Leetcode #97: Interleaving String
In this guide, we solve Leetcode #97 Interleaving String 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
Given strings s1, s2, and s3, find whether s3 is formed by an interleaving of s1 and s2. An interleaving of two strings s and t is a configuration where s and t are divided into n and m substrings respectively, such that: s = s1 + s2 + ...
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 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true
Explanation: One way to obtain s3 is:
Split s1 into s1 = "aa" + "bc" + "c", and s2 into s2 = "dbbc" + "a".
Interleaving the two splits, we get "aa" + "dbbc" + "bc" + "a" + "c" = "aadbbcbcac".
Since s3 can be obtained by interleaving s1 and s2, we return true.
Python Solution
class Solution:
def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
def dfs(i: int, j: int) -> bool:
if i >= m and j >= n:
return True
k = i + j
if i < m and s1[i] == s3[k] and dfs(i + 1, j):
return True
if j < n and s2[j] == s3[k] and dfs(i, j + 1):
return True
return False
m, n = len(s1), len(s2)
if m + n != len(s3):
return False
return dfs(0, 0)
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.