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Minimum Operations to Transform Binary String - Solution & Explanation

MediumStringDynamic ProgrammingGreedy3 min read
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Problem Statement

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'.

Approach Overview

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.

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Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Brute Force SimulationO(n²)O(1)Useful for understanding operations or very small inputs
Greedy Transition CountingO(n)O(1)Best general solution for interviews and production code
Prefix State TrackingO(n)O(1)When previous operations affect future positions

Video Solution

Leetcode 3980 | Minimum Operations to Transform Binary String | Biweekly contest 186 • CodeWithMeGuys • 493 views views

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Frequently Asked Questions

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.

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