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Minimum Swaps to Make Strings Equal

Medium

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Strings

You are given two strings s1 and s2 of equal length consisting of letters "x" and "y" only. Your task is to make these two strings equal to each other. You can swap any two characters that belong to different strings, which means: swap s1[i] and s2[j].

Return the minimum number of swaps required to make s1 and s2 equal, or return -1 if it is impossible to do so.

Example 1:

Input: s1 = "xx", s2 = "yy"
Output: 1
Explanation: Swap s1[0] and s2[1], s1 = "yx", s2 = "yx".

Example 2:

Input: s1 = "xy", s2 = "yx"
Output: 2
Explanation: Swap s1[0] and s2[0], s1 = "yy", s2 = "xx".
Swap s1[0] and s2[1], s1 = "xy", s2 = "xy".
Note that you cannot swap s1[0] and s1[1] to make s1 equal to "yx", cause we can only swap chars in different strings.

Example 3:

Input: s1 = "xx", s2 = "xy"
Output: -1

Constraints:

1 <= s1.length, s2.length <= 1000
s1.length == s2.length
s1, s2 only contain 'x' or 'y'.

Minimum Swaps to Make Strings Equal

Medium

Asked by:

2 views

Topics:

Strings

Return the minimum number of swaps required to make s1 and s2 equal, or return -1 if it is impossible to do so.

Example 1:

Input: s1 = "xx", s2 = "yy"
Output: 1
Explanation: Swap s1[0] and s2[1], s1 = "yx", s2 = "yx".

Example 2:

Input: s1 = "xy", s2 = "yx"
Output: 2
Explanation: Swap s1[0] and s2[0], s1 = "yy", s2 = "xx".
Swap s1[0] and s2[1], s1 = "xy", s2 = "xy".
Note that you cannot swap s1[0] and s1[1] to make s1 equal to "yx", cause we can only swap chars in different strings.

Example 3:

Input: s1 = "xx", s2 = "xy"
Output: -1

Constraints:

1 <= s1.length, s2.length <= 1000
s1.length == s2.length
s1, s2 only contain 'x' or 'y'.

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 expected behavior if the input strings, `s1` and `s2`, have different lengths?
Can the input strings, `s1` and `s2`, contain characters other than 'x' and 'y'?
If it's impossible to make the strings equal through swaps, what value should I return?
Can I assume that both input strings `s1` and `s2` will be non-null and non-empty?
If there are multiple valid sequences of minimum swaps, is any one acceptable?

Brute Force Solution

Approach

The brute force method involves exploring every possible combination of swaps between the two strings. We will systematically generate each possible sequence of swaps and evaluate whether the strings become equal. This guarantees finding the solution, although it may take a very long time.

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

Start by considering no swaps at all. Check if the strings are equal already.
If they aren't, consider swapping the first characters of the two strings and see if that makes them equal. If it does, we are done. If not, undo the swap.
Then, try swapping the first character of the first string with the second character of the second string. Again, check for equality and undo the swap if necessary.
Continue swapping the first character of the first string with every other character of the second string, each time checking if the strings are equal and undoing the swap.
After you've tried all swaps involving the first character, move on to the second character of the first string and repeat the same process of swapping it with every character of the second string.
Keep repeating this process, trying all possible single swaps, then all possible pairs of swaps, then triplets, and so on. With each series of swaps, check if the strings are equal.
Remember that each time you try a swap (or a series of swaps), you need to undo it to explore the other possibilities.
Continue this exhaustive process until you have tried all possible combinations of swaps, or you find a sequence of swaps that makes the strings equal. Keep track of the minimum number of swaps needed.
If, after exploring all possibilities, you haven't found any swaps that make the strings equal, then it's impossible to make them equal with any number of swaps.

Code Implementation

def minimum_swaps_to_make_strings_equal_brute_force(first_string, second_string):
    minimum_swaps = float('inf')
    string_length = len(first_string)

    def strings_are_equal(current_first_string, current_second_string):
        return current_first_string == current_second_string

    def explore_swaps(current_first_string, current_second_string, swap_count):
        nonlocal minimum_swaps

        if strings_are_equal(current_first_string, current_second_string):
            minimum_swaps = min(minimum_swaps, swap_count)
            return

        if swap_count >= minimum_swaps:
            return

        # Limit the search if we already found better result

        for first_string_index in range(string_length):
            for second_string_index in range(string_length):
                # Try swapping characters at these indices
                first_string_list = list(current_first_string)
                second_string_list = list(current_second_string)

                first_string_list[first_string_index], second_string_list[second_string_index] = \
                    second_string_list[second_string_index], first_string_list[first_string_index]

                new_first_string = "".join(first_string_list)
                new_second_string = "".join(second_string_list)

                explore_swaps(new_first_string, new_second_string, swap_count + 1)

    # Start the recursive exploration with no swaps
    explore_swaps(first_string, second_string, 0)

    if minimum_swaps == float('inf'):
        return -1
    else:
        return minimum_swaps

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n!) – The brute force approach considers all possible combinations of swaps. In the worst-case scenario, we might need to explore all possible permutations of swaps, which grows factorially. Specifically, if the string has length n, and we consider up to n swaps, the number of combinations becomes incredibly large. Thus, the time complexity is dominated by the factorial growth of possible swap combinations, approximating O(n!).

