Problem Description

You have a binary string s containing only '0's and '1's. Every second, all occurrences of the substring "01" in the string are simultaneously replaced with "10". This replacement happens to all "01" patterns at the same time in each second, and the process continues until there are no more "01" substrings left in the string.

Your task is to find how many seconds it takes for this process to complete.

For example, if s = "0110101":

• After 1 second: "0110101" → "1011001" (the "01" at positions 0-1 and 4-5 are replaced)
• After 2 seconds: "1011001" → "1101010" (the "01" at position 1-2 is replaced)
• After 3 seconds: "1101010" → "1110100" (the "01" at position 3-4 is replaced)
• After 4 seconds: "1110100" → "1111000" (the "01" at position 4-5 is replaced)

The process stops when no "01" patterns remain, and you return the total number of seconds (4 in this case).

The key point is that all "01" patterns are replaced simultaneously in each second, not one by one. The process naturally moves all '1's to the left and all '0's to the right until they are completely separated.

Intuition

The pattern "01" represents a '0' followed by a '1'. When we replace "01" with "10", we're essentially swapping these two digits - the '1' moves one position to the left while the '0' moves one position to the right.

Think of this process like bubbles rising in water. Each '1' wants to move leftward (like bubbles rising up), and each '0' wants to move rightward (like heavier particles sinking down). Every second, each '1' that has a '0' directly to its left will swap positions with that '0'.

The process continues until all '1's have "bubbled up" to the left side of the string and all '0's have "sunk" to the right side. At this point, the string looks like "111...000" with all '1's grouped on the left and all '0's grouped on the right. Since there are no more "01" patterns, the process stops.

The simulation approach directly models what the problem describes: repeatedly find and replace all "01" patterns with "10" until no such patterns exist. Each iteration represents one second of time. We count how many iterations (seconds) it takes until the string no longer contains any "01" substring.

This works because:

1. Each replacement operation happens simultaneously across the entire string
2. The string stabilizes when '1's and '0's are completely separated
3. We can directly check if "01" exists using string operations and perform the replacement using built-in string methods

Learn more about Dynamic Programming patterns.

Solution Approach

The solution uses a straightforward simulation approach that directly implements the problem's requirements:

1. Initialize a counter: Start with ans = 0 to track the number of seconds elapsed.

2. Check for pattern existence: Use s.count('01') to check if any "01" patterns exist in the string. This method returns the number of non-overlapping occurrences of "01" in the string. If it returns 0, the while loop terminates.

3. Perform simultaneous replacement: When "01" patterns exist, use s.replace('01', '10') to replace ALL occurrences of "01" with "10" simultaneously. This built-in string method handles the simultaneous replacement requirement perfectly - it finds all occurrences first, then replaces them all at once.

4. Increment the time counter: After each replacement operation, increment ans by 1 to represent one second passing.

5. Repeat until stable: Continue the loop until s.count('01') returns 0, meaning no more "01" patterns exist in the string.

The algorithm's key insight is that Python's replace() method naturally handles simultaneous replacements. When we call s.replace('01', '10'), it:

• First identifies all positions where "01" occurs
• Then replaces all of them in one operation
• This ensures that replacements don't interfere with each other

For example, if s = "0101", calling replace('01', '10') gives us "1010" in one step (both the "01" at position 0-1 and position 2-3 are replaced simultaneously), rather than replacing them sequentially which would give a different result.

The time complexity is O(n²) in the worst case, where n is the length of the string, since we might need up to O(n) iterations and each iteration involves O(n) string operations.

Example Walkthrough

Let's trace through the solution with s = "00110":

Initial state: s = "00110", ans = 0

Step 1: Check for "01" patterns

• s.count('01') returns 1 (found "01" at positions 1-2)
• Since count > 0, we enter the loop

Step 2: First replacement (Second 1)

• s.replace('01', '10') transforms "00110" → "01010"
  • The "01" at positions 1-2 becomes "10"
• Increment ans = 1

Step 3: Check again

• s.count('01') returns 2 (found "01" at positions 0-1 and 2-3)
• Continue loop

Step 4: Second replacement (Second 2)

