Interleaving String
Detailed guide and Python implementation for the 'Interleaving String' problem.
1. Concept Overview
The 'Interleaving String' problem is a key challenge in the 2D DP section.
This implementation focuses on medium-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Interleaving String.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Interleaving String carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
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 they are divided into non-empty substrings such that s = s1 + s2 + ... + sn, t = t1 + t2 + ... + tm, |n - m| <= 1, and the interleaved string is s1 + t1 + s2 + t2 + ... or t1 + s1 + t2 + s2 + ...
Write a function isInterleave(s1: str, s2: str, s3: str) -> bool.
- •0 <= len(s1), len(s2) <= 100
- •0 <= len(s3) <= 200
- •s1, s2, and s3 consist of lowercase English letters
Examples
s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
True
aadbbcbcac can be formed by interleaving "aabcc" and "dbbca".
s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
False
It is impossible to interleave s1 and s2 to obtain s3.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
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Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def is_interleave_opt(s1, s2, s3):
if len(s1) + len(s2) != len(s3): return False
dp = [[False for j in range(len(s2) + 1)] for i in range(len(s1) + 1)]
dp[len(s1)][len(s2)] = True
for i in range(len(s1), -1, -1):
for j in range(len(s2), -1, -1):
if i < len(s1) and s1[i] == s3[i + j] and dp[i + 1][j]: dp[i][j] = True
if j < len(s2) and s2[j] == s3[i + j] and dp[i][j + 1]: dp[i][j] = True
return dp[0][0]Brute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def is_interleave_brute(s1, s2, s3):
if len(s1) + len(s2) != len(s3): return False
def solve(i, j, k):
if k == len(s3): return True
res = False
if i < len(s1) and s1[i] == s3[k]: res = res or solve(i + 1, j, k + 1)
if j < len(s2) and s2[j] == s3[k]: res = res or solve(i, j + 1, k + 1)
return res
return solve(0, 0, 0)Algorithm Pattern Checklist
When dealing with 2D DP data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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