Space Complexity

O(N!) – The brute force method described explores every possible combination of swaps. While the algorithm may not explicitly create large auxiliary data structures like arrays or hash maps to store all swap combinations simultaneously, the recursive calls required to explore all possible swap sequences lead to a significant space overhead. In the worst-case, the recursion depth can grow factorially with the input size N, where N represents the length of the strings. Each recursive call consumes memory on the call stack for function arguments, local variables, and the return address. Therefore, the auxiliary space complexity is dominated by the recursion depth, resulting in O(N!).

Optimal Solution

Approach

The most efficient approach recognizes patterns in the mismatched characters to minimize the number of swaps. Instead of blindly swapping, we count how many times certain mismatched pairs appear and use that information to swap strategically.

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

First, count the number of times we see a pair where the first string has a specific character and the second string has a different specific character. Let's say we count the occurrences of 'xy' pairs (first string has x, second has y) and 'yx' pairs (first string has y, second has x).
If the number of 'xy' pairs is even, we can fix all of these mismatches with half that many swaps. Each swap corrects two 'xy' pairs.
If the number of 'yx' pairs is also even, we can fix all of these mismatches with half that many swaps. Each swap corrects two 'yx' pairs.
Now, if either the 'xy' or 'yx' counts are odd, we'll need an extra swap to connect an 'xy' pair to a 'yx' pair.
Finally, add up all the swaps we need from the even counts and the extra swap (if any) from the odd counts. This gives you the minimum number of swaps.

Code Implementation

def minimum_swaps_to_make_strings_equal(string1, string2):
    xy_count = 0
    yx_count = 0

    for i in range(len(string1)):
        if string1[i] == 'x' and string2[i] == 'y':
            xy_count += 1
        elif string1[i] == 'y' and string2[i] == 'x':
            yx_count += 1

    # If the sum of mismatched pairs is odd, it is impossible.
    if (xy_count + yx_count) % 2 != 0:
        return -1

    swaps = 0

    # Count swaps needed for even xy and yx pairs.
    swaps += xy_count // 2
    swaps += yx_count // 2

    # Add an extra swap if both counts are odd.
    if xy_count % 2 != 0 and yx_count % 2 != 0: 
        swaps += 1

    return swaps

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the strings once to count the occurrences of 'xy' and 'yx' pairs, where n is the length of the strings. Counting 'xy' and 'yx' pairs involves checking each character once. The subsequent calculations (division, checking for odd numbers, and addition) take constant time. Therefore, the time complexity is determined by the single linear pass through the strings, making it O(n).

Space Complexity

O(1) – The algorithm uses a constant amount of extra space. It only needs to store the counts of 'xy' and 'yx' pairs. These counts are stored in variables that require constant space, regardless of the length of the input strings (N). Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty strings s1 and s2	Return -1 if either string is null or empty as no comparison or swaps can be performed.
Strings s1 and s2 have different lengths	Return -1 if the strings have different lengths as they cannot be made equal with swaps.
Strings s1 and s2 are identical	Return 0 because no swaps are required to make them equal.
Input strings are very large (close to memory limits)	The solution's linear time and constant space complexity should handle this efficiently if within memory constraints, but consider alternative streaming solutions for extremely large input.
Strings contain characters other than 'x' and 'y'	The solution should validate input strings to contain only 'x' and 'y' characters; otherwise return an error code or throw exception.
Uneven distribution of 'x' and 'y' such that the total number of 'x's and 'y's are not both even.	If the count of 'x' and 'y' mismatches are both odd, return -1 as a solution is not possible.
Integer overflow when calculating swaps if inputs are excessively large	Use data types that can accommodate large numbers, or implement checks to prevent overflow during calculations.
Cases with extremely skewed distributions of 'x' and 'y' differences between the strings	The constant space solution ensures that performance isn't affected dramatically, as it avoids excessive memory usage regardless of skewed distribution.

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 expected behavior if the input strings, `s1` and `s2`, have different lengths?
Can the input strings, `s1` and `s2`, contain characters other than 'x' and 'y'?
If it's impossible to make the strings equal through swaps, what value should I return?
Can I assume that both input strings `s1` and `s2` will be non-null and non-empty?
If there are multiple valid sequences of minimum swaps, is any one acceptable?

