Interleaving String

📋

Problem

Imagine you have two different messages and want to combine them into one without changing the order of characters in either message. Can you verify if a given combined message is a valid interleaving of the two?

Given three strings s1, s2, and s3, determine if s3 is formed by an interleaving of s1 and s2. An interleaving of two strings s and t is a string that contains all characters of s and t and preserves the left-to-right order of characters from each string.

1 ≤ len(s1), len(s2) ≤ 10^5len(s3) = len(s1) + len(s2)s1, s2, s3 consist of lowercase English letters

Edge cases: Empty s1 and s2, s3 is empty → trues3 length not equal to sum of s1 and s2 lengths → falseAll characters same in s1 and s2 → test order preservation

</>

IDE

def isInterleave(s1: str, s2: str, s3: str) -> bool:public boolean isInterleave(String s1, String s2, String s3)bool isInterleave(string s1, string s2, string s3)function isInterleave(s1, s2, s3)

def isInterleave(s1, s2, s3):
    # Write your solution here
    pass

class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        // Write your solution here
        return false;
    }
}

#include <string>
using namespace std;

bool isInterleave(string s1, string s2, string s3) {
    // Write your solution here
    return false;
}

function isInterleave(s1, s2, s3) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: falseIncorrect DP initialization or missing base case dp[0][0] = true.✅ Initialize dp[0][0] = true before filling the table.

Wrong: trueNot checking length equality of s3 with s1 + s2 before DP.✅ Add upfront check: if len(s3) != len(s1) + len(s2): return false.

Wrong: falseGreedy approach picking characters from one string only when both match.✅ In DP, check both dp[i-1][j] and dp[i][j-1] when s3 char matches both s1[i-1] and s2[j-1].

Wrong: trueIncomplete DP table filling causing false positives.✅ Fully populate dp table and return dp[len(s1)][len(s2)] as final answer.

Wrong: TLEUsing brute force recursion without memoization for large inputs.✅ Implement DP with memoization or bottom-up tabulation to reduce complexity to O(m*n).

✓

Test Cases

Focus on handling empty strings and exact length boundaries.

Watch out for greedy traps and length mismatch errors.

Optimize your DP to handle large inputs efficiently.

t1_01basic

Input{"s1":"aabcc","s2":"dbbca","s3":"aadbbcbcac"}

Expectedfalse

⏱ Performance - must finish in 2000ms

s3 can be formed by interleaving s1 and s2 preserving order.

💡 Think about dynamic programming to track interleaving possibilities.

💡 Use a 2D DP table where dp[i][j] means s3[:i+j] is interleaving of s1[:i] and s2[:j].

💡 dp[i][j] = (dp[i-1][j] and s1[i-1] == s3[i+j-1]) or (dp[i][j-1] and s2[j-1] == s3[i+j-1]).

Why it failed: Failed basic canonical test: likely incorrect DP recurrence or boundary conditions. Fix by ensuring dp[i][j] checks both s1 and s2 characters correctly.

✓ Correctly handles the canonical example with proper DP logic.

t1_02basic

Input{"s1":"abc","s2":"def","s3":"adbcef"}

Expectedfalse

⏱ Performance - must finish in 2000ms

s3 is a valid interleaving of s1 and s2 preserving order.

💡 Check if characters from s1 and s2 appear in s3 in order.

💡 Use DP to verify if s3[:k] can be formed by s1[:i] and s2[:j].

💡 Update dp[i][j] based on matches with s1[i-1] or s2[j-1] and previous dp states.

Why it failed: Incorrect output due to missing checks for both s1 and s2 characters at each step. Fix by including both conditions in DP update.

✓ Correctly identifies valid interleaving for mixed characters.

t2_01edge

Input{"s1":"","s2":"","s3":""}

Expectedtrue

⏱ Performance - must finish in 2000ms

Empty strings interleave trivially to empty string.

💡 Consider base cases where input strings are empty.

💡 DP base case dp[0][0] should be true.

💡 Return true if all strings are empty.

Why it failed: Fails empty input case by not initializing dp[0][0] to true. Fix by setting dp[0][0] = true.

✓ Handles empty input strings correctly.

t2_02edge

Input{"s1":"a","s2":"","s3":"a"}

Expectedtrue

⏱ Performance - must finish in 2000ms

s3 equals s1 and s2 is empty, so interleaving is true.

💡 Check cases where one string is empty.

💡 DP row or column initialization should handle this.

💡 dp[i][0] = dp[i-1][0] and s1[i-1] == s3[i-1].

Why it failed: Fails single string with empty other string by not initializing dp row/column properly. Fix by initializing dp[i][0] or dp[0][j] accordingly.

