2914. Minimum Number of Changes to Make Binary String Beautiful - Detailed Explanation

Problem Statement

Description:
You are given a binary string s. A binary string is defined as beautiful if it does not contain the substring "010". In one operation, you can flip a character from '0' to '1' or from '1' to '0'. Your goal is to return the minimum number of operations required to make the string beautiful.

Examples:

1. Example 1:

   • Input: s = "010"
   • Output: 1
   • Explanation:
     The string "010" is not beautiful because it contains the substring "010". By flipping the last character from '0' to '1', we obtain "011", which does not contain "010". Hence, only 1 operation is required.

2. Example 2:

   • Input: s = "01010"
   • Output: 1
   • Explanation:
     Although the string appears to have overlapping "010" patterns, a single flip is sufficient.
     Greedy walkthrough:
     • Scan the string from left to right.
     • At index 0–2, we see "010". Flip the character at index 2 to get "01110".
     • The resulting string "01110" does not contain "010".
       Therefore, only 1 operation is needed.

3. Example 3:

   • Input: s = "101"
   • Output: 0
   • Explanation:
     The string "101" does not contain the substring "010", so it is already beautiful and no changes are required.

Constraints:

• The string consists only of the characters '0' and '1'.
• The length of s is at least 1.

Hints

• Hint 1:
  Focus on detecting the exact substring "010". Once you find such a pattern, consider which character to flip so that you do not inadvertently create a new "010" pattern in overlapping regions.

• Hint 2:
  A greedy strategy works well here: When you encounter "010", flip the last '0' to '1'. This choice helps to “fix” the current pattern while preventing potential overlapping issues from affecting subsequent portions of the string.

Approach

Greedy Approach (One-Pass Scan)

Explanation

The idea is to traverse the string from left to right and whenever the exact substring "010" is found, perform an operation to “fix” the pattern. The greedy decision is to flip the last '0' in the pattern to '1'. This choice is effective because:

• It removes the current occurrence of "010".
• It minimizes interference with adjacent characters, thereby reducing the risk of creating new "010" patterns that overlap.

Step-by-Step Walkthrough:

1. Convert to a Mutable Structure:
   Since strings in many languages (like Python and Java) are immutable, convert the string to a character array (or list) for easy in-place modifications.

2. Initialize a Counter:
   Set a counter (count) to 0 to track the number of changes made.

3. Traverse the String:
   Start iterating from index 2 (the first possible ending index for a substring of length 3) until the end of the string.

   • For every index i, check if the substring formed by characters at positions i-2, i-1, and i is "010".
   • If yes, increment the count and flip the character at index i (i.e., change it from '0' to '1').

4. Skip Overlapping Checks:
   After flipping a character, the modified value may affect overlapping patterns. In our greedy solution, since we always flip the third character, the already processed region is fixed, and subsequent checks continue normally.

5. Return the Result:
   After scanning through the string, return the counter as the minimum number of operations required.

Pseudocode

function minChanges(s):
    Convert s to a mutable array, say charArray
    count = 0
    for i from 2 to length(charArray) - 1:
        if charArray[i-2] == '0' and charArray[i-1] == '1' and charArray[i] == '0':
            count += 1
            charArray[i] = '1'  // Flip the third character
    return count

Code Examples

Python Code