Problem Description

You have two binary strings s1 and s2 of the same length n (containing only '0's and '1's), and a positive integer x. Your goal is to make s1 equal to s2 using the minimum total cost.

You can perform two types of operations on s1:

1. Operation 1: Choose any two indices i and j, then flip both s1[i] and s1[j]. This costs x.
2. Operation 2: Choose an index i where i < n - 1, then flip both s1[i] and s1[i + 1] (two adjacent characters). This costs 1.

Flipping means changing '0' to '1' or '1' to '0'.

The problem asks you to find the minimum cost needed to transform s1 into s2. If it's impossible to make them equal, return -1.

For example, if you have positions where s1 and s2 differ, you need to flip those positions in s1. Since each operation flips exactly two characters, if there's an odd number of positions where the strings differ, it's impossible to make them equal (you can't flip an odd number of positions using operations that always flip pairs).

The key insight is that you need to identify all positions where s1[i] ≠ s2[i], then find the optimal way to pair up these positions for flipping using either the expensive operation (cost x for any two positions) or the cheap operation (cost 1 for adjacent positions, but you might need multiple operations to connect positions that aren't originally adjacent).

Intuition

Let's first observe what positions actually need to be changed. If s1[i] ≠ s2[i], then position i needs to be flipped. Since each operation flips exactly two positions, we need an even number of differing positions - otherwise it's impossible.

Once we identify all differing positions and store them in an array idx, the problem becomes: how do we optimally pair up these positions for flipping?

Consider we have positions [idx[0], idx[1], ..., idx[m-1]] that need flipping. We have two ways to flip any pair:

• Use Operation 1 (cost x) to directly flip any two positions
• Use Operation 2 multiple times to "connect" two positions through adjacent flips

For Operation 2, if we want to flip positions idx[i] and idx[j] where i < j, we can achieve this by performing adjacent flips from idx[i] to idx[j]. The cost would be idx[j] - idx[i] (the distance between them).

Now comes the key insight: for any subarray of positions idx[i..j], we need to decide how to pair them up. For the endpoints idx[i] and idx[j], we have three choices:

1. Pair them together: Use Operation 1 with cost x to flip idx[i] and idx[j] directly. This leaves us with the subproblem idx[i+1..j-1].

2. Pair idx[i] with idx[i+1]: Use Operation 2 to connect these adjacent positions in our list with cost idx[i+1] - idx[i]. This leaves us with the subproblem idx[i+2..j].

3. Pair idx[j-1] with idx[j]: Use Operation 2 to connect these adjacent positions with cost idx[j] - idx[j-1]. This leaves us with the subproblem idx[i..j-2].

This recursive structure naturally leads to a dynamic programming solution where dfs(i, j) represents the minimum cost to handle all positions from index i to j in our idx array. The base case is when i > j, meaning no positions left to handle, so the cost is 0.

The memoization helps avoid recalculating the same subproblems, as different choices might lead to the same subproblem configuration.

Learn more about Dynamic Programming patterns.

Solution Approach

First, we extract all positions where s1[i] ≠ s2[i] and store them in an array idx. Let m be the length of this array. If m is odd (checked using m & 1), we immediately return -1 since it's impossible to make the strings equal.

The core of the solution is a recursive function dfs(i, j) that computes the minimum cost to flip all positions in idx[i..j]. We use Python's @cache decorator for memoization to avoid redundant calculations.

Base Case:

• When i > j, there are no positions to flip, so we return 0.

Recursive Cases: For each subproblem dfs(i, j), we consider three options:

1. Option A - Pair the endpoints:

   • Flip idx[i] and idx[j] using Operation 1
   • Cost: x + dfs(i + 1, j - 1)
   • This removes both endpoints and recursively solves the inner subarray

2. Option B - Pair the first two positions:

   • Flip idx[i] and idx[i + 1] using Operation 2
   • Cost: (idx[i + 1] - idx[i]) + dfs(i + 2, j)
   • The distance idx[i + 1] - idx[i] represents the number of Operation 2 moves needed
   • This removes the first two positions and recursively solves the remaining right portion

3. Option C - Pair the last two positions:

   • Flip idx[j - 1] and idx[j] using Operation 2
   • Cost: (idx[j] - idx[j - 1]) + dfs(i, j - 2)
   • This removes the last two positions and recursively solves the remaining left portion

The function returns the minimum of these three options: min(a, b, c).

Implementation Details:

@cache
def dfs(i: int, j: int) -> int:
    if i > j:
        return 0
    a = dfs(i + 1, j - 1) + x
    b = dfs(i + 2, j) + idx[i + 1] - idx[i]
    c = dfs(i, j - 2) + idx[j] - idx[j - 1]
    return min(a, b, c)

The final answer is dfs(0, m - 1), which represents the minimum cost to handle all differing positions from the first to the last in our idx array.

The memoization ensures that each unique subproblem (i, j) is computed only once, giving us a time complexity of O(m²) where m is the number of differing positions.

Example Walkthrough

Let's walk through a concrete example to understand the solution approach.

Example:

• s1 = "10110"
• s2 = "00101"
• x = 4

Step 1: Identify differing positions

Compare each position:

• Position 0: s1[0]='1', s2[0]='0' → differs
• Position 1: s1[1]='0', s2[1]='0' → same
• Position 2: s1[2]='1', s2[2]='1' → same
• Position 3: s1[3]='1', s2[3]='0' → differs
• Position 4: s1[4]='0', s2[4]='1' → differs

So idx = [0, 3, 4] (positions that need flipping)

Since we have 3 differing positions (odd number), it's impossible to make the strings equal. Return -1.

