Let's analyze the problem. We are given two binary strings, s1 and s2, of length n, and an integer x. We want to find the minimum cost to make s1 equal to s2 using two types of operations:

1. Flip two bits in s1 at a cost of x.
2. Flip two adjacent bits in s1 at a cost of 1.

If it's impossible to make the strings equal, return -1.

Naive Approach

A brute-force approach would involve trying all possible combinations of operations to transform s1 into s2. This would involve exploring a large search space and would quickly become inefficient, especially given the constraints that 1 <= n, x <= 500.

The time complexity would be exponential, and it's not a practical solution.

Optimal Approach

The key insight is to realize that we only care about the indices where s1[i] != s2[i]. Let's call these indices diff_indices. If the number of such indices is odd, it's impossible to make the strings equal because each operation flips an even number of bits. In this case, we should return -1.

If the number of differing indices is even, we can use dynamic programming to find the minimum cost. Let dp[i] be the minimum cost to make the first i differing bits equal. We can either flip the i-th and (i-1)-th bits using adjacent flips, or flip the i-th bit with some other bit j < i where j is also a differing index.

1. Consider all possible pairings of differing bits.
2. If we consider that we only use pair flips of cost x, then we only need to consider all pairs.
3. Use adjacent flips of cost 1 to handle adjacent differing bits.

Here's the algorithm:

1. Find all indices where s1[i] != s2[i] and store them in diff_indices.
2. If len(diff_indices) is odd, return -1.
3. Initialize min_cost = infinity.
4. Iterate through all possible pairs of indices in diff_indices. For each pair (i, j) with i < j, calculate the cost. The cost is x if we directly flip them or abs(diff_indices[i] - diff_indices[j]) if it's cheaper to use adjacent flips.
5. For the minimum cost, consider adjacent differences.

Code:

def min_cost_to_equalize(s1: str, s2: str, x: int) -> int:
    n = len(s1)
    diff_indices = [i for i in range(n) if s1[i] != s2[i]]

    if len(diff_indices) % 2 != 0:
        return -1

    if not diff_indices:
        return 0

    dp = [float('inf')] * (len(diff_indices) + 1)
    dp[0] = 0

    for i in range(1, len(diff_indices) + 1):
        # Option 1: Pair with the previous differing index
        if i > 1:
            dp[i] = min(dp[i], dp[i - 2] + min(x, (diff_indices[i-1] - diff_indices[i-2])))

        # Option 2: Leave the current differing index unpaired (this is not possible but we keep track of it for consistency. It will lead to infinity.

        dp[i] = min(dp[i], float('inf'))

    
    #Special case to consider adjacent swaps
    min_cost = float('inf')

    #Consider directly swapping with cost x
    import itertools
    for pair in itertools.combinations(diff_indices, 2):
      min_cost = min(min_cost, x) #directly swapping
      min_cost = min(min_cost, abs(pair[0] - pair[1])) #cost of adjacent flips
    
    if (len(diff_indices) == 2):
      min_cost = min(x,abs(diff_indices[0]-diff_indices[1]))
    
    #Now consider cases where diff_indices are adjacent
    
    if (len(diff_indices)>1):
      adj_cost = 0
      for i in range(0,len(diff_indices),2):
        if i + 1 < len(diff_indices):
          adj_cost+= min(x, abs(diff_indices[i] - diff_indices[i+1]))
      min_cost = min(min_cost,adj_cost)


    
    if(len(diff_indices) == 0):
      return 0
    else:
      return min_cost

Time Complexity: O(n + (k choose 2)) where n is the length of the string and k is number of differing bits Space Complexity: O(k), where k is the number of differing bits.

Edge Cases:

• If s1 and s2 are already equal, the cost is 0.
• If the number of differing bits is odd, it's impossible to make them equal, so return -1.
• If x is very large, the cost of flipping two bits may be larger than using adjacent flips to move the bits closer and then flipping them.
• Consider edge cases where the differing bits are adjacent.