Problem Description

You have a binary string s (containing only '0's and '1's) with an even length, and it's 0-indexed.

A string is considered beautiful when you can split it into one or more parts where:

• Every part has an even length (2, 4, 6, etc.)
• Every part contains only '0's or only '1's (all characters in a part must be the same)

For example:

• "0000" is beautiful (one part with all 0's, length 4)
• "1100" is beautiful (two parts: "11" and "00", each length 2)
• "0101" is NOT beautiful as-is (would need parts like "01" which has mixed characters)

You can change any character in the string from '0' to '1' or from '1' to '0'.

The task is to find the minimum number of changes needed to make the string beautiful.

The solution recognizes that to minimize changes, we should partition the string into the smallest possible even-length substrings (length 2). For each pair of adjacent characters at positions [0,1], [2,3], [4,5], etc., we need both characters in the pair to be the same. If s[i-1] and s[i] (where i is odd) are different, we need to change one of them. The code counts these mismatches by checking all odd indices against their preceding even indices.

Intuition

To make the string beautiful, we need to partition it into substrings where each substring has even length and contains only identical characters.

The key insight is: what's the optimal way to partition the string to minimize changes?

Consider that we have flexibility in choosing partition sizes - we could use substrings of length 2, 4, 6, etc. However, using larger partitions means more characters need to be identical, which potentially requires more changes.

For example, if we have "0110":

• Partitioning as one substring of length 4 would require all 4 characters to be the same, needing at least 2 changes
• Partitioning as two substrings of length 2 ("01" and "10") would require only 2 changes total (one per pair)

This leads us to realize that using the smallest possible partition size (length 2) gives us the most flexibility and minimizes changes. With length-2 partitions, we only need to ensure pairs of adjacent characters are identical.

So the problem simplifies to: divide the string into consecutive pairs [s[0], s[1]], [s[2], s[3]], [s[4], s[5]], etc., and for each pair, make both characters the same.

For any pair where the two characters differ, we need exactly one change (we can change either character to match the other). If they're already the same, no change is needed.

Therefore, we just need to count how many pairs have different characters. This is why the solution checks positions 1, 3, 5, ... (all odd indices) against their preceding positions 0, 2, 4, ... (even indices) and counts the mismatches.

Solution Approach

The implementation follows a straightforward counting approach based on our intuition that we should partition the string into pairs of length 2.

Algorithm Steps:

1. Traverse odd indices: We iterate through positions 1, 3, 5, ... up to len(s) - 1 using range(1, len(s), 2). This gives us the second position of each pair.

2. Check each pair: For each odd index i, we compare s[i] with s[i-1] (the character at the previous even index). These two characters form a pair that needs to be identical in the beautiful string.

3. Count mismatches: If s[i] != s[i-1], the pair has different characters and needs one change. We add 1 to our count.

4. Return the total: The sum of all mismatches gives us the minimum number of changes needed.

Implementation Details:

The solution uses a generator expression with sum() to count mismatches in a single line:

sum(s[i] != s[i - 1] for i in range(1, len(s), 2))

• range(1, len(s), 2) generates odd indices: 1, 3, 5, ...
• s[i] != s[i - 1] evaluates to True (counted as 1) when characters differ, False (counted as 0) when they're the same
• sum() adds up all the True values to get the total number of required changes

Time Complexity: O(n) where n is the length of the string, as we visit each odd index once.

Space Complexity: O(1) as we only use a constant amount of extra space for the counting.

This greedy approach works because each pair is independent - changing a character in one pair doesn't affect the decision for any other pair.

Example Walkthrough

Let's walk through the solution with the string s = "10110010".

Step 1: Identify the pairs We divide the string into consecutive pairs of length 2:

• Pair 1: s[0:2] = "10" (positions 0 and 1)
• Pair 2: s[2:4] = "11" (positions 2 and 3)
• Pair 3: s[4:6] = "00" (positions 4 and 5)
• Pair 4: s[6:8] = "10" (positions 6 and 7)

Step 2: Check each pair for mismatches We iterate through odd indices (1, 3, 5, 7) and compare with the previous even index:

• Index 1: Compare s[1]='0' with s[0]='1' → Different → Need 1 change
• Index 3: Compare s[3]='1' with s[2]='1' → Same → No change needed
• Index 5: Compare s[5]='0' with s[4]='0' → Same → No change needed
• Index 7: Compare s[7]='0' with s[6]='1' → Different → Need 1 change

