Is Minimum Operations to Transform Binary String easy or hard?

This problem is generally considered medium difficulty because the brute force idea is straightforward but the optimal observation is less obvious. Recognizing that consecutive mismatches can often be handled together is the main challenge.

Minimum Operations to Transform Binary String Python/Java solution

Python solutions typically use a single loop with mismatch tracking for concise O(n) code. Java implementations follow the same greedy logic using character arrays or String.charAt for efficient traversal. Both versions achieve linear runtime and constant extra space.

How to solve Minimum Operations to Transform Binary String in O(n)?

Traverse both binary strings simultaneously and track mismatch regions. Whenever a new mismatch segment starts, increment the operation count or update the transformation state. By avoiding repeated rescans and direct string updates, the algorithm stays linear.

What is the best approach for Minimum Operations to Transform Binary String?

The best approach is usually a greedy linear scan that groups consecutive mismatches into a single operation whenever possible. Instead of modifying every character independently, you count transition points between matching and non-matching segments. This reduces the complexity to O(n) time with O(1) extra space.

Is Minimum Operations to Transform Binary String asked at Google/Amazon/Meta?

Binary string transformation and greedy optimization problems are common in interviews at Google, Amazon, Meta, and similar companies. Variants often test your ability to recognize patterns, minimize operations, and optimize brute force simulations into linear-time solutions.

What data structure is used in Minimum Operations to Transform Binary String?

Most optimal solutions do not require advanced data structures. Simple counters, index pointers, and boolean state variables are enough. Some implementations use arrays or strings directly while processing mismatch segments.

What is the time complexity of Minimum Operations to Transform Binary String?

The optimal solution runs in O(n) time because each character is processed once during a left-to-right traversal. Most implementations also use O(1) auxiliary space since only counters or state flags are required.

Description Hints 5 Video 12:09

3980. Minimum Operations to Transform Binary String

Medium37.0% acceptance

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:

Choose an index i such that s1[i] == '0', and change it to '1'.
Choose an index 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:

Change index 0 from '0' to '1', so "01" becomes "11".
Change indices 0 and 1 from '1' to '0', so "11" becomes "00".
Change index 0 from '0' to '1', so "00" becomes "10".
Thus, the answer is 3.

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 <= 10⁵
s1 and s2 consist only of '0' and '1'.

Topics

Hints open one at a time

Hint 1

The first operation is the only way to change a '0' into a '1'.

Hint 2

The second operation is the only way to change a '1' into a '0', but it affects two adjacent positions at once.

Hint 3

Use dynamic programming from left to right. When processing position i, keep enough information to know whether position i has already been affected by an operation involving position i - 1.

Hint 4

For each position, either make it match directly, or use the second operation on the pair (i, i + 1) after making both characters '1' if necessary.

Hint 5

Handle the case n == 1 separately, since the second operation cannot be used.

LanguageLoading editor…

1class Solution:
2    def minOperations(self, s1: str, s2: str) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

"11"
"00"

Input

"01"
"10"

Input

"1"
"0"

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