Interview AiBox logo
Interview AiBox
#2896
Medium
auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Apply Operations to Make Two Strings Equal

Apply Operations to Make Two Strings Equal involves transforming binary strings with a cost-based dynamic programming approach.

stringdynamic programming
category

2

Topics

code_blocks

5

Code langs

hub

3

Related

play_arrowStart Solutionarrow_forward
speedComplexityopen_in_newLeetCode

Practice Focus

Medium · State transition dynamic programming

bolt

Answer-first summary

Apply Operations to Make Two Strings Equal involves transforming binary strings with a cost-based dynamic programming approach.

Interview AiBox logo

Ace coding interviews with Interview AiBox guidance for State transition dynamic programming

Try AiBox Copilotarrow_forward

The task is to find the minimum cost required to make two binary strings equal, using operations on mismatched characters. The problem can be efficiently tackled by focusing on mismatched positions and applying state transition dynamic programming to minimize the cost of changes. The approach involves iterating over indices with different characters and using operations that minimize the transformation cost.

Problem Statement

You are given two binary strings, s1 and s2, both of length n, and a positive integer x. You can perform operations on string s1 to transform it into string s2. The allowed operations are as follows: Change a single bit in s1 to match s2 at a cost of 1 or flip a pair of adjacent bits at a cost of x. Your goal is to determine the minimum cost to make s1 equal to s2. If it is impossible to do so, return -1.

To solve this problem, you must focus on the positions where s1 and s2 differ. By saving these indices into a list and applying the operations accordingly, you can minimize the cost. If there are no mismatches, the cost is zero. If any mismatched pairs cannot be transformed due to the nature of the operations, return -1.

Examples

Example 1

Input: s1 = "1100011000", s2 = "0101001010", x = 2

Output: 4

We can do the following operations:

  • Choose i = 3 and apply the second operation. The resulting string is s1 = "1101111000".
  • Choose i = 4 and apply the second operation. The resulting string is s1 = "1101001000".
  • Choose i = 0 and j = 8 and apply the first operation. The resulting string is s1 = "0101001010" = s2. The total cost is 1 + 1 + 2 = 4. It can be shown that it is the minimum cost possible.

Example 2

Input: s1 = "10110", s2 = "00011", x = 4

Output: -1

It is not possible to make the two strings equal.

Constraints

  • n == s1.length == s2.length
  • 1 <= n, x <= 500
  • s1 and s2 consist only of the characters '0' and '1'.

Solution Approach

State Transition Dynamic Programming

Use dynamic programming to track the state transitions of mismatched positions. Iterate over the positions where s1 and s2 differ, considering the possible operations and their associated costs. This minimizes the cost of transforming s1 into s2.

Work Only With Mismatched Positions

By focusing only on the indices where the characters of s1 and s2 differ, you reduce the complexity of the problem. Save the mismatched positions and use this list to apply the operations efficiently.

Minimizing the Transformation Cost

Apply the flipping operation on pairs of adjacent mismatches when the cost of flipping (x) is less than the cost of changing individual bits. This minimizes the total cost of transforming s1 into s2.

Complexity Analysis

Metric Value
Time Depends on the final approach
Space Depends on the final approach

The time and space complexity depend on the number of mismatched positions. In the worst case, when all positions differ, the complexity is O(n), where n is the length of the strings.

What Interviewers Usually Probe

  • Look for an understanding of dynamic programming and state transitions.
  • Check the candidate's ability to reduce problem complexity by focusing on mismatched positions.
  • Evaluate how efficiently the candidate applies operations to minimize transformation cost.

Common Pitfalls or Variants

Common pitfalls

  • Failing to focus on mismatched positions, leading to inefficient solutions.
  • Misapplying the flipping operation when it's not cost-effective.
  • Overcomplicating the problem by not leveraging dynamic programming.

Follow-up variants

  • What if the strings are of different lengths?
  • How can you optimize the approach for larger values of x?
  • How does this problem relate to other dynamic programming challenges?

FAQ

What is the minimum cost to make two binary strings equal?

The minimum cost is determined by applying state transition dynamic programming to optimize the operations needed to transform one string into another.

What is the main pattern behind the Apply Operations to Make Two Strings Equal problem?

The problem follows a state transition dynamic programming pattern, where the key is to focus on mismatched positions and apply operations optimally.

How does dynamic programming apply to the Apply Operations to Make Two Strings Equal problem?

Dynamic programming is used to track the states of mismatched positions and minimize the total cost by choosing the best operations at each step.

Is there any condition under which the problem is unsolvable?

Yes, if it is impossible to transform the strings using the allowed operations, the function should return -1.

What is the time complexity of solving the Apply Operations to Make Two Strings Equal problem?

The time complexity depends on the number of mismatched positions and is generally O(n), where n is the length of the strings.

terminal

Solution

Solution 1: Memoization

We notice that since each operation reverses two characters, if the number of different characters in the two strings is odd, it is impossible to make them equal, and we directly return $-1$. Otherwise, we store the indices of the different characters in the two strings in an array $idx$, and let $m$ be the length of $idx$.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
class Solution:
    def minOperations(self, s1: str, s2: str, x: int) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if i > j:
                return 0
            a = dfs(i + 1, j - 1) + x
            b = dfs(i + 2, j) + idx[i + 1] - idx[i]
            c = dfs(i, j - 2) + idx[j] - idx[j - 1]
            return min(a, b, c)

        n = len(s1)
        idx = [i for i in range(n) if s1[i] != s2[i]]
        m = len(idx)
        if m & 1:
            return -1
        return dfs(0, m - 1)

Solution 2

#### Java

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
class Solution:
    def minOperations(self, s1: str, s2: str, x: int) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if i > j:
                return 0
            a = dfs(i + 1, j - 1) + x
            b = dfs(i + 2, j) + idx[i + 1] - idx[i]
            c = dfs(i, j - 2) + idx[j] - idx[j - 1]
            return min(a, b, c)

        n = len(s1)
        idx = [i for i in range(n) if s1[i] != s2[i]]
        m = len(idx)
        if m & 1:
            return -1
        return dfs(0, m - 1)

Continue Practicing

#2900 longest unequal adjacent groups subsequence i#2901 longest unequal adjacent groups subsequence ii#2911 minimum changes to make k semi palindromes
category

Continue Topic

string

Practice more edge cases under the same topic.

arrow_forwardauto_awesome

Continue Pattern

State transition dynamic programming

Expand the same solving frame across more problems.

arrow_forwardsignal_cellular_alt

Same Difficulty Track

Medium

Stay on this level to stabilize interview delivery.

arrow_forward
Apply Operations to Make Two Strings Equal Solution: State transition dynamic programming | LeetCode #2896 Medium