Problem Description

The problem presents us with a binary string s which consists only of the characters '0' and '1'. A monotone increasing binary string is defined as one where all '0's appear before any '1's. This means that there is no '1' that comes before a '0' in the string.

However, the input string s may not be monotone increasing already, so we need to transform it into one by flipping some of its characters. Flipping a character means changing a '0' to a '1' or a '1' to a '0'. The objective is to perform the minimum number of such flips so that the resulting string is monotone increasing.

We need to return the smallest number of flips necessary to convert the string s into a monotone increasing string.

Intuition

To solve this problem, we can use a dynamic programming approach to keep track of the minimum flips required to make prefixes and suffixes of the string monotone increasing. The core of the solution revolves around recognizing that for any position in the string, we can split the string into two parts:

• the prefix (all the characters before this position), which we want to turn into all '0's,
• and the suffix (all the characters from this position), which we want to turn into all '1's.

Now, if we know how many flips it would take to turn the prefix into all '0's and the suffix into all '1's, the total flips for the whole string would be the sum of both.

To compute these efficiently, we build two arrays, left and right.

• The left array at position i contains the number of flips to turn all characters from 0 to i-1 into '0's (note this includes the character at position i-1).
• The right array at position i contains the number of flips to turn all characters from i to n-1 into '1's.

With these arrays, we iterate through each position in the string, representing all possible split points, compute the sum of flips from the left and right arrays for that position, and keep track of the minimum sum seen. After going through all the split points, the minimum sum found will be our answer, the least number of flips required to make the string monotone increasing.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution uses dynamic programming to minimize the number of flips needed to make the string monotone increasing. Here’s a step-by-step breakdown of how the provided code accomplishes this task:

1. Initialize two arrays, left and right, both of size n+1, where n is the length of the string s. These arrays will be used to store the cumulative number of '1's in the left array (from the start of the string to the current index) and the cumulative number of '0's in the right array (from the current index to the end of the string), respectively.

2. Populate the left array by iterating from 1 to n+1 (inclusive). For each index i in left, increment the current value by 1 if the corresponding character in the string s[i - 1] is '1'. This gives us the number of '1's that would need to be flipped to '0's if the split is made after index i-1.

3. Populate the right array by iterating from n-1 down to 0 (inclusive). For each index i in right, increment the current value by 1 if the corresponding character in the string s[i] is '0'. This gives us the number of '0's that would need to be flipped to '1's if the split is made before index i.

4. Now, we must determine the optimal split point. Set an initial ans value to a large number, representing the upper bound of the flips (the code uses 0x3F3F3F3F as this value).

5. Iterate through each possible split point (from 0 to n+1 inclusive). At each split point i, compute the total flips required by adding left[i] (flips required for the prefix to become all '0's) and right[i] (flips required for the suffix to become all '1's).

6. Update ans with the minimum between the current ans and the total flips for the current split position i.

7. After completing the iteration through all split points, ans will contain the minimum number of flips required to make the original string s monotone increasing.

The algorithm efficiently uses dynamic programming to compute the answer in O(n) time, where n is the length of the string. It avoids recalculation of the cumulative sums by storing them in the left and right arrays, which is a common technique used in dynamic programming to optimize for time complexity. The code does not use any complicated data structures, relying only on arrays for storage, which makes for an elegant and efficient solution.

Example Walkthrough

Let's consider an example to illustrate the solution approach. Assume we have the binary string s = "00110" which we want to convert into a monotone increasing string using the minimum number of flips.

Step-by-Step Process:

1. Initialize Arrays:

   • Our string s has n = 5 characters, so we initialize left and right arrays of size n+1, which is 6 in this case.

2. Populate the left Array:

   • The left array tracks the cumulative number of flips to turn all characters to '0's. We iterate through the string and our left array will be built as follows:

     Initialize left = [0, 0, 0, 0, 0, 0]
     After processing '0': left = [0, 0, 0, 0, 0, 0]
     After processing '0': left = [0, 0, 0, 0, 0, 0]
     After processing '1': left = [0, 0, 1, 1, 1, 1]
     After processing '1': left = [0, 0, 1, 2, 2, 2]
     After processing '0': left = [0, 0, 1, 2, 2, 2]

3. Populate the right Array:

   • The right array tracks the cumulative number of flips to turn all characters to '1's. We iterate from the end of the string and form our right array as:

     Initialize right = [0, 0, 0, 0, 0, 0]
     After processing '0': right = [1, 1, 1, 1, 1, 0]
     After processing '1': right = [1, 1, 1, 1, 0, 0]
     After processing '1': right = [1, 1, 1, 0, 0, 0]
     After processing '0': right = [1, 1, 0, 0, 0, 0]
     After processing '0': right = [1, 0, 0, 0, 0, 0]

4. Determine Optimal Split Point:

   • We set our initial answer as ans = Infinity (or a very large value) for comparison purposes.
   • We then iterate through the possible split points to calculate the minimum flips:

     At split 0: flips = left[0] + right[0] = 0 + 1 = 1
     At split 1: flips = left[1] + right[1] = 0 + 0 = 0
     At split 2: flips = left[2] + right[2] = 0 + 0 = 0
     At split 3: flips = left[3] + right[3] = 1 + 0 = 1
     At split 4: flips = left[4] + right[4] = 2 + 0 = 2
     At split 5: flips = left[5] + right[5] = 2 + 0 = 2

5. Find Minimum Flips:

   • The minimum number of flips required to make the string monotone increasing occurs at split positions 1 and 2, with 0 flips.
   • Thus, ans = 0.

Therefore, the given string s = "00110" requires 1 flip to become a monotone increasing string. The correct answer is 1.

This example clearly demonstrates that by keeping track of the number of flips for making all characters to the left '0's and all characters to the right '1's for every position, we can easily compute the minimum total number of flips needed by considering all possible split points.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code computes the minimum number of flips needed to make a binary string monotonically increasing using dynamic programming. The main operations are iterating through the string twice and computing the minimum in another pass. Let's break it down:

1. Initializing two arrays, left and right, each of size n + 1, where n is the length of the string s. This takes O(1) time as it's just initializing the lists with zeros.

2. A single for-loop to fill the left array. The loop runs n times (where n is the length of the string s), resulting in a complexity of O(n).

3. Another single for-loop in reverse to fill the right array, which is also O(n).

4. A final loop that goes from 0 to n to find the ans (minimum flips needed). This loop also runs n + 1 times and the min operation inside it takes O(1) time. The total time complexity for this step is O(n).

Therefore, the overall time complexity is the sum of the individual complexities: O(n) + O(n) + O(n) = O(3n). Since constant factors are ignored in big O notation, the final time complexity simplifies to O(n).

Space Complexity

The space complexity is the amount of additional memory space required to execute the code, which depends on the size of the input:

1. Two arrays left and right each of length n + 1 are created. Hence, the space taken by these arrays is 2 * (n + 1).

2. Constant space is used for variables n, ans, and the loop indices, which is O(1).

Thus, the total space complexity is O(2n + 2) which simplifies to O(n) as lower order terms and constant factors are dropped in big O notation.

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