Minimum Number of Flips to Make the Binary String Alternating

Medium

Asked by:

20 views

Topics:

StringsGreedy Algorithms

You are given a binary string s. You are allowed to perform two types of operations on the string in any sequence:

Type-1: Remove the character at the start of the string s and append it to the end of the string.
Type-2: Pick any character in s and flip its value, i.e., if its value is '0' it becomes '1' and vice-versa.

Return the minimum number of type-2 operations you need to perform such that s becomes alternating.

The string is called alternating if no two adjacent characters are equal.

For example, the strings "010" and "1010" are alternating, while the string "0100" is not.

Example 1:

Input: s = "111000"
Output: 2
Explanation: Use the first operation two times to make s = "100011".
Then, use the second operation on the third and sixth elements to make s = "101010".

Example 2:

Input: s = "010"
Output: 0
Explanation: The string is already alternating.

Example 3:

Input: s = "1110"
Output: 1
Explanation: Use the second operation on the second element to make s = "1010".

Constraints:

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

Try coding it in LeetCode

Minimum Number of Flips to Make the Binary String Alternating

Medium

Asked by:

20 views

Topics:

StringsGreedy Algorithms

You are given a binary string s. You are allowed to perform two types of operations on the string in any sequence:

Type-1: Remove the character at the start of the string s and append it to the end of the string.
Type-2: Pick any character in s and flip its value, i.e., if its value is '0' it becomes '1' and vice-versa.

Return the minimum number of type-2 operations you need to perform such that s becomes alternating.

The string is called alternating if no two adjacent characters are equal.

For example, the strings "010" and "1010" are alternating, while the string "0100" is not.

Example 1:

Input: s = "111000"
Output: 2
Explanation: Use the first operation two times to make s = "100011".
Then, use the second operation on the third and sixth elements to make s = "101010".

Example 2:

Input: s = "010"
Output: 0
Explanation: The string is already alternating.

Example 3:

Input: s = "1110"
Output: 1
Explanation: Use the second operation on the second element to make s = "1010".

Constraints:

1 <= s.length <= 10⁵
s[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 is the maximum possible length of the input string?
Can the input string be empty or null? If so, what should be returned?
Is the input string guaranteed to contain only '0' and '1' characters?
If there are multiple ways to achieve the minimum number of flips, is any one acceptable, or is there a specific order I should follow when exploring possibilities?
By 'alternating', do you strictly mean '0101...' or '1010...' or are both acceptable alternating patterns?

Brute Force Solution

Approach

The brute force method for this problem involves exploring every possible way to make the binary string alternate. We will consider two target alternating patterns and count the flips needed for each.

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

Imagine the string should start with a '0'.
Go through the string, and for each position, check if the digit matches what it should be in the alternating pattern (starting with '0').
If it doesn't match, count it as a flip.
Now, imagine the string should start with a '1'.
Again, go through the string, and for each position, check if the digit matches what it should be in this new alternating pattern (starting with '1').
If it doesn't match, count it as a flip.
Compare the number of flips needed for both patterns (starting with '0' and starting with '1').
The smaller of these two counts is the minimum number of flips needed to make the string alternating.

Code Implementation

def min_flips_to_make_alternating(binary_string):
    string_length = len(binary_string)

    # Calculate flips needed starting with '0'
    flips_starting_with_zero = 0
    for index in range(string_length):
        expected_char = '0' if index % 2 == 0 else '1'
        if binary_string[index] != expected_char:

            # Increment count because the character is different
            flips_starting_with_zero += 1

    # Calculate flips needed starting with '1'
    flips_starting_with_one = 0
    for index in range(string_length):
        expected_char = '1' if index % 2 == 0 else '0'
        if binary_string[index] != expected_char:

            # Increment count because the character is different
            flips_starting_with_one += 1

    # Return the minimum of the two flip counts
    return min(flips_starting_with_zero, flips_starting_with_one)

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the string twice. The first pass simulates the alternating pattern starting with '0', and the second pass simulates the pattern starting with '1'. Each pass involves checking each of the n characters in the input string 's' against the expected alternating pattern character at that position. Because we perform a constant number of operations for each character in each pass, the total number of operations is proportional to n. Thus, the time complexity is O(n).

Space Complexity

O(1) – The algorithm only uses a few constant space variables: counters for flips needed when starting with '0' and starting with '1'. No auxiliary data structures like arrays or hash maps are created. The space used does not scale with the input string's length, N. Therefore, the space complexity is constant.

Optimal Solution

Approach

The key is to realize we only have two possible alternating patterns. We can efficiently calculate the number of flips needed for each pattern and then take the minimum. No complex search is needed.

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

Imagine the ideal alternating string starts with a zero. Count how many positions in the given string don't match this ideal pattern.
Now, imagine the ideal alternating string starts with a one. Count how many positions in the given string don't match this pattern.
The minimum of these two counts is the fewest number of flips required.

Code Implementation

def min_flips(binary_string):
    string_length = len(binary_string)

    flips_starting_with_zero = 0
    flips_starting_with_one = 0

    # Iterate through the string and count flips for each pattern.
    for i in range(string_length):
        if i % 2 == 0:
            # Checking for the pattern starting with 0.
            if binary_string[i] == '1':
                flips_starting_with_zero += 1

            # Checking for the pattern starting with 1.
            if binary_string[i] == '0':
                flips_starting_with_one += 1
        else:
            # Checking for the pattern starting with 0.
            if binary_string[i] == '0':
                flips_starting_with_zero += 1

            # Checking for the pattern starting with 1.
            if binary_string[i] == '1':
                flips_starting_with_one += 1

    # Compare number of flips for each case.
    return min(flips_starting_with_zero, flips_starting_with_one)

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input string of length n twice. The first iteration calculates the number of flips needed if the alternating string starts with '0', and the second iteration calculates the number of flips needed if the alternating string starts with '1'. Since both iterations are independent and each visits every element in the string once, the time complexity is proportional to n. Therefore, the overall time complexity is O(n).

Space Complexity

O(1) – The algorithm calculates mismatches against two fixed alternating patterns. It only uses a constant number of integer variables (specifically, two counters to track the number of flips needed for each alternating pattern) irrespective of the input string's length, N. No auxiliary data structures like arrays or hash maps are created to store intermediate results related to N. Therefore, the auxiliary space required remains constant, leading to O(1) space complexity.

Edge Cases

Case	How to Handle
Null or empty input string	Return 0 if the string is null or empty, as there are no flips needed to make an empty string alternating.
String of length 1	Return 0 if the string has only one character, as it is already considered alternating.
String with all 0s or all 1s	The algorithm correctly calculates the minimum flips needed to make the string alternating even with identical characters.
Very long input string (performance)	The linear time complexity O(n) of the algorithm makes it efficient for large input strings.
String already alternating (best case)	The algorithm correctly identifies that no flips are needed and returns 0 in the best-case scenario.
Integer overflow if the string is extremely long and requires many flips	Since the number of flips cannot exceed the length of the string, standard integer types are sufficient.
String contains non-binary characters	Throw an IllegalArgumentException if the string contains non-binary characters to maintain input validity.
Equal number of flips needed for both alternating patterns.	The algorithm will return one of the optimal solutions, and the problem asks for the minimum flips not all minimum flips.