Is Minimum Cost to Make Two Binary Strings Equal easy or hard?

The problem is usually rated Medium because the core logic is simple once the mismatch observation is made, but identifying the correct greedy pairing strategy can take time. Candidates often start with a brute force pairing idea before optimizing to the linear solution.

Minimum Cost to Make Two Binary Strings Equal Python/Java solution

Implementations typically scan both strings, push mismatch indices into a list, and then apply the greedy pairing logic to accumulate the minimum cost. The same algorithm translates directly across Python, Java, C++, Go, and TypeScript because it only requires basic loops and arrays.

How to solve Minimum Cost to Make Two Binary Strings Equal in O(n)?

First iterate through both strings and collect indices where the characters differ. Then process these mismatch indices in order while greedily deciding whether to pair the current mismatch with the previous one or resolve them separately. Each index is handled once, giving an O(n) time algorithm.

What is the best approach for Minimum Cost to Make Two Binary Strings Equal?

The optimal approach uses a greedy strategy that processes all mismatch indices in order. By pairing nearby mismatches when it reduces the total operation cost, the algorithm minimizes the overall cost with a single linear pass. This reduces the time complexity to O(n) while using O(n) space to track mismatch positions.

Is Minimum Cost to Make Two Binary Strings Equal asked at Google/Amazon/Meta?

Problems involving binary string transformations and minimum-cost operations frequently appear in interviews at companies like Amazon, Google, and Meta. They test greedy reasoning, string processing, and the ability to optimize from brute force to linear-time solutions.

What data structure is used in Minimum Cost to Make Two Binary Strings Equal?

The main data structure is a simple array or list that stores indices where the two strings differ. This list allows the algorithm to process mismatches sequentially and apply greedy pairing decisions efficiently.

What is the time complexity of Minimum Cost to Make Two Binary Strings Equal?

The optimal greedy solution runs in O(n) time where n is the length of the binary strings. Each index is checked once to detect mismatches, and the mismatch list is processed sequentially. Space complexity is O(n) in the worst case if every position differs.

3800. Minimum Cost to Make Two Binary Strings Equal - Practice Online

You are given two binary strings s and t, both of length n, and three positive integers flipCost, swapCost, and crossCost.

You are allowed to apply the following operations any number of times (in any order) to the strings s and t:

Choose any index i and flip s[i] or t[i] (change '0' to '1' or '1' to '0'). The cost of this operation is flipCost.
Choose two distinct indices i and j, and swap either s[i] and s[j] or t[i] and t[j]. The cost of this operation is swapCost.
Choose an index i and swap s[i] with t[i]. The cost of this operation is crossCost.

Return an integer denoting the minimum total cost needed to make the strings s and t equal.

Example 1:

Input: s = "01000", t = "10111", flipCost = 10, swapCost = 2, crossCost = 2

Output: 16

Explanation:

We can perform the following operations:

Swap s[0] and s[1] (swapCost = 2). After this operation, s = "10000" and t = "10111".
Cross swap s[2] and t[2] (crossCost = 2). After this operation, s = "10100" and t = "10011".
Swap s[2] and s[3] (swapCost = 2). After this operation, s = "10010" and t = "10011".
Flip s[4] (flipCost = 10). After this operation, s = t = "10011".

The total cost is 2 + 2 + 2 + 10 = 16.

Example 2:

Input: s = "001", t = "110", flipCost = 2, swapCost = 100, crossCost = 100

Output: 6

Explanation:

Flipping all the bits of s makes the strings equal, and the total cost is 3 * flipCost = 3 * 2 = 6.

Example 3:

Input: s = "1010", t = "1010", flipCost = 5, swapCost = 5, crossCost = 5

Output: 0

Explanation:

The strings are already equal, so no operations are required.

Constraints:

n == s.length == t.length
1 <= n <= 10⁵
1 <= flipCost, swapCost, crossCost <= 10⁹
s and t consist only of the characters '0' and '1'.