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Minimum Number of Swaps to Make the Binary String Alternating

Medium

Amazon

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Topics:

StringsGreedy Algorithms

Given a binary string s, return the minimum number of character swaps to make it alternating, or -1 if it is impossible. The string is called alternating if no two adjacent characters are equal. Any two characters may be swapped, even if they are not adjacent.

For example:

s = "111000" should return 1. We can swap positions 1 and 4: "111000" -> "101010".
s = "010" should return 0. The string is already alternating.
s = "1110" should return -1. It's not possible to make this string alternating.

Write a function to solve this problem efficiently. Consider edge cases and explain the time and space complexity of your solution.

Minimum Number of Swaps to Make the Binary String Alternating

Medium

Amazon

5 views

Topics:

StringsGreedy Algorithms

For example:

s = "111000" should return 1. We can swap positions 1 and 4: "111000" -> "101010".
s = "010" should return 0. The string is already alternating.
s = "1110" should return -1. It's not possible to make this string alternating.

Write a function to solve this problem efficiently. Consider edge cases and explain the time and space complexity of your solution.

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 string? Is it guaranteed to contain only '0' and '1'?
What is the maximum length of the binary string? Is there a limit on the number of characters?
If the input string is already alternating, should I return 0?
If it's impossible to make the string alternating through swaps, what value should I return? For example, should I return -1?
By 'alternating', do you mean that adjacent characters must be different? For example, either '0101...' or '1010...' are considered alternating?

Brute Force Solution

Approach

The brute force method for this problem involves checking every single possible arrangement of the binary string to see if it alternates. We want to find out, for each possible arrangement, how many swaps it takes to get to the alternating pattern and then pick the arrangement that takes the fewest swaps.

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

Consider all possible arrangements for alternating binary strings: one starting with 0 and the other starting with 1.
For the arrangement that starts with 0, compare it to our original string and count how many positions are different. These positions are where we'll need to swap numbers.
Similarly, for the arrangement that starts with 1, compare it to our original string and count how many positions are different. These positions need swaps.
The number of swaps is half the number of different positions we found in each case because each swap corrects two positions.
Find the minimum number of swaps between the two arrangements (starting with 0 and starting with 1). That's our answer!

Code Implementation

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

    # Generate alternating strings starting with 0 and 1
    start_zero_string = "".join(['0' if i % 2 == 0 else '1' for i in range(string_length)])
    start_one_string = "".join(['1' if i % 2 == 0 else '0' for i in range(string_length)])

    # Count differences for string starting with 0
    diff_zero_count = 0
    for i in range(string_length):
        if binary_string[i] != start_zero_string[i]:
            diff_zero_count += 1

    # Count differences for string starting with 1
    diff_one_count = 0
    for i in range(string_length):
        if binary_string[i] != start_one_string[i]:
            diff_one_count += 1

    # Calculate swaps (half the differences)
    swaps_zero = diff_zero_count // 2
    swaps_one = diff_one_count // 2

    # Return the minimum swaps
    return min(swaps_zero, swaps_one)

Big(O) Analysis

Time Complexity

O(n) – The algorithm considers two possible alternating binary strings (one starting with 0, the other with 1). For each of these strings, it iterates through the original string of length n, comparing each character to the corresponding character in the alternating string. This comparison takes O(1) time per character. Therefore, each comparison operation takes O(n) time. Since we perform this comparison twice (once for each alternating string), the total time complexity remains O(n).

Space Complexity

O(1) – The described algorithm calculates the number of differences between the input string and two possible alternating strings (starting with 0 and starting with 1). It does this by iterating through the input string, comparing each character without storing any additional data structures dependent on the input size N. Only a few constant-size variables are used to keep track of the number of differences, which are integers. Therefore, the auxiliary space used does not depend on the length of the input string, resulting in constant space complexity.

Optimal Solution

Approach

To find the fewest swaps to make a binary string alternating, we avoid checking every possibility. Instead, we consider only two alternating patterns and count how many swaps each would need. The approach then picks the pattern requiring the fewer swaps.

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

Imagine the string should start with a zero, creating the pattern '010101...'.
Count how many digits are in the wrong place to match this '010101...' pattern. This tells us how many swaps it would take to achieve this pattern.
Now, imagine the string should start with a one, creating the pattern '101010...'.
Count how many digits are in the wrong place to match this '101010...' pattern. This tells us how many swaps it would take to achieve this pattern.
Compare the number of swaps needed for the '010101...' pattern versus the '101010...' pattern.
Choose the smaller number of swaps, as that represents the fewest changes needed to make the string alternating.

Code Implementation

def min_swaps_to_make_alternating(binary_string):
    string_length = len(binary_string)
    swaps_starting_with_zero = 0
    swaps_starting_with_one = 0

    # Count swaps needed to make the string start with '0'.
    for i in range(string_length):
        if i % 2 == 0:
            if binary_string[i] == '1':
                swaps_starting_with_zero += 1
        else:
            if binary_string[i] == '0':
                swaps_starting_with_zero += 1

    # Count swaps needed to make the string start with '1'.
    for i in range(string_length):
        if i % 2 == 0:
            if binary_string[i] == '0':
                swaps_starting_with_one += 1
        else:
            if binary_string[i] == '1':
                swaps_starting_with_one += 1

    # Find the minimum swaps between the two possible alternating strings
    return min(swaps_starting_with_zero, swaps_starting_with_one)

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the string twice, once to compare it against the '010101...' pattern and once to compare it against the '101010...' pattern. Each of these iterations involves checking each character in the string of length n. Therefore, the dominant operation is proportional to n, resulting in a linear time complexity of O(n).

