Apply Operations to Make Two Strings Equal

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StringsDynamic Programming

You are given two 0-indexed binary strings s1 and s2, both of length n, and a positive integer x.

You can perform any of the following operations on the string s1 any number of times:

Choose two indices i and j, and flip both s1[i] and s1[j]. The cost of this operation is x.
Choose an index i such that i < n - 1 and flip both s1[i] and s1[i + 1]. The cost of this operation is 1.

Return the minimum cost needed to make the strings s1 and s2 equal, or return -1 if it is impossible.

Note that flipping a character means changing it from 0 to 1 or vice-versa.

Example 1:

Input: s1 = "1100011000", s2 = "0101001010", x = 2
Output: 4
Explanation: We can do the following operations:
- Choose i = 3 and apply the second operation. The resulting string is s1 = "1101111000".
- Choose i = 4 and apply the second operation. The resulting string is s1 = "1101001000".
- Choose i = 0 and j = 8 and apply the first operation. The resulting string is s1 = "0101001010" = s2.
The total cost is 1 + 1 + 2 = 4. It can be shown that it is the minimum cost possible.

Example 2:

Input: s1 = "10110", s2 = "00011", x = 4
Output: -1
Explanation: It is not possible to make the two strings equal.

Constraints:

n == s1.length == s2.length
1 <= n, x <= 500
s1 and s2 consist only of the characters '0' and '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 characters can the strings contain; are they limited to the English alphabet, or can they include Unicode characters or numbers?
Are the two input strings guaranteed to be of the same length, or can they have different lengths?
If the strings are not equal and no series of operations can make them equal, what should I return?
What constitutes an operation? Are there any constraints on what kinds of character changes are allowed, or is it simply changing one character to another?
Are there any limits to the number of operations that can be performed to make the two strings equal?

Brute Force Solution

Approach

The brute force approach to making two strings equal involves trying every possible combination of operations to see if we can transform one string into the other. We essentially explore every single path, even the inefficient ones, until we find a solution. This approach guarantees finding the answer if it exists, but it might take a very long time.

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

Consider all the possible positions where we could apply an operation.
For each position, try applying the operation, and see how that changes the first string.
After applying an operation, compare the modified first string to the second string.
If they are equal, we've found a solution!
If they are not equal, try applying another operation at a different position (or even the same position again) in the modified string.
Keep trying different combinations of operations until we either find a solution (the strings are equal) or we've exhausted all possible combinations.
If we've tried every single possible operation combination and the strings are still not equal, then it's impossible to make them equal.

Code Implementation

def apply_operations_brute_force(
    first_string, second_string
):    if len(first_string) != len(second_string):        return False    queue = [(first_string, [])]
    visited = {first_string}    while queue:
        current_string, operations = queue.pop(0)        if current_string == second_string:
            return True        for index in range(len(current_string) - 1):
            modified_string = (                current_string[:index]                + ('1' if current_string[index] == '0' else '0')                + ('1' if current_string[index + 1] == '0' else '0')                + current_string[index + 2:]            )
            # Only explore if this string hasn't been visited yet            if modified_string not in visited:
                visited.add(modified_string)
                queue.append((modified_string, operations + [(index, index + 1)]))
    # If the queue is empty, we have exhausted every possibility    return False

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach involves exploring all possible combinations of operations at each position in the string of length n. At each position, we have the option to apply an operation or not, resulting in 2 choices per position. Therefore, the total number of possible combinations of operations we need to consider is 2 * 2 * ... * 2 (n times), which equals 2^n. In the worst case, we might need to try all these combinations before finding a solution or determining that no solution exists. Thus, the time complexity is O(2^n).

Space Complexity

O(N!) – The brute force approach explores all possible combinations of operations. In the worst-case scenario, the algorithm may need to maintain a recursion stack proportional to the number of possible operation combinations. Since we are considering all possible operations at each position, the space used by the call stack could potentially grow factorially with respect to the length of the string, N. Although this is an overestimate, due to the difficulty of precisely quantifying the number of possible operation sequences before termination (if any) is reached, we err on the side of caution. The algorithm will continue to modify the strings, and try again until it finds the solution, this leads to O(N!) space complexity.

