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Flip String to Monotone Increasing

Medium

Flip String to Monotone Increasing

Medium

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum length of the binary string `s`?
Can the input string `s` be empty or null?
Are there any specific performance considerations or constraints beyond optimal time complexity?
If there are multiple ways to achieve the minimum flips, is any valid solution acceptable?
The problem states flipping characters, is it an in-place operation or am I allowed to create a new string?

Brute Force Solution

Approach

The problem wants us to change a string of 0s and 1s so it becomes 'monotone increasing', meaning all the 0s come before all the 1s. The brute force way to solve this is to try every possible 'split point' in the string and count the changes needed for each.

Here's how the algorithm would work step-by-step:

Imagine the string is made of tiles, and we are testing different places to put a divider.
First, try putting the divider at the very beginning, meaning everything is 1s and count how many flips it takes.
Next, shift the divider one tile over, so the first tile is now 0 and everything else is 1. Count the needed flips.
Keep shifting the divider one tile at a time, each time counting the number of flips needed to make all the left tiles 0 and all the right tiles 1.
After you've tried every possible place for the divider, see which divider position required the fewest flips.
That's your answer: the fewest flips needed to make the string monotone increasing.

Code Implementation

def flip_string_to_monotone_increasing_brute_force(binary_string):
    minimum_flips = float('inf')

    # Iterate through all possible split points.
    for split_index in range(len(binary_string) + 1):

        flips_needed = 0

        # Count flips needed for the left side to be all 0s.
        for index_left in range(split_index):
            if binary_string[index_left] == '1':
                flips_needed += 1

        # Count flips needed for the right side to be all 1s.
        for index_right in range(split_index, len(binary_string)):
            if binary_string[index_right] == '0':
                flips_needed += 1

        # Update the minimum flips if necessary.
        minimum_flips = min(minimum_flips, flips_needed)

    return minimum_flips

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible split points in the string, which takes O(n) time. For each split point, it counts the number of flips needed to make the left side all 0s and the right side all 1s. Counting the flips requires iterating through the entire string again, which also takes O(n) time. Because we have O(n) iterations of checking split points, and for each iteration, we do O(n) work counting flips, the total time complexity is O(n * n) which simplifies to O(n²).

Space Complexity

O(1) – The provided algorithm iterates through the string and counts flips at each potential split point. It does not use any auxiliary data structures like lists, maps, or sets to store intermediate results. The only space used is for a few integer variables to keep track of the minimum flips seen so far and potentially the index of the best split point. Thus, the space used remains constant, irrespective of the input string's length N.

Optimal Solution

Approach

The goal is to make the string monotone increasing, meaning all the zeros come before all the ones. We want to find the fewest number of flips to achieve this. The key idea is to consider all possible split points between zeros and ones and minimize the number of flips needed for each split.

Here's how the algorithm would work step-by-step:

Imagine drawing a line somewhere in the string. Everything to the left of the line will be forced to be zeros, and everything to the right will be forced to be ones.
For each possible line position, count how many characters on the left side need to be flipped to become zero, and how many characters on the right side need to be flipped to become one.
Add those two counts together to get the total number of flips needed for that line position.
Do this for every possible line position in the string.
Find the line position that gives you the smallest total number of flips. That's your answer.

Code Implementation

def flip_string_to_monotone_increasing(s):
    minimum_flips = float('inf')
    
    for i in range(len(s) + 1):
        #For each split, calculate flips
        left_flips = 0

        for j in range(i):
            if s[j] == '1':
                left_flips += 1

        right_flips = 0

        for j in range(i, len(s)):
            if s[j] == '0':
                right_flips += 1

        total_flips_needed = left_flips + right_flips
        
        #Update the overall minimum
        minimum_flips = min(minimum_flips, total_flips_needed)

    return minimum_flips

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through all possible split points in the string of length n. For each split point, it counts the number of flips needed on the left and right sides. Counting flips on each side takes O(n) time since we examine all the characters on either side of the split. However, a better implementation can compute the prefix sum of ones and suffix sum of zeros beforehand in O(n) time. Then, for each split point, the number of flips can be determined in O(1) time by accessing the precomputed prefix and suffix sums. Since there are n possible split points and calculating the number of flips per split takes O(1) with the precomputed sums, the overall time complexity is dominated by the initial prefix and suffix computation and the iteration through split points, resulting in O(n) + O(n)*O(1) = O(n).

