output: 1

We change s[1] to 1 to get string "11". It can be seen that the string "11" is beautiful because we can partition it into "11". It can be proven that 1 is the minimum number of changes needed to make the string beautiful.

Example 3

Input: s = "0000"

Output: 0

We don't need to make any changes as the string "0000" is beautiful already.

Constraints

2 <= s.length <= 105
s has an even length.
s[i] is either '0' or '1'.

Solution Approach

Understand the Structure of Beautiful String

First, understand that a beautiful string can be split into substrings where each part contains an even number of identical characters. For each partition, ensure that the adjacent characters match the requirement, making it easy to transform the string by adjusting minimal characters.

Iterate Through the String Efficiently

You can iterate through the string and check for every consecutive pair of characters. If the two characters differ, a change is necessary. Efficiently track the number of changes required to satisfy the condition of even parts.

Optimize with Greedy Approach

Since the string's length is even, you can take advantage of a greedy strategy by modifying characters only when necessary. Use a linear scan to count the required changes and keep track of the minimum number of alterations needed.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(1)

The time complexity of this solution is O(n), as we only need to iterate through the string once to count the necessary changes. The space complexity is O(1), as we only use a constant amount of extra space.

What Interviewers Usually Probe

Candidate should focus on understanding the partition structure of the string and identifying the minimal changes needed.
Look for the candidate's ability to implement the greedy approach efficiently, ensuring they avoid unnecessary operations.
Assess if the candidate correctly handles edge cases, such as strings that are already beautiful.

Common Pitfalls or Variants

Common pitfalls

Failing to recognize that every part of the partition must have an even number of identical characters.
Overcomplicating the problem by making unnecessary changes to characters when fewer are sufficient.
Not handling cases where the string is already beautiful and no changes are needed.

Follow-up variants

Modify the problem to handle odd-length strings, exploring how the partition strategy changes.
Increase the complexity by requiring multiple partitions with different size constraints for each segment.
Extend the problem to allow for partitions based on different grouping strategies, such as alternating blocks of 0s and 1s.

FAQ

What is the pattern used to solve the Minimum Number of Changes to Make Binary String Beautiful?

This problem follows a string-driven solution pattern, leveraging partitioning techniques and greedy strategies to minimize changes.

What is the time complexity of the solution for Minimum Number of Changes to Make Binary String Beautiful?

The time complexity of the solution is O(n), as it requires a single pass through the string.

Can the string be modified in place for the Minimum Number of Changes to Make Binary String Beautiful?

Yes, the string can be modified in place with minimal space overhead, maintaining O(1) space complexity.

How do I handle edge cases in Minimum Number of Changes to Make Binary String Beautiful?

Edge cases include strings that are already beautiful, where no changes are needed, and strings with alternating characters where each change matters.

What is a beautiful string in the context of Minimum Number of Changes to Make Binary String Beautiful?

A beautiful string is one that can be partitioned into substrings, each containing an even number of identical characters, allowing minimal changes to achieve this structure.

terminal

Solution

Solution 1: Counting

We only need to traverse all odd indices $1, 3, 5, \cdots$ of the string $s$. If the current odd index is different from the previous index, i.e., $s[i] \ne s[i - 1]$, we need to modify the current character so that $s[i] = s[i - 1]$. Therefore, the answer needs to be incremented by $1$.

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class Solution:
    def minChanges(self, s: str) -> int:
        return sum(s[i] != s[i - 1] for i in range(1, len(s), 2))

Continue Practicing

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