Optimal Solution

Approach

The core idea is to focus on identifying the differences between the two strings. We aim to minimize the number of operations by grouping adjacent differing positions together, since a single operation can correct two adjacent differences.

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

First, identify all positions where the characters in the two strings are different.
Then, try to pair up adjacent differing positions. If you find two differences next to each other, you can correct them with one operation, so that's the most efficient thing to do.
If there's a difference that cannot be paired with an adjacent difference (because it's at the beginning or end, or surrounded by matching characters), then that difference will require its own operation.
The minimum number of operations is equal to the number of pairs you formed, plus the number of unpaired differences.

Code Implementation

def apply_operations(string1, string2):
    differences = []
    for index in range(len(string1)):
        if string1[index] != string2[index]:
            differences.append(index)

    operations_count = 0
    index = 0
    while index < len(differences):
        # Try to pair adjacent differing indices
        if index + 1 < len(differences) and differences[index + 1] == differences[index] + 1:
            operations_count += 1
            index += 2

        else:
            # If a difference cannot be paired, it needs its own operation.
            operations_count += 1
            index += 1

    return operations_count

Big(O) Analysis

Time Complexity

O(n) – The primary operation involves iterating through the strings of length n to identify differing positions. The pairing of adjacent differing positions also takes at most O(n) time since each position is examined at most a constant number of times. Therefore, the dominant factor is the single pass through the strings, resulting in a time complexity of O(n).

Space Complexity

O(N) – The space complexity is primarily determined by the need to store the indices where the two strings differ. In the worst-case scenario, where all characters in the two strings are different, we would need to store all N indices in a list. Therefore, the auxiliary space required grows linearly with the input size N, where N is the length of the strings. No other significant data structures are used.

Edge Cases

Case	How to Handle
Strings s and t are null or empty	Return an empty list immediately as no operations are possible; handle nulls gracefully to avoid NullPointerExceptions.
Strings s and t have different lengths	Return an empty list immediately, as equality is impossible with the given operations if the lengths differ.
Strings s and t are identical	Return an empty list immediately since no operations are needed.
Strings s and t differ in only one position	Return an empty list immediately since adjacent swaps are needed and only adjacent swaps are allowed.
Strings s and t have maximum length (e.g., near integer limit) and a large number of differences	Ensure the algorithm's time complexity remains within reasonable bounds to prevent timeouts (e.g., avoid O(n^2) or worse solutions).
Differences between s and t are all concentrated at the end of the string	The algorithm must still correctly identify and apply the necessary operations to the latter portion of the string.
No valid solution exists because differences require non-adjacent swaps	The algorithm should efficiently detect and return an empty list when adjacent swaps cannot create equality.
Algorithm is inefficient for large strings with many differences	Optimise search for swappable pairs to prevent quadratic or worse time complexity.

Strings s and t are null or empty

How to Handle:

Return an empty list immediately as no operations are possible; handle nulls gracefully to avoid NullPointerExceptions.

Strings s and t have different lengths

How to Handle:

Return an empty list immediately, as equality is impossible with the given operations if the lengths differ.

Strings s and t are identical

How to Handle:

Return an empty list immediately since no operations are needed.

Strings s and t differ in only one position

How to Handle:

Return an empty list immediately since adjacent swaps are needed and only adjacent swaps are allowed.

Strings s and t have maximum length (e.g., near integer limit) and a large number of differences

How to Handle:

Ensure the algorithm's time complexity remains within reasonable bounds to prevent timeouts (e.g., avoid O(n^2) or worse solutions).

Differences between s and t are all concentrated at the end of the string

How to Handle:

The algorithm must still correctly identify and apply the necessary operations to the latter portion of the string.

No valid solution exists because differences require non-adjacent swaps

How to Handle:

The algorithm should efficiently detect and return an empty list when adjacent swaps cannot create equality.

Algorithm is inefficient for large strings with many differences

How to Handle:

Optimise search for swappable pairs to prevent quadratic or worse time complexity.