Space Complexity

O(1) – The described algorithm iterates through the string, calculating flips for each possible split point. It only requires a few integer variables to store the minimum number of flips found so far and the current number of flips for a given split. No auxiliary data structures like arrays or hash maps are created, and the depth of any recursion stack is not dependent on the size of the input string (N). Therefore, the auxiliary space used is constant, regardless of the input size N.

Edge Cases

Case	How to Handle
Null or empty input string	Return 0 since an empty string is already monotone increasing.
String of length 1 or 2	Evaluate all possible flips and return the minimum, which can be computed directly.
String with all '0's	Return 0 as the string is already monotone increasing.
String with all '1's	Return 0 as the string is already monotone increasing.
String with alternating '0's and '1's (e.g., '010101')	Calculate flips needed to make it '000...' and '111...' and take the minimum and calculate flips to convert to 000... or 111... and take the minimum.
Maximum length string (scalability)	The solution should have linear time complexity (O(n)), making it efficient for large inputs; dynamic programming or prefix sums can ensure this.
Integer overflow when counting flips	Use an integer type large enough to hold the maximum possible number of flips (e.g., long in Java/C++).
String with a very long sequence of 0s followed by a very long sequence of 1s or vice versa.	DP or prefix sum based solutions efficiently handle such distributions by accurately accumulating the number of flips needed at each step.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum length of the binary string `s`?
Can the input string `s` be empty or null?
Are there any specific performance considerations or constraints beyond optimal time complexity?
If there are multiple ways to achieve the minimum flips, is any valid solution acceptable?
The problem states flipping characters, is it an in-place operation or am I allowed to create a new string?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Imagine the string is made of tiles, and we are testing different places to put a divider.
First, try putting the divider at the very beginning, meaning everything is 1s and count how many flips it takes.
Next, shift the divider one tile over, so the first tile is now 0 and everything else is 1. Count the needed flips.
Keep shifting the divider one tile at a time, each time counting the number of flips needed to make all the left tiles 0 and all the right tiles 1.
After you've tried every possible place for the divider, see which divider position required the fewest flips.
That's your answer: the fewest flips needed to make the string monotone increasing.

Code Implementation

def flip_string_to_monotone_increasing_brute_force(binary_string):
    minimum_flips = float('inf')

    # Iterate through all possible split points.
    for split_index in range(len(binary_string) + 1):

        flips_needed = 0

        # Count flips needed for the left side to be all 0s.
        for index_left in range(split_index):
            if binary_string[index_left] == '1':
                flips_needed += 1

        # Count flips needed for the right side to be all 1s.
        for index_right in range(split_index, len(binary_string)):
            if binary_string[index_right] == '0':
                flips_needed += 1

        # Update the minimum flips if necessary.
        minimum_flips = min(minimum_flips, flips_needed)

    return minimum_flips

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Imagine drawing a line somewhere in the string. Everything to the left of the line will be forced to be zeros, and everything to the right will be forced to be ones.
For each possible line position, count how many characters on the left side need to be flipped to become zero, and how many characters on the right side need to be flipped to become one.
Add those two counts together to get the total number of flips needed for that line position.
Do this for every possible line position in the string.
Find the line position that gives you the smallest total number of flips. That's your answer.

Code Implementation

def flip_string_to_monotone_increasing(s):
    minimum_flips = float('inf')
    
    for i in range(len(s) + 1):
        #For each split, calculate flips
        left_flips = 0

        for j in range(i):
            if s[j] == '1':
                left_flips += 1

        right_flips = 0

        for j in range(i, len(s)):
            if s[j] == '0':
                right_flips += 1

        total_flips_needed = left_flips + right_flips
        
        #Update the overall minimum
        minimum_flips = min(minimum_flips, total_flips_needed)

    return minimum_flips

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input string	Return 0 since an empty string is already monotone increasing.
String of length 1 or 2	Evaluate all possible flips and return the minimum, which can be computed directly.
String with all '0's	Return 0 as the string is already monotone increasing.
String with all '1's	Return 0 as the string is already monotone increasing.
String with alternating '0's and '1's (e.g., '010101')	Calculate flips needed to make it '000...' and '111...' and take the minimum and calculate flips to convert to 000... or 111... and take the minimum.
Maximum length string (scalability)	The solution should have linear time complexity (O(n)), making it efficient for large inputs; dynamic programming or prefix sums can ensure this.
Integer overflow when counting flips	Use an integer type large enough to hold the maximum possible number of flips (e.g., long in Java/C++).
String with a very long sequence of 0s followed by a very long sequence of 1s or vice versa.	DP or prefix sum based solutions efficiently handle such distributions by accurately accumulating the number of flips needed at each step.