2896. Apply Operations to Make Two Strings Equal

Problem Description

In this problem, we are given two binary strings of equal length, s1 and s2, as well as an integer cost x. Our goal is to transform s1 into s2 through a series of operations, each of which incurs some cost. There are two types of operations we can perform:

1. We can flip (change from 0 to 1 or from 1 to 0) the bits at two chosen indices i and j in s1. This operation has a fixed cost of x.
2. We can flip the bits at consecutive indices i and i+1 in s1, provided i < n - 1. This operation has a cost of 1.

Our objective is to calculate the minimum cost required to make s1 identical to s2. If it's not possible to make the strings identical, we should return -1.

An important observation is that an even number of differences between s1 and s2 is required to make the transformation feasible because each flip operation changes the state of exactly two bits.

Intuition

The intuition behind the solution begins with the understanding that we are only able to change two characters at a time. If there is an odd number of differing characters between s1 and s2, it's impossible to achieve equality, and the function returns -1. When the number of differing characters is even, there is potential to reach the goal.

We use an array idx to store the indices of s1 where the characters differ from those in s2. The first step in our approach is a quick check of the length of idx: if it is odd, we return -1 right away as it would be impossible to make the strings equal.

Next, we define a function dfs(i, j) to represent the minimum cost needed to flip the characters in the sub-array idx[i..j]. We're looking for the answer to dfs(0, len(idx) - 1).

The calculation for dfs(i, j) considers three possible scenarios:

1. Flipping endpoints (i and j): We flip characters at the two endpoints, using the first operation type at a cost of x. Then we recursively call dfs to handle the sub-problem for idx[i+1] to idx[j-1].
2. Flipping near i: We flip the characters at idx[i] and idx[i+1] using the second operation type. We add the distance between idx[i] and idx[i+1] to the cost of the recursive dfs call from idx[i+2] to idx[j].
3. Flipping near j: We flip the characters at idx[j-1] and idx[j] in similar fashion, adding the distance between these indices to the cost of the recursive dfs call for idx[i] to idx[j-2].

We seek the minimum cost of these three approaches to be the result for dfs(i, j).

Memoization is used to cache the results of dfs(i, j) to avoid recomputing sub-problems. This is done by storing previously calculated results of dfs(i, j) and retrieving them instead of recalculating when they are requested again.

This solution approach uses depth-first search combined with memoization (caching) to efficiently solve the problem.

Learn more about Dynamic Programming patterns.

Solution Approach

The given solution utilizes Depth-First Search (DFS) with memoization as its core algorithm. The DFS explores all combinations of pairs that can be flipped to transform s1 into s2, while memoization ensures that the minimum cost for any given range of indices is calculated only once and then reused. The use of memoization is a common pattern when there is a possibility of computing the same values repeatedly, as seen in many dynamic programming problems.

The code defines a recursive function dfs(i, j), which calculates the minimum cost of making subarrays of idx, specifically idx[i..j], identical. The Python @cache decorator is used to store the results of dfs(i, j) calls, so identical sub-problems are not re-calculated from scratch—a typical memoization pattern.

There are three potential scenarios covered by the recursive function:

1. Flipping endpoints (i and j): The cost is calculated by flipping the characters at i and j, which consumes a fixed cost of x.

   1a = dfs(i + 1, j - 1) + x

2. Flipping near the start index (i and i + 1): The cost is the cost of flipping consecutive characters starting from i which costs 1. However, since 1 unit cost is specific to two adjacent characters, the actual cost is the distance between idx[i] and idx[i + 1]. This distance is then added to the cost of the recursive call starting from i + 2.

   1b = dfs(i + 2, j) + idx[i + 1] - idx[i]

3. Flipping near the end index (j - 1 and j): This is similar to the previous case, but here we start flipping from j.

   1c = dfs(i, j - 2) + idx[j] - idx[j - 1]

In each call of dfs(i, j), the function computes the minimum cost among the three scenarios (a, b, and c).

The index array idx helps to keep track of the indices where the two strings differ, emphasizing that we only care about the positions that need transformation. This can be seen as a space optimization over considering the entire string for each recursive call.

The condition if m & 1 checks whether we have an odd number of differing bits by using bitwise AND with 1. If m is odd, it immediately returns -1, as an odd number of differing bits cannot be solved with the given operations.

Lastly, return dfs(0, m - 1) kicks off the recursive calls with the full range of differing indices.

This method provides a balance between the brute force approach (which would be too slow) and a more optimized approach that keeps the time complexity in check by only re-calculating necessary sub-problems when needed. Exploring all combinations through DFS ensures that all possibilities are considered, while memoization prevents unnecessary work, thus optimizing the solution.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach:

Suppose we have the binary strings s1 = "1101" and s2 = "1000" with an integer cost x = 3. We want to transform s1 into s2 with the minimum cost using the operations described.

First, we identify the indices where s1 and s2 differ, which are idx = [1, 3]. There are two bits different, so it's an even number, and thus, the problem is solvable.

Let's apply the Depth-First Search (DFS) approach with memoization step by step:

1. The function dfs(0, 1) is called to solve the entire range of differing indices.
2. Inside dfs(0, 1), we calculate the cost of flipping endpoints (i = 0, j = 1). The cost for flipping idx[0] and idx[1] is x = 3.

   1a = dfs(1 - 1, 1 + 1) + x
   2a = dfs(0, 2) + 3  # Out of bounds, but assume dfs(0, 2) = 0
   3a = 0 + 3
   4a = 3

3. Then we compute the cost of flipping near i (i = 0 and i + 1 = 1). The cost of flipping idx[0] and idx[1] is distance idx[1] - idx[0], which is 3 - 1 = 2.

   1b = dfs(0 + 2, 1) + idx[1] - idx[0]
   2b = dfs(2, 1)  # Out of bounds, but assume dfs(2, 1) = 0
   3b = 0 + 2
   4b = 2

4. Since there are only two indices, there's no distinct cost computation for flipping near j, as it would be the same as flipping near i.

The minimum cost among the scenarios is min(a, b), which is min(3, 2). Thus, the minimum cost is 2.

However, we realize that since we used a hypothetical out-of-bounds result for the recursive dfs(0, 2) and dfs(2, 1), in a real situation, the actual function would need to check for boundaries and return appropriate values (usually 0 if there are no more characters to flip). This is handled in the implementation where each recursive call checks for the validity of the indices.

In this case, only the operation of flipping bits at consecutive indices i and i+1 (the second type of operation) is applied, resulting in a total minimum cost of 2 to transform s1 into s2.

This example demonstrates how the algorithm applies DFS with memoization to find the minimum cost to transform one binary string into another efficiently.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(n^2) where n is the length of idx, which derives from the number of mismatched characters between s1 and s2. For each call to dfs(i, j), there are up to three recursive calls, but the function is memoized using the @cache decorator. This ensures that each possible state (i, j) is calculated only once. The state (i, j) represents a subproblem of finding the minimum operations for the substring from idx[i] to idx[j], and there are O(n^2) such states because i can range from 0 to n-1 and j can range from i to n-1.

The space complexity of the code is also O(n^2) due to the memoization cache that stores the computed result for each unique (i, j) pair. Each entry in the cache takes constant space, and since there are O(n^2) pairs, the overall space used by the cache is O(n^2), plus the space used for the idx list and the recursion stack.

Learn more about how to find time and space complexity quickly using problem constraints.