Space Complexity

O(1) – The algorithm calculates the number of swaps required for two alternating patterns ('010101...' and '101010...') by iterating through the input string. It uses a fixed number of integer variables to store the counts of misplaced digits for each pattern. The space used by these variables does not depend on the input string's length (N), thus the auxiliary space complexity is constant.

Edge Cases

Case	How to Handle
Empty string input	Return 0 since an empty string is already considered alternating.
String with only one character	Return 0 because a single-character string is already alternating.
String with all '0' characters	Calculate swaps to make it '1010...' and '0101...' then return the minimum.
String with all '1' characters	Calculate swaps to make it '1010...' and '0101...' then return the minimum.
String already alternating ('101010...')	Return 0 as no swaps are needed.
Very long string (performance considerations)	Ensure the algorithm has O(n) time complexity to efficiently handle long strings.
String of even length that cannot be perfectly alternated	The algorithm calculates the minimum swaps required for both alternating patterns, providing a best-case result even if perfect alternation is impossible.
String with a mix of '0' and '1' where one character is dominant	The solution considers both '0101...' and '1010...' patterns, accommodating skewed distributions.

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 string? Is it guaranteed to contain only '0' and '1'?
What is the maximum length of the binary string? Is there a limit on the number of characters?
If the input string is already alternating, should I return 0?
If it's impossible to make the string alternating through swaps, what value should I return? For example, should I return -1?
By 'alternating', do you mean that adjacent characters must be different? For example, either '0101...' or '1010...' are considered alternating?

Brute Force Solution

Approach

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

Consider all possible arrangements for alternating binary strings: one starting with 0 and the other starting with 1.
For the arrangement that starts with 0, compare it to our original string and count how many positions are different. These positions are where we'll need to swap numbers.
Similarly, for the arrangement that starts with 1, compare it to our original string and count how many positions are different. These positions need swaps.
The number of swaps is half the number of different positions we found in each case because each swap corrects two positions.
Find the minimum number of swaps between the two arrangements (starting with 0 and starting with 1). That's our answer!

Code Implementation

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

    # Generate alternating strings starting with 0 and 1
    start_zero_string = "".join(['0' if i % 2 == 0 else '1' for i in range(string_length)])
    start_one_string = "".join(['1' if i % 2 == 0 else '0' for i in range(string_length)])

    # Count differences for string starting with 0
    diff_zero_count = 0
    for i in range(string_length):
        if binary_string[i] != start_zero_string[i]:
            diff_zero_count += 1

    # Count differences for string starting with 1
    diff_one_count = 0
    for i in range(string_length):
        if binary_string[i] != start_one_string[i]:
            diff_one_count += 1

    # Calculate swaps (half the differences)
    swaps_zero = diff_zero_count // 2
    swaps_one = diff_one_count // 2

    # Return the minimum swaps
    return min(swaps_zero, swaps_one)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Imagine the string should start with a zero, creating the pattern '010101...'.
Count how many digits are in the wrong place to match this '010101...' pattern. This tells us how many swaps it would take to achieve this pattern.
Now, imagine the string should start with a one, creating the pattern '101010...'.
Count how many digits are in the wrong place to match this '101010...' pattern. This tells us how many swaps it would take to achieve this pattern.
Compare the number of swaps needed for the '010101...' pattern versus the '101010...' pattern.
Choose the smaller number of swaps, as that represents the fewest changes needed to make the string alternating.

Code Implementation

def min_swaps_to_make_alternating(binary_string):
    string_length = len(binary_string)
    swaps_starting_with_zero = 0
    swaps_starting_with_one = 0

    # Count swaps needed to make the string start with '0'.
    for i in range(string_length):
        if i % 2 == 0:
            if binary_string[i] == '1':
                swaps_starting_with_zero += 1
        else:
            if binary_string[i] == '0':
                swaps_starting_with_zero += 1

    # Count swaps needed to make the string start with '1'.
    for i in range(string_length):
        if i % 2 == 0:
            if binary_string[i] == '0':
                swaps_starting_with_one += 1
        else:
            if binary_string[i] == '1':
                swaps_starting_with_one += 1

    # Find the minimum swaps between the two possible alternating strings
    return min(swaps_starting_with_zero, swaps_starting_with_one)

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty string input	Return 0 since an empty string is already considered alternating.
String with only one character	Return 0 because a single-character string is already alternating.
String with all '0' characters	Calculate swaps to make it '1010...' and '0101...' then return the minimum.
String with all '1' characters	Calculate swaps to make it '1010...' and '0101...' then return the minimum.
String already alternating ('101010...')	Return 0 as no swaps are needed.
Very long string (performance considerations)	Ensure the algorithm has O(n) time complexity to efficiently handle long strings.
String of even length that cannot be perfectly alternated	The algorithm calculates the minimum swaps required for both alternating patterns, providing a best-case result even if perfect alternation is impossible.
String with a mix of '0' and '1' where one character is dominant	The solution considers both '0101...' and '1010...' patterns, accommodating skewed distributions.