LeetCode 2896: Apply Operations to Make Two Strings Equal

Explanation:

To solve this problem, we need to find the minimum cost needed to make two binary strings s1 and s2 equal by applying specific operations. We can achieve this by iterating over both strings and comparing character by character. We consider two scenarios for each mismatch:

1. If s1[i] is different from s2[i]:

   • If we can flip adjacent characters in s1, we choose the operation with the minimum cost.
   • If we can flip characters at arbitrary indices in s1, we choose the operation with the minimum cost.
   • If neither operation is possible, we return -1.

2. If s1[i] is the same as s2[i], we move to the next character.

Finally, we return the total cost required to make the strings equal or -1 if it is impossible.

Time Complexity: O(n) where n is the length of the strings s1 and s2. Space Complexity: O(1)

:

class Solution {
    public int getMinCost(String s1, String s2, int x) {
        int cost = 0;
        for (int i = 0; i < s1.length(); i++) {
            if (s1.charAt(i) != s2.charAt(i)) {
                if (i + 1 < s1.length() && s1.charAt(i + 1) == s2.charAt(i + 1)) {
                    cost += Math.min(x, 2);
                    i++;
                } else if (x < 2) {
                    cost += x;
                } else {
                    return -1;
                }
            }
        }
        return cost;
    }
}