What is the time complexity of Interleaving String?

The optimal solution has a time complexity of O(m*n).

What data structures are used to solve Interleaving String?

This problem uses String, Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Interleaving String?

Insight 1: Dynamic Programming (DP) is suitable for this problem because it exhibits optimal substructure. The problem can be broken down into smaller overlapping subproblems: can a substring of `s3` be formed by interleaving substrings of `s1` and `s2`? Insight 2: A 2D DP table (which we optimize to a 1D array) can efficiently store the results of these subproblems. The DP table represents whether a prefix of `s3` can be formed by interleaving a prefix of `s1` and a prefix of `s2`.

Which companies ask LeetCode 97. Interleaving String in interviews?

This problem is frequently asked at Amazon, Apple, DE Shaw, Goldman Sachs, Microsoft and 7 more companies.

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Interleaving String - LeetCode 97 Solution

Q: How to solve Interleaving String?

The provided code uses dynamic programming with space optimization. A 1D boolean array `dp` of size `len2 + 1` is used. `dp[j]` represents whether the prefix of length `i+j` of `s3` can be formed by interleaving the prefix of length `i` of `s1` and the prefix of length `j` of `s2`. The algorithm iterates through all possible lengths of prefixes of `s1` and `s2`, updating the `dp` array based on whether the current characters of `s1` or `s2` match the corresponding character of `s3`. The final result is stored in `dp[len2]`, which indicates whether the entire string `s3` can be formed by interleaving `s1` and `s2`.

Interleaving String - Complete Solution Guide

Interleaving String is LeetCode problem 97, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 = s 1 + s 2 + ... + s n t = t 1 + t 2 + ... + t m |n - m| <= 1 The interleaving is s 1 + t 1 + s 2 + t 2 + s 3 + t 3 + ... or t 1 + s 1 + t 2 + s 2 + t 3 + s 3 + ... Note: a + b is the concatenation of strings a and b . Example 1: Input: s1 = "aabcc", s2 = "dbbca", s3 = "

Detailed Explanation

The problem asks whether a string `s3` can be formed by interleaving two other strings, `s1` and `s2`. Interleaving means combining `s1` and `s2` such that their characters appear in `s3` while maintaining their original order within themselves. For example, if `s1` is "abc" and `s2` is "def", a possible interleaving `s3` would be "adbecf", "abdecf", or "abcdef". The lengths of `s1` and `s2` must sum up to the length of `s3`.

Solution Approach

The provided code uses dynamic programming with space optimization. A 1D boolean array `dp` of size `len2 + 1` is used. `dp[j]` represents whether the prefix of length `i+j` of `s3` can be formed by interleaving the prefix of length `i` of `s1` and the prefix of length `j` of `s2`. The algorithm iterates through all possible lengths of prefixes of `s1` and `s2`, updating the `dp` array based on whether the current characters of `s1` or `s2` match the corresponding character of `s3`. The final result is stored in `dp[len2]`, which indicates whether the entire string `s3` can be formed by interleaving `s1` and `s2`.

Step-by-Step Algorithm

Step 1: Check if the sum of the lengths of `s1` and `s2` equals the length of `s3`. If not, return `false` because interleaving is impossible.
Step 2: Initialize a 1D boolean array `dp` of size `len2 + 1` with all elements set to `false`. `dp[j]` represents if `s3[0...i+j-1]` is an interleaving of `s1[0...i-1]` and `s2[0...j-1]`.
Step 3: Iterate through all possible lengths `i` of the prefix of `s1` from 0 to `len1` (inclusive).
Step 4: For each `i`, iterate through all possible lengths `j` of the prefix of `s2` from 0 to `len2` (inclusive).
Step 5: Calculate the value of `dp[j]` based on the following cases:
- If `i == 0` and `j == 0`: Set `dp[j]` to `true` (empty strings interleave to form an empty string).
- If `i == 0`: Set `dp[j]` to `dp[j-1] && s2[j-1] == s3[j-1]` (only `s2` contributes to `s3`).
- If `j == 0`: Set `dp[j]` to `dp[j] && s1[i-1] == s3[i-1]` (only `s1` contributes to `s3`).
- Otherwise: Set `dp[j]` to `(dp[j] && s1[i-1] == s3[i+j-1]) || (dp[j-1] && s2[j-1] == s3[i+j-1])`. This means `s3[i+j-1]` can come from either the last character of `s1` or the last character of `s2`.
Step 6: After the iterations, return `dp[len2]`. This value indicates whether `s3` can be formed by interleaving `s1` and `s2`.

Key Insights

Insight 1: Dynamic Programming (DP) is suitable for this problem because it exhibits optimal substructure. The problem can be broken down into smaller overlapping subproblems: can a substring of `s3` be formed by interleaving substrings of `s1` and `s2`?
Insight 2: A 2D DP table (which we optimize to a 1D array) can efficiently store the results of these subproblems. The DP table represents whether a prefix of `s3` can be formed by interleaving a prefix of `s1` and a prefix of `s2`.
Insight 3: The current state of the DP table depends only on the previous state (the row or column above/left). This allows us to reduce the space complexity from O(m*n) to O(min(m, n)) by using only one row/column for DP calculation, thereby addressing the follow up.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(n)

Topics

This problem involves: String, Dynamic Programming.

Companies

Asked at: Amazon, Apple, DE Shaw, Goldman Sachs, Microsoft, Nuro, TikTok, Uber, Walmart Labs, Yahoo, Zeta, thoughtspot.