Brute Force Solution

Approach

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

Start by considering no swaps at all. Check if the strings are equal already.
If they aren't, consider swapping the first characters of the two strings and see if that makes them equal. If it does, we are done. If not, undo the swap.
Then, try swapping the first character of the first string with the second character of the second string. Again, check for equality and undo the swap if necessary.
Continue swapping the first character of the first string with every other character of the second string, each time checking if the strings are equal and undoing the swap.
After you've tried all swaps involving the first character, move on to the second character of the first string and repeat the same process of swapping it with every character of the second string.
Keep repeating this process, trying all possible single swaps, then all possible pairs of swaps, then triplets, and so on. With each series of swaps, check if the strings are equal.
Remember that each time you try a swap (or a series of swaps), you need to undo it to explore the other possibilities.
Continue this exhaustive process until you have tried all possible combinations of swaps, or you find a sequence of swaps that makes the strings equal. Keep track of the minimum number of swaps needed.
If, after exploring all possibilities, you haven't found any swaps that make the strings equal, then it's impossible to make them equal with any number of swaps.

Code Implementation

def minimum_swaps_to_make_strings_equal_brute_force(first_string, second_string):
    minimum_swaps = float('inf')
    string_length = len(first_string)

    def strings_are_equal(current_first_string, current_second_string):
        return current_first_string == current_second_string

    def explore_swaps(current_first_string, current_second_string, swap_count):
        nonlocal minimum_swaps

        if strings_are_equal(current_first_string, current_second_string):
            minimum_swaps = min(minimum_swaps, swap_count)
            return

        if swap_count >= minimum_swaps:
            return

        # Limit the search if we already found better result

        for first_string_index in range(string_length):
            for second_string_index in range(string_length):
                # Try swapping characters at these indices
                first_string_list = list(current_first_string)
                second_string_list = list(current_second_string)

                first_string_list[first_string_index], second_string_list[second_string_index] = \
                    second_string_list[second_string_index], first_string_list[first_string_index]

                new_first_string = "".join(first_string_list)
                new_second_string = "".join(second_string_list)

                explore_swaps(new_first_string, new_second_string, swap_count + 1)

    # Start the recursive exploration with no swaps
    explore_swaps(first_string, second_string, 0)

    if minimum_swaps == float('inf'):
        return -1
    else:
        return minimum_swaps

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, count the number of times we see a pair where the first string has a specific character and the second string has a different specific character. Let's say we count the occurrences of 'xy' pairs (first string has x, second has y) and 'yx' pairs (first string has y, second has x).
If the number of 'xy' pairs is even, we can fix all of these mismatches with half that many swaps. Each swap corrects two 'xy' pairs.
If the number of 'yx' pairs is also even, we can fix all of these mismatches with half that many swaps. Each swap corrects two 'yx' pairs.
Now, if either the 'xy' or 'yx' counts are odd, we'll need an extra swap to connect an 'xy' pair to a 'yx' pair.
Finally, add up all the swaps we need from the even counts and the extra swap (if any) from the odd counts. This gives you the minimum number of swaps.

Code Implementation

def minimum_swaps_to_make_strings_equal(string1, string2):
    xy_count = 0
    yx_count = 0

    for i in range(len(string1)):
        if string1[i] == 'x' and string2[i] == 'y':
            xy_count += 1
        elif string1[i] == 'y' and string2[i] == 'x':
            yx_count += 1

    # If the sum of mismatched pairs is odd, it is impossible.
    if (xy_count + yx_count) % 2 != 0:
        return -1

    swaps = 0

    # Count swaps needed for even xy and yx pairs.
    swaps += xy_count // 2
    swaps += yx_count // 2

    # Add an extra swap if both counts are odd.
    if xy_count % 2 != 0 and yx_count % 2 != 0: 
        swaps += 1

    return swaps

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty strings s1 and s2	Return -1 if either string is null or empty as no comparison or swaps can be performed.
Strings s1 and s2 have different lengths	Return -1 if the strings have different lengths as they cannot be made equal with swaps.
Strings s1 and s2 are identical	Return 0 because no swaps are required to make them equal.
Input strings are very large (close to memory limits)	The solution's linear time and constant space complexity should handle this efficiently if within memory constraints, but consider alternative streaming solutions for extremely large input.
Strings contain characters other than 'x' and 'y'	The solution should validate input strings to contain only 'x' and 'y' characters; otherwise return an error code or throw exception.
Uneven distribution of 'x' and 'y' such that the total number of 'x's and 'y's are not both even.	If the count of 'x' and 'y' mismatches are both odd, return -1 as a solution is not possible.
Integer overflow when calculating swaps if inputs are excessively large	Use data types that can accommodate large numbers, or implement checks to prevent overflow during calculations.
Cases with extremely skewed distributions of 'x' and 'y' differences between the strings	The constant space solution ensures that performance isn't affected dramatically, as it avoids excessive memory usage regardless of skewed distribution.