• s.replace('01', '10') transforms "01010" → "10100"
  • Both "01" patterns are replaced simultaneously:
    • "01" at positions 0-1 → "10"
    • "01" at positions 2-3 → "10"
• Increment ans = 2

Step 5: Check again

• s.count('01') returns 1 (found "01" at positions 2-3)
• Continue loop

Step 6: Third replacement (Second 3)

• s.replace('01', '10') transforms "10100" → "11000"
  • The "01" at positions 2-3 becomes "10"
• Increment ans = 3

Step 7: Final check

• s.count('01') returns 0 (no "01" patterns found)
• Exit loop

Result: Return ans = 3

The string evolved as: "00110" → "01010" → "10100" → "11000"

Notice how the '1's gradually "bubble" leftward while '0's move rightward, until they're completely separated. The simultaneous replacement in Step 4 is crucial - both "01" patterns are replaced in the same second, demonstrating why replace() perfectly models the problem's requirement.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n²) where n is the length of the string.

The algorithm simulates the process of moving all '1's to the right by repeatedly replacing "01" with "10". In the worst case, we need O(n) iterations (when all '1's are at the beginning and need to move to the end), and each iteration involves:

• s.count('01'): O(n) to count occurrences
• s.replace('01', '10'): O(n) to create a new string with replacements

Therefore, the overall time complexity is O(n) × O(n) = O(n²).

Space Complexity: O(n) where n is the length of the string.

The space complexity comes from:

• The string s is reassigned in each iteration with s.replace(), which creates a new string of length n
• Python strings are immutable, so each replace() operation creates a new string object
• However, only one version of the string is kept in memory at a time (the old one is garbage collected)

Thus, the space complexity is O(n) for storing the modified string.

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

Common Pitfalls

1. Attempting Sequential Replacement Instead of Simultaneous

A common mistake is trying to manually iterate through the string and replace "01" patterns one at a time, which would give incorrect results since the problem requires ALL "01" patterns to be replaced simultaneously in each second.

Incorrect Approach:

def secondsToRemoveOccurrences(self, s: str) -> int:
    seconds = 0
    while "01" in s:
        # Wrong: This might replace patterns one by one
        for i in range(len(s) - 1):
            if s[i:i+2] == "01":
                s = s[:i] + "10" + s[i+2:]
                break  # This breaks after first replacement!
        seconds += 1
    return seconds

Why it's wrong: This replaces only one "01" per iteration, not all simultaneously. For example, with "0101", this would take 2 seconds instead of 1.

Solution: Use Python's built-in replace() method which naturally handles simultaneous replacements, or if implementing manually, collect all positions first, then replace them all at once.

2. String Immutability Performance Issues

In languages where strings are immutable (like Python), creating new strings repeatedly can be inefficient for very long strings. Each replace() operation creates a new string object.

Better Alternative for Large Inputs:

def secondsToRemoveOccurrences(self, s: str) -> int:
    seconds = 0
    chars = list(s)  # Convert to mutable list
  
    while True:
        found = False
        i = 0
        while i < len(chars) - 1:
            if chars[i] == '0' and chars[i + 1] == '1':
                chars[i], chars[i + 1] = '1', '0'
                found = True
                i += 2  # Skip the swapped pair
            else:
                i += 1
      
        if not found:
            break
        seconds += 1
  
    return seconds

3. Not Handling Edge Cases

Failing to consider strings that are already sorted or contain only one type of character.

Edge Cases to Consider:

• Empty string: s = "" → Should return 0
• Single character: s = "0" or s = "1" → Should return 0
• Already sorted: s = "1110000" → Should return 0
• All same character: s = "0000" or s = "1111" → Should return 0

4. Mathematical Optimization Opportunity Missed

The simulation approach works but misses a potential O(n) mathematical solution. The number of seconds needed equals the maximum "delay" any '1' experiences reaching its final position.

Optimized O(n) Solution:

def secondsToRemoveOccurrences(self, s: str) -> int:
    zeros = 0
    seconds = 0
  
    for char in s:
        if char == '0':
            zeros += 1
        elif zeros > 0:  # char is '1' and there are zeros before it
            seconds = max(seconds + 1, zeros)
  
    return seconds

This tracks how many positions each '1' needs to move left and calculates the maximum time needed.