Interleaving String | LeetCode 97

Medium

String Dynamic Programming

Time: O(m*n)Space: O(n)

Solution Code

Available in:

class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        int len1 = s1.length();
        int len2 = s2.length();
        int len3 = s3.length();

        if (len1 + len2 != len3) {
            return false;
        }

        boolean[] dp = new boolean[len2 + 1];

        for (int i = 0; i <= len1; i++) {
            for (int j = 0; j <= len2; j++) {
                if (i == 0 && j == 0) {
                    dp[j] = true;
                } else if (i == 0) {
                    dp[j] = dp[j-1] && s2.charAt(j-1).equals(s3.charAt(j-1));
                } else if (j == 0) {
                    dp[j] = dp[j] && s1.charAt(i-1).equals(s3.charAt(i-1));
                } else {
                    dp[j] = (dp[j] && s1.charAt(i-1).equals(s3.charAt(i+j-1))) ||
                            (dp[j-1] && s2.charAt(j-1).equals(s3.charAt(i+j-1)));
                }
            }
        }

        return dp[len2];
    }
}

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Problem Info

Description

Problem 97: Interleaving String - Detailed explanation and solution approaches.

Topics

String Dynamic Programming

Companies

AmazonAppleDE ShawGoldman SachsMicrosoftNuroTikTokUber

Stats

Difficulty:Medium

Languages:4

Companies:12

Problem #:97

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Check if the sum of the lengths of `s1` and `s2` equals the length of `s3`. If not, return `false` because interleaving is impossible.
2
Step 2: Initialize a 1D boolean array `dp` of size `len2 + 1` with all elements set to `false`. `dp[j]` represents if `s3[0...i+j-1]` is an interleaving of `s1[0...i-1]` and `s2[0...j-1]`.
3
Step 3: Iterate through all possible lengths `i` of the prefix of `s1` from 0 to `len1` (inclusive).
4
Step 4: For each `i`, iterate through all possible lengths `j` of the prefix of `s2` from 0 to `len2` (inclusive).
5
Step 5: Calculate the value of `dp[j]` based on the following cases:
6
- If `i == 0` and `j == 0`: Set `dp[j]` to `true` (empty strings interleave to form an empty string).
7
- If `i == 0`: Set `dp[j]` to `dp[j-1] && s2[j-1] == s3[j-1]` (only `s2` contributes to `s3`).
8
- If `j == 0`: Set `dp[j]` to `dp[j] && s1[i-1] == s3[i-1]` (only `s1` contributes to `s3`).
9
- Otherwise: Set `dp[j]` to `(dp[j] && s1[i-1] == s3[i+j-1]) || (dp[j-1] && s2[j-1] == s3[i+j-1])`. This means `s3[i+j-1]` can come from either the last character of `s1` or the last character of `s2`.
10
Step 6: After the iterations, return `dp[len2]`. This value indicates whether `s3` can be formed by interleaving `s1` and `s2`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity arises from the nested loops iterating `m` and `n` times, respectively. The space complexity is linear because we're using a single array of size `n+1` to store intermediate results. A naive DP solution would be O(m*n) space.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursion with Memoization

A recursive approach can be used to solve this problem by exploring all possible interleavings. Memoization can be used to avoid redundant calculations.

Time: O(m*n)Space: O(m*n) due to the memoization table and the call stack in worst case

Trade-offs: While conceptually simpler for some, recursion with memoization often has a larger constant factor overhead than the iterative DP approach due to function call overhead and can potentially lead to stack overflow issues for very large inputs. The iterative DP with space optimization is generally preferred.

✨

Key Insights

1Insight 1: Dynamic Programming (DP) is suitable for this problem because it exhibits optimal substructure. The problem can be broken down into smaller overlapping subproblems: can a substring of `s3` be formed by interleaving substrings of `s1` and `s2`?
2Insight 2: A 2D DP table (which we optimize to a 1D array) can efficiently store the results of these subproblems. The DP table represents whether a prefix of `s3` can be formed by interleaving a prefix of `s1` and a prefix of `s2`.
3Insight 3: The current state of the DP table depends only on the previous state (the row or column above/left). This allows us to reduce the space complexity from O(m*n) to O(min(m, n)) by using only one row/column for DP calculation, thereby addressing the follow up.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty strings. If `s1`, `s2`, and `s3` are all empty, the code correctly returns `true` because an empty string is considered an interleaving of two empty strings.
Edge case 2: One of the strings is empty. The code handles this case correctly by checking if `i == 0` or `j == 0` and updating the `dp` array accordingly. This allows interleaving when one of the strings contributes nothing.
Edge case 3: When `len(s1) + len(s2) != len(s3)`. This is a fundamental constraint; interleaving is impossible if the combined length doesn't match `s3`'s length. The algorithm checks this at the very beginning and returns `false` immediately.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking the base cases correctly. Especially the cases where `s1` or `s2` are empty strings.

✗Common mistake 2: Incorrect indexing. It's crucial to ensure the indices used to access characters in `s1`, `s2`, and `s3` are correct and consistent with the current state of the DP array.

✗Common mistake 3: Not handling the length constraint: failing to check that len(s1) + len(s2) == len(s3) at the beginning will lead to incorrect results and possibly index out of bounds errors.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Text editing and merging. Determining if a merged document can be created by interleaving changes from two separate versions of the original document.

🏦Real-world scenario 2: Bioinformatics: Sequence alignment. Identifying if a DNA sequence could have resulted from interleaving two other DNA sequences after some insertions/deletions.

🎮Real-world scenario 3: Network packet analysis. Checking if a combined network stream can be formed by interleaving packets from two separate connections based on their sequence numbers.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 6, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Interleaving String"?

For this medium-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?