Problem Description

You're given a binary string s consisting of only '0's and '1's. A binary string is considered monotone increasing if it follows a specific pattern: first, there are some number of '0's (possibly zero), followed by some number of '1's (also possibly zero).

Examples of monotone increasing strings:

• "0000" (all zeros)
• "1111" (all ones)
• "0011" (zeros followed by ones)
• "01" (zero followed by one)
• "" (empty string)

Examples of strings that are NOT monotone increasing:

• "010" (has a '0' after a '1')
• "10" (starts with '1' then has '0')

Your task is to transform the given string s into a monotone increasing string by performing flips. A flip operation means changing a character at position i from '0' to '1' or from '1' to '0'.

The goal is to find the minimum number of flips needed to make the string monotone increasing.

For example:

• If s = "00110", you could flip the '1's at positions 2 and 3 to get "00000" (2 flips)
• Or you could flip the last '0' at position 4 to get "00111" (1 flip)
• The minimum is 1 flip

The solution approach considers each possible "split point" where all characters to the left should be '0' and all characters to the right should be '1', calculating the cost for each split and finding the minimum.

Intuition

To make a binary string monotone increasing, we need to ensure all '0's come before all '1's. This means we need to pick a "split point" in the string where everything to the left should be '0' and everything to the right should be '1'.

The key insight is that we can try every possible split point and calculate the cost for each. For any split point i:

• All '1's to the left of (and including) position i need to be flipped to '0'
• All '0's to the right of position i need to be flipped to '1'

Let's think about what information we need to calculate this cost efficiently:

1. At each position i, we need to know how many '1's are to the left (these need flipping to '0')
2. We need to know how many '0's are to the right (these need flipping to '1')

Since the number of '1's to the left equals (i+1) - number_of_zeros_to_left, and the number of '0's to the right equals total_zeros - number_of_zeros_to_left, we can track just the cumulative count of '0's as we traverse the string.

The formula for the cost at position i becomes:

• Number of '1's to flip to '0' = (i+1) - cur where cur is count of '0's up to position i
• Number of '0's to flip to '1' = tot - cur where tot is total '0's in the string
• Total cost = (i+1) - cur + tot - cur

We also consider the edge case where we flip all '0's to '1's (making the entire string "1111..."), which costs tot flips.

By trying all possible split points and keeping track of the minimum cost, we find the optimal solution.

Learn more about Dynamic Programming patterns.

Solution Approach

The implementation uses a prefix sum technique combined with enumeration of all possible split points.

Step 1: Count total zeros

tot = s.count("0")

First, we count the total number of '0's in the string. This gives us the cost of one extreme case: flipping all '0's to '1's to make the string all ones.

Step 2: Initialize variables

ans, cur = tot, 0

• ans is initialized to tot, representing the cost of making the entire string consist of only '1's
• cur will track the cumulative count of '0's as we traverse the string

Step 3: Enumerate each position

for i, c in enumerate(s, 1):
    cur += int(c == "0")
    ans = min(ans, i - cur + tot - cur)

For each position i (using 1-based indexing for easier calculation):

• Update cur by adding 1 if the current character is '0'
• Calculate the cost if we make position i the last '0' in our monotone string:
  • i - cur: Number of '1's in positions [1, i] that need to be flipped to '0'
  • tot - cur: Number of '0's in positions [i+1, n] that need to be flipped to '1'
  • Total cost: i - cur + tot - cur
• Update ans with the minimum cost found so far

Why this works:

• By using 1-based indexing (enumerate(s, 1)), position i represents the count of characters we've seen so far
• i - cur gives us the number of '1's seen so far (total characters minus '0's)
• tot - cur gives us the remaining '0's after position i
• The formula i - cur + tot - cur calculates the exact number of flips needed if we split at position i

Time Complexity: O(n) where n is the length of the string - we traverse the string twice (once for counting, once for enumeration)

Space Complexity: O(1) - we only use a constant amount of extra space

Example Walkthrough

Let's walk through the solution with s = "00110".

