Minimum Suffix Flips

Medium

Asked by:

109 views

Topics:

StringsGreedy Algorithms

You are given a 0-indexed binary string target of length n. You have another binary string s of length n that is initially set to all zeros. You want to make s equal to target.

In one operation, you can pick an index i where 0 <= i < n and flip all bits in the inclusive range [i, n - 1]. Flip means changing '0' to '1' and '1' to '0'.

Return the minimum number of operations needed to make s equal to target.

Example 1:

Input: target = "10111"
Output: 3
Explanation: Initially, s = "00000".
Choose index i = 2: "00000" -> "00111"
Choose index i = 0: "00111" -> "11000"
Choose index i = 1: "11000" -> "10111"
We need at least 3 flip operations to form target.

Example 2:

Input: target = "101"
Output: 3
Explanation: Initially, s = "000".
Choose index i = 0: "000" -> "111"
Choose index i = 1: "111" -> "100"
Choose index i = 2: "100" -> "101"
We need at least 3 flip operations to form target.

Example 3:

Input: target = "00000"
Output: 0
Explanation: We do not need any operations since the initial s already equals target.

Constraints:

n == target.length
1 <= n <= 10⁵
target[i] is either '0' or '1'.

Try coding it in LeetCode

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 are the possible characters in the input string `S`, and what's the maximum length of `S`?
Can the input string `S` be empty or null?
Is the target string always '0' or could it be another character?
If multiple sequences of flips achieve the minimum, is any of them acceptable, or is there a specific criteria for choosing one?
Could you provide an example with the expected output to ensure my understanding of the problem?

Brute Force Solution

Approach

The brute force approach to this puzzle means we'll try every possible combination of flips until we find the minimum. We'll explore each possibility by flipping sections from the end and see what results in the fewest changes.

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

Start by looking at the original arrangement without making any changes.
Then, try flipping the entire last section.
Next, try flipping only the last two sections, and then only the last three, and so on, all the way up to flipping the entire arrangement.
For each of these arrangements, check if it matches our desired target.
If it does, count the number of flips it took to get there.
Keep track of the arrangement that matches the target with the fewest number of flips.
If we've gone through all the possibilities and haven't found one, then there's no solution. Otherwise, the fewest flips we found is our answer.

Code Implementation

def minimum_suffix_flips_brute_force(initial_arrangement, target_arrangement):
    minimum_flips = float('inf')

    # Iterate through all possible suffix flips
    for number_of_flips in range(len(initial_arrangement) + 1):
        current_arrangement = list(initial_arrangement)
        flips_count = 0

        # Perform the suffix flips
        for flip_length in range(number_of_flips):

            # This simulates the flips
            suffix = current_arrangement[len(current_arrangement) - flip_length - 1:]
            suffix.reverse()
            current_arrangement = current_arrangement[:len(current_arrangement) - flip_length - 1] + suffix
            flips_count += 1

        # Check if the arrangement matches the target after each flip
        if current_arrangement == list(target_arrangement):

            # Keep track of the minimum flips
            minimum_flips = min(minimum_flips, flips_count)

    # If no solution is found, return -1
    if minimum_flips == float('inf'):
        return -1
    else:
        return minimum_flips

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through possible suffix flips. For each suffix length from 1 to n, it performs a flip operation and checks if the resulting state matches the target. The flip operation for a suffix of length k takes O(k) time, but since we analyze the highest order behavior, consider it as O(n). Since we iterate through n different suffix lengths and perform an O(n) operation for each, the overall time complexity is approximately n * n. Therefore, the time complexity is O(n²).

Space Complexity

O(N) – The brute force approach described involves creating copies of the input arrangement while exploring different flip combinations. In the worst case, we might create up to N copies, where N is the length of the original arrangement. Each copy represents a potential state after applying a specific sequence of flips from the end. Therefore, the algorithm requires O(N) auxiliary space to store these potentially modified arrangements.

Optimal Solution

Approach

We want to find the fewest switches needed to make all characters in a string the same. The clever trick is to realize that you only need to flip a suffix (the end part) of the string. So, we check for differences and make the necessary flips.

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

Start by looking at the very first character of the string.
Now, go through the string, character by character, comparing each character to the one before it.
Whenever you find a character that's different from the previous one, it means you need to perform a suffix flip.
Count each of these differences. The total count is the minimum number of suffix flips you need to do to make the entire string uniform.

Code Implementation

def minimum_suffix_flips(binary_string):
    flips_needed = 0
    
    # Iterate through the string, tracking differences
    for i in range(1, len(binary_string)): 
        #Check for differences to determine flips
        if binary_string[i] != binary_string[i-1]:

            flips_needed += 1

    # Plus 0 or 1 depending on what is needed to be changed
    if binary_string[-1] == '0':
        flips_needed+=1

    return flips_needed

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the string once, comparing each character to the previous one. The number of operations is directly proportional to the length of the input string (n), where n is the number of characters in the string. Therefore, the time complexity is linear.

Space Complexity

O(1) – The algorithm iterates through the input string, comparing adjacent characters. It uses a single counter to keep track of the number of flips. No additional data structures like arrays, hash maps, or recursion are utilized. Therefore, the space used is constant and independent of the input string's length, N.

Edge Cases

Case	How to Handle
Null or empty string input	Return 0 since no flips are needed for an empty string.
String of length 1	Return 0 since a single character needs no flips.
String with all identical characters ('0000' or '1111')	Return 0 because the string is already uniform.
String with alternating characters ('010101')	Return n-1 if the string alternates between 0 and 1, where n is the string length.
String with leading zeros ('00110')	The algorithm should correctly handle leading zeros by flipping the suffix starting from the first '1'.
String with trailing zeros ('11000')	Trailing zeros might require an extra flip to make the suffix uniform.
Maximum string length to consider potential integer overflow if calculating flips using a running count.	Ensure the data type used for counting flips can handle the maximum string length without overflow; alternatively avoid direct counting by comparing characters.
Very long input string to check time complexity (efficient O(n) solution is required)	Use an algorithm with O(n) time complexity to handle very large strings within the time constraints.

Null or empty string input

How to Handle:

Return 0 since no flips are needed for an empty string.

String of length 1

How to Handle:

Return 0 since a single character needs no flips.

String with all identical characters ('0000' or '1111')

How to Handle:

Return 0 because the string is already uniform.

String with alternating characters ('010101')

How to Handle:

Return n-1 if the string alternates between 0 and 1, where n is the string length.

String with leading zeros ('00110')

How to Handle:

The algorithm should correctly handle leading zeros by flipping the suffix starting from the first '1'.

String with trailing zeros ('11000')

How to Handle:

Trailing zeros might require an extra flip to make the suffix uniform.

Maximum string length to consider potential integer overflow if calculating flips using a running count.

How to Handle:

Ensure the data type used for counting flips can handle the maximum string length without overflow; alternatively avoid direct counting by comparing characters.

Very long input string to check time complexity (efficient O(n) solution is required)

How to Handle:

Use an algorithm with O(n) time complexity to handle very large strings within the time constraints.