How do you solve Apply Operations to Make Two Strings Equal on LeetCode?

Apply Operations to Make Two Strings Equal (LeetCode #2896) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the cost of operations is crucial to minimize the total cost.

What is the brute force solution for Apply Operations to Make Two Strings Equal?

The brute force approach for Apply Operations to Make Two Strings Equal is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible pairs of indices to flip in order to make the two strings equal. This method is straightforward but inefficient, as it may require many operations to find the solution. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Apply Operations to Make Two Strings Equal?

Apply Operations to Make Two Strings Equal uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Apply Operations to Make Two Strings Equal?

Always clarify the problem constraints and edge cases before diving into coding. Think about how to break down the problem into smaller parts that can be solved independently. Discuss your thought process and approach with the interviewer to showcase your problem-solving skills.

What are common mistakes when solving Apply Operations to Make Two Strings Equal?

Not considering the cost of adjacent flips versus pair flips properly. Failing to handle cases where it's impossible to make the strings equal.

#2896

Apply Operations to Make Two Strings Equal

Medium

StringDynamic ProgrammingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages dynamic programming to minimize the cost of operations by focusing on the differing indices. This approach efficiently calculates the minimum cost by considering the costs of flipping adjacent bits and pairs.

⚙️

Algorithm

3 steps

1Step 1: Identify all indices where s1 and s2 differ and store them in a list.
2Step 2: Use dynamic programming to calculate the minimum cost to make the strings equal based on the differing indices.
3Step 3: For each differing index, decide whether to use the cost of flipping pairs or adjacent bits based on the minimum cost calculated so far.

solution.py11 lines

1def minCost(s1, s2, x):
2    n = len(s1)
3    diff = [i for i in range(n) if s1[i] != s2[i]]
4    m = len(diff)
5    dp = [float('inf')] * (m + 1)
6    dp[0] = 0
7    for i in range(m):
8        dp[i + 1] = min(dp[i + 1], dp[i] + 1)  # flip adjacent
9        if i + 1 < m:
10            dp[i + 2] = min(dp[i + 2], dp[i] + x)  # flip pair
11    return dp[m] if dp[m] != float('inf') else -1

ℹ

Complexity note: This complexity is linear because we only traverse the differing indices and maintain a dynamic programming array proportional to the number of differences.

1Understanding the cost of operations is crucial to minimize the total cost.
2Dynamic programming can significantly reduce the complexity by breaking the problem into smaller subproblems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.