Step 1: Count total zeros

• tot = 3 (there are three '0's in the string)
• Initialize ans = 3 (cost of flipping all '0's to make "11111")
• Initialize cur = 0 (cumulative count of '0's)

Step 2: Process each position

Position i=1, character '0':

• cur = 0 + 1 = 1 (found one '0' so far)
• Cost if we split here: 1 - 1 + 3 - 1 = 0 + 2 = 2
  • Left side [0]: 0 ones to flip to '0'
  • Right side [0110]: 2 zeros to flip to '1'
• ans = min(3, 2) = 2

Position i=2, character '0':

• cur = 1 + 1 = 2 (found two '0's so far)
• Cost if we split here: 2 - 2 + 3 - 2 = 0 + 1 = 1
  • Left side [00]: 0 ones to flip to '0'
  • Right side [110]: 1 zero to flip to '1'
• ans = min(2, 1) = 1

Position i=3, character '1':

• cur = 2 + 0 = 2 (still two '0's total)
• Cost if we split here: 3 - 2 + 3 - 2 = 1 + 1 = 2
  • Left side [001]: 1 one to flip to '0'
  • Right side [10]: 1 zero to flip to '1'
• ans = min(1, 2) = 1

Position i=4, character '1':

• cur = 2 + 0 = 2
• Cost if we split here: 4 - 2 + 3 - 2 = 2 + 1 = 3
  • Left side [0011]: 2 ones to flip to '0'
  • Right side [0]: 1 zero to flip to '1'
• ans = min(1, 3) = 1

Position i=5, character '0':

• cur = 2 + 1 = 3
• Cost if we split here: 5 - 3 + 3 - 3 = 2 + 0 = 2
  • Left side [00110]: 2 ones to flip to '0'
  • Right side []: 0 zeros to flip to '1'
• ans = min(1, 2) = 1

Result: The minimum number of flips is 1. This corresponds to flipping the last '0' to get "00111".

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the string s. This is because:

• The count("0") operation traverses the entire string once, taking O(n) time
• The enumerate() loop iterates through each character in the string exactly once, performing constant-time operations (int(c == "0"), addition, and min() comparison) in each iteration, taking O(n) time
• Total: O(n) + O(n) = O(n)

The space complexity is O(1). The algorithm only uses a fixed number of variables (tot, ans, cur, i, c) regardless of the input size, requiring constant extra space.

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

Common Pitfalls

1. Off-by-One Error with Split Point Interpretation

A common mistake is misunderstanding what the split point represents. Developers often confuse whether position i is the last '0' or the first '1' in the final monotone string.

Incorrect Approach:

def minFlipsMonoIncr(self, s: str) -> int:
    total_zeros = s.count("0")
    min_flips = total_zeros
    zeros_seen = 0
  
    for i in range(len(s)):  # Using 0-based indexing
        if s[i] == "0":
            zeros_seen += 1
        # Wrong calculation - mixing 0-based index with count logic
        min_flips = min(min_flips, i - zeros_seen + total_zeros - zeros_seen)
  
    return min_flips

Why it fails: Using 0-based indexing while trying to calculate the number of characters processed leads to incorrect flip counts.

Correct Solution: Use 1-based enumeration to ensure position represents the count of characters processed, making the arithmetic intuitive.

2. Forgetting Edge Cases

Another pitfall is not considering the cases where the optimal solution is to make the entire string all '0's or all '1's.

Incomplete Approach:

def minFlipsMonoIncr(self, s: str) -> int:
    total_zeros = s.count("0")
    min_flips = float('inf')  # Missing initialization with edge case
    zeros_seen = 0
  
    for position, char in enumerate(s, 1):
        if char == "0":
            zeros_seen += 1
        min_flips = min(min_flips, position - zeros_seen + total_zeros - zeros_seen)
  
    return min_flips

Why it fails: If the string is already all '1's, we never update min_flips properly. Also doesn't consider making everything '0's.

Correct Solution: Initialize min_flips with total_zeros (cost of making everything '1's). The algorithm naturally handles making everything '0's when it processes the last position.

3. Incorrect Formula for Calculating Flips

A subtle error is using the wrong formula when calculating the number of flips needed at each split point.

Wrong Formula:

def minFlipsMonoIncr(self, s: str) -> int:
    total_zeros = s.count("0")
    min_flips = total_zeros
    zeros_seen = 0
  
    for position, char in enumerate(s, 1):
        if char == "0":
            zeros_seen += 1
        # Wrong: double-counting or using wrong variables
        min_flips = min(min_flips, zeros_seen + (len(s) - position - (total_zeros - zeros_seen)))
  
    return min_flips

Why it fails: The formula doesn't correctly calculate the number of '1's to flip before the split and '0's to flip after the split.

Correct Formula:

• '1's before position i that need flipping: position - zeros_seen
• '0's after position i that need flipping: total_zeros - zeros_seen
• Total: (position - zeros_seen) + (total_zeros - zeros_seen)

4. Not Handling Empty String or Single Character

While the algorithm naturally handles these cases, explicitly checking can prevent unexpected behavior in some implementations.

Better Practice:

def minFlipsMonoIncr(self, s: str) -> int:
    if not s or len(s) <= 1:
        return 0
  
    # Continue with main algorithm...

This makes the code more readable and explicitly shows that edge cases have been considered.