Leetcode #2896: Apply Operations to Make Two Strings Equal
In this guide, we solve Leetcode #2896 Apply Operations to Make Two Strings Equal in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given two 0-indexed binary strings s1 and s2, both of length n, and a positive integer x. You can perform any of the following operations on the string s1 any number of times: Choose two indices i and j, and flip both s1[i] and s1[j].
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: String, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: s1 = "1100011000", s2 = "0101001010", x = 2
Output: 4
Explanation: 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.
Python Solution
class Solution:
def minOperations(self, s1: str, s2: str, x: int) -> int:
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)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.