Let's modify the example for a solvable case:

• s1 = "10110"
• s2 = "00111"
• x = 4

Differing positions: idx = [0, 3] (even number, so it's solvable)

Step 2: Apply dynamic programming

We need to find dfs(0, 1) - the minimum cost to handle positions 0 through 1 in our idx array.

For dfs(0, 1), we have three options:

1. Option A - Pair endpoints (idx[0]=0 and idx[1]=3):

   • Use Operation 1 to flip positions 0 and 3
   • Cost: x + dfs(1, 0) = 4 + 0 = 4
   • (dfs(1, 0) returns 0 since i > j)

2. Option B - Pair first two positions:

   • This would be idx[0] and idx[1], which are positions 0 and 3
   • Use Operation 2 to connect them: need to flip (0,1), then (1,2), then (2,3)
   • Cost: (idx[1] - idx[0]) + dfs(2, 1) = (3 - 0) + 0 = 3

3. Option C - Pair last two positions:

   • Same as Option B since we only have two positions total
   • Cost: (idx[1] - idx[0]) + dfs(0, -1) = (3 - 0) + 0 = 3

The minimum is min(4, 3, 3) = 3.

Result: The minimum cost is 3, achieved by using Operation 2 three times:

• Flip positions (0,1): s1 becomes "01110"
• Flip positions (1,2): s1 becomes "00010"
• Flip positions (2,3): s1 becomes "00111" = s2 ✓

Another example with more positions:

• idx = [1, 2, 5, 6] and x = 3

For dfs(0, 3):

1. Option A: Pair idx[0]=1 with idx[3]=6

   • Cost: 3 + dfs(1, 2)
   • Need to solve dfs(1, 2) for positions [2, 5]
   • dfs(1, 2) = min(3 + 0, 3 + 0, 3 + 0) = 3
   • Total: 3 + 3 = 6

2. Option B: Pair idx[0]=1 with idx[1]=2

   • Cost: (2-1) + dfs(2, 3) = 1 + dfs(2, 3)
   • dfs(2, 3) for positions [5, 6] = min(3, 1, 1) = 1
   • Total: 1 + 1 = 2

3. Option C: Pair idx[2]=5 with idx[3]=6

   • Cost: (6-5) + dfs(0, 1) = 1 + dfs(0, 1)
   • dfs(0, 1) for positions [1, 2] = min(3, 1, 1) = 1
   • Total: 1 + 1 = 2

The minimum is min(6, 2, 2) = 2, achieved by pairing adjacent positions in our idx array.

Solution Implementation

Time and Space Complexity

The time complexity is O(m^2) where m is the number of mismatched positions between s1 and s2. Since m ≤ n where n is the length of the strings, and in the worst case m = n, the time complexity can be expressed as O(n^2).

The recursive function dfs(i, j) explores all possible states where i and j represent indices in the idx array. There are O(m^2) possible states since i can range from 0 to m-1 and j can range from 0 to m-1. Each state is computed once due to memoization using @cache, and each computation takes O(1) time.

The space complexity is O(m^2) which is also O(n^2) in the worst case. This comes from:

• The memoization cache storing up to O(m^2) states
• The recursion stack depth which can go up to O(m) in the worst case
• The idx array which takes O(m) space

Since O(m^2) dominates O(m), the overall space complexity is O(m^2) or O(n^2) when considering the worst case where all positions are mismatched.

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

Common Pitfalls

1. Integer Division vs Float Division Pitfall

A common mistake is using float division when calculating costs for consecutive positions. Some developers might write:

# WRONG - creates floating point result
pair_left = dfs(left + 2, right) + (diff_positions[left + 1] - diff_positions[left]) / 2

This happens when misunderstanding that the cost between adjacent positions is their distance (number of Operation 2 moves needed), not half their distance. The correct cost is simply the difference between indices.

Solution: Use the actual distance without division:

# CORRECT
pair_left = dfs(left + 2, right) + diff_positions[left + 1] - diff_positions[left]

2. Incorrect Base Case Handling

Another pitfall is not properly handling edge cases when the recursion reaches boundaries:

# WRONG - doesn't check array bounds
def dfs(left: int, right: int) -> int:
    if left > right:
        return 0
    # This will cause IndexError when left + 1 >= len(diff_positions)
    pair_left = dfs(left + 2, right) + diff_positions[left + 1] - diff_positions[left]

While the memoized solution handles this implicitly (since left + 1 won't exceed bounds when left <= right), it's still a conceptual pitfall.

Solution: The current implementation is actually safe because when left > right, we return immediately, and when left == right, we have a single element which is invalid (odd count). The recursion structure ensures we never access out-of-bounds indices.

3. Not Caching/Memoizing the Recursive Function

Forgetting to use memoization leads to exponential time complexity:

# WRONG - no memoization, causes timeout
def dfs(left: int, right: int) -> int:
    if left > right:
        return 0
    # ... rest of the logic

This creates overlapping subproblems that get recalculated multiple times.

Solution: Always use @cache decorator or manual memoization:

# CORRECT
from functools import cache

@cache
def dfs(left: int, right: int) -> int:
    # ... implementation

4. Misunderstanding the Cost Calculation for Adjacent Flips

Some might think that flipping adjacent positions always costs 1:

# WRONG - assumes all adjacent flips cost 1
pair_left = dfs(left + 2, right) + 1

However, if positions are at indices 2 and 5, you need 3 Operation 2 moves to connect them (flip 2-3, then 3-4, then 4-5).

Solution: Calculate the actual distance between positions:

# CORRECT - uses actual distance
pair_left = dfs(left + 2, right) + diff_positions[left + 1] - diff_positions[left]