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You are given two binary strings s1 and s2 of the same length n.
You can perform the following operations on s1 any number of times, in any order:
i such that s1[i] == '0', and change it to '1'.i such that 0 <= i < n - 1, and both s1[i] and s1[i + 1] are '1'. Change both characters to '0'.Return the minimum number of operations required to make s1 equal to s2. If it is impossible, return -1.
Example 1:
Input: s1 = "11", s2 = "00"
Output: 1
Explanation:
Change indices 0 and 1 from '1' to '0' in one operation, so "11" becomes "00". Thus, the answer is 1.
Example 2:
Input: s1 = "01", s2 = "10"
Output: 3
Explanation:
'0' to '1', so "01" becomes "11".'1' to '0', so "11" becomes "00".'0' to '1', so "00" becomes "10".Example 3:
Input: s1 = "1", s2 = "0"
Output: -1
Explanation:
The first operation cannot change '1' to '0', and the second operation requires two adjacent characters. Therefore, it is impossible.
Constraints:
1 <= n == s1.length == s2.length <= 105s1 and s2 consist only of '0' and '1'.Problem Overview: You need to transform a binary string into a target configuration using the minimum number of allowed operations. The challenge is identifying which changes can be grouped together instead of flipping characters independently.
Approach 1: Brute Force Simulation (O(n2) time, O(1) space)
The direct approach simulates every valid operation and repeatedly updates the string until it matches the target. For each mismatch, you scan neighboring positions or possible ranges to determine the cheapest modification sequence. This works for small inputs because it mirrors the problem statement exactly, but repeated rescanning makes it slow for large strings. Use this version mainly to verify correctness before optimizing.
Approach 2: Greedy Transition Counting (O(n) time, O(1) space)
The optimal solution tracks mismatch segments instead of individual characters. Iterate through both strings once and detect where mismatch regions start and end. Consecutive mismatches can often be fixed using a single operation, so the key insight is counting transitions rather than counting every differing index. This greedy strategy avoids unnecessary operations and gives linear performance with constant extra memory. Problems involving greedy algorithms and string processing commonly use this pattern.
Approach 3: State Tracking with Prefix Logic (O(n) time, O(1) space)
Another efficient method maintains the current transformation state while iterating from left to right. Instead of modifying the string directly, you track whether previous operations already affected the current bit. This reduces repeated updates and keeps every position processed once. The technique is closely related to prefix-style state accumulation where earlier operations influence future decisions.
Recommended for interviews: Interviewers typically expect the greedy linear-time solution because it demonstrates that you can recognize structural patterns in binary transformations. Starting with brute force shows understanding of the mechanics, but optimizing to O(n) by grouping mismatch segments is what usually separates a complete solution from an average one.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Simulation | O(n²) | O(1) | Useful for understanding operations or very small inputs |
| Greedy Transition Counting | O(n) | O(1) | Best general solution for interviews and production code |
| Prefix State Tracking | O(n) | O(1) | When previous operations affect future positions |