Step 3: Count total changes Total mismatches = 2 (at pairs 1 and 4)

Step 4: Apply changes to verify We need 2 changes minimum. For example:

• Change s[0] from '1' to '0' → Pair 1 becomes "00"
• Change s[6] from '1' to '0' → Pair 4 becomes "00"

Result: "00110000" which is beautiful (splits into "00", "11", "00", "00")

The algorithm correctly identifies that we need 2 changes to make the string beautiful.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the string s. The algorithm iterates through the string once using a range that steps through odd indices from 1 to len(s) with step size 2, which means it performs approximately n/2 iterations. Each iteration performs a constant-time comparison operation (s[i] != s[i-1]). Therefore, the overall time complexity is O(n/2) which simplifies to O(n).

The space complexity is O(1). The algorithm uses only a constant amount of extra space regardless of the input size. The sum() function with a generator expression doesn't create an intermediate list but instead evaluates the comparisons one at a time and accumulates the result in a single variable. No additional data structures are created that scale with the input size.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Misunderstanding the Problem - Trying to Find Optimal Partition Sizes

The Mistake: Many people initially think they need to find the optimal way to partition the string into variable-length segments (like choosing between length 2, 4, 6, etc.) to minimize changes. They might try dynamic programming or complex algorithms to determine whether grouping as "0000" (length 4) or "00" + "00" (two length-2 parts) would be better.

Why It Happens: The problem statement mentions that parts can have any even length (2, 4, 6, etc.), which suggests there might be an optimization problem in choosing partition sizes.

The Reality: Always partitioning into length-2 segments gives the optimal result. Here's why:

• If we have a segment of length 4 like "0101", we need 2 changes to make it all same
• If we split it into two length-2 segments "01" and "01", we still need 2 changes (one per pair)
• The number of changes required is the same regardless of how we group the pairs
• Therefore, the simplest approach (always using length 2) is optimal

Pitfall 2: Off-by-One Errors in Index Handling

The Mistake:

# Wrong: Starting from index 0
return sum(s[i] != s[i + 1] for i in range(0, len(s), 2))

# Wrong: Going beyond bounds
return sum(s[i] != s[i - 1] for i in range(1, len(s)))

Why It Happens: Confusion about whether to iterate through even or odd indices, and which direction to compare (i with i+1 vs i with i-1).

The Correct Approach:

• Use range(1, len(s), 2) to get odd indices (1, 3, 5, ...)
• Compare each odd index with the previous even index: s[i] != s[i-1]
• This ensures we check each pair exactly once without index errors

Pitfall 3: Overthinking Which Character to Change

The Mistake: Trying to determine which specific character in a mismatched pair should be changed to '0' or '1', perhaps considering the surrounding context or trying to create longer uniform segments.

Why It Happens: The problem asks for the minimum number of changes, which might lead to thinking we need to specify exactly which changes to make.

The Reality:

• We only need to count the number of changes, not specify which ones
• For each mismatched pair, exactly one change is needed regardless of which character we change
• The choice doesn't affect other pairs since each pair is independent

Pitfall 4: Forgetting Edge Cases

The Mistake: Not considering or testing edge cases properly:

# Might fail on empty string or single character (though problem guarantees even length)
def minChanges(self, s: str) -> int:
    if not s:  # Unnecessary check given problem constraints
        return 0
    return sum(s[i] != s[i - 1] for i in range(1, len(s), 2))

The Reality: The problem guarantees the string has even length, so we don't need special handling for odd lengths or empty strings. However, it's good practice to verify the smallest valid input (length 2) works correctly.

Solution to Avoid These Pitfalls:

1. Understand the key insight: Each pair of consecutive characters is independent, and we always want pairs of identical characters
2. Use clear variable names and comments if expanding the one-liner:

def minChanges(self, s: str) -> int:
    changes_needed = 0
    # Process each pair: (0,1), (2,3), (4,5), etc.
    for pair_start in range(0, len(s), 2):
        first_char = s[pair_start]
        second_char = s[pair_start + 1]
        if first_char != second_char:
            changes_needed += 1
    return changes_needed

3. Test with simple examples to verify understanding:
   • "01" → 1 change needed
   • "0011" → 0 changes needed
   • "0101" → 2 changes needed