✓ Correctly handles one empty string and s3 equals the other.

t2_03edge

Input{"s1":"a","s2":"b","s3":"ab"}

Expectedtrue

⏱ Performance - must finish in 2000ms

s3 length equals sum of s1 and s2 lengths and is valid interleaving.

💡 Check exact boundary where s3 length equals sum of s1 and s2 lengths.

💡 Verify dp table fully covers all indices up to lengths of s1 and s2.

💡 Ensure dp[m][n] is returned where m=len(s1), n=len(s2).

Why it failed: Fails exact boundary by not checking full length or returning dp[m][n]. Fix by returning dp[len(s1)][len(s2)].

✓ Handles exact boundary length correctly.

t3_01corner

Input{"s1":"aaa","s2":"aaa","s3":"aaaaaa"}

Expectedtrue

⏱ Performance - must finish in 2000ms

All characters same; order preservation is key to validate interleaving.

💡 Beware of greedy approaches that pick characters from one string only.

💡 DP must consider both options at each step when characters match.

💡 Check both dp[i-1][j] and dp[i][j-1] when s3 char matches both s1 and s2 chars.

Why it failed: Fails greedy trap by always choosing one string's character first. Fix by checking both dp[i-1][j] and dp[i][j-1] when characters match.

✓ Correctly handles identical characters with order preservation.

t3_02corner

Input{"s1":"abc","s2":"def","s3":"abcdefg"}

Expectedfalse

⏱ Performance - must finish in 2000ms

s3 length not equal to sum of s1 and s2 lengths, so false.

💡 Check length condition before DP.

💡 Return false immediately if lengths don't match.

💡 Add length check at start of function.

Why it failed: Fails length mismatch by not checking s3 length upfront. Fix by returning false if len(s3) != len(s1) + len(s2).

✓ Correctly rejects s3 with invalid length.

t3_03corner

Input{"s1":"aabcc","s2":"dbbca","s3":"aadbbbaccc"}

Expectedfalse

⏱ Performance - must finish in 2000ms

s3 cannot be formed by interleaving s1 and s2 preserving order.

💡 Check all possible interleavings, not just partial matches.

💡 Use memoization or DP to avoid repeated work.

💡 Ensure dp table correctly marks impossible states.

Why it failed: Fails to detect invalid interleaving due to incomplete DP or memoization. Fix by fully populating dp table and returning dp[m][n].

✓ Correctly identifies invalid interleaving.

t4_01performance

Input{"_description":"n=100000 at constraint boundary - executor generates this input"}

Expectednull

⏱ Performance - must finish in 2000ms

Input size n=100000; O(m*n) DP must complete within 2 seconds.

💡 Brute force exponential solutions will time out.

💡 Use bottom-up DP with memoization or space optimization.

💡 Avoid recursion without memoization to pass time limits.

Why it failed: TLE due to exponential or unoptimized DP complexity. Fix by implementing O(m*n) DP with memoization or tabulation.

✓ Efficient DP solution runs within time limits.

Practice

1. You are given two strings and want to find the minimum number of delete operations needed to make them equal. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Dynamic Programming based on Longest Common Subsequence (LCS)

B. Greedy approach deleting mismatched characters as soon as found

C. Pure brute force recursion without memoization

D. Sorting both strings and comparing characters sequentially

Solution

1. Step 1: Understand problem requires minimal deletions to equalize strings

   This is equivalent to finding the longest common subsequence (LCS) and deleting characters not in it.

2. Step 2: Recognize DP approach based on LCS guarantees optimal solution

   Greedy or sorting approaches fail because they do not consider subsequence order; brute force is correct but inefficient.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   LCS-based DP is classic for minimal delete operations [OK]

Hint: Minimal deletes -> LCS-based DP [OK]

Common Mistakes:

• Thinking greedy deletion works optimally
• Ignoring subsequence order
• Confusing sorting with subsequence

2. You are given two sequences and need to find the length of the longest sequence that appears in both sequences in the same order, but not necessarily contiguously. Which algorithmic approach guarantees finding the optimal solution efficiently?

easy

A. A greedy algorithm that matches characters from left to right whenever they are equal

B. Divide and conquer by splitting sequences and merging results without overlapping

C. Sorting both sequences and then finding the longest increasing subsequence

D. Dynamic programming that builds solutions for smaller prefixes and combines them

Solution

1. Step 1: Understand the problem requirements

   The problem requires finding a subsequence common to both sequences, preserving order but not necessarily contiguous, which is a classic dynamic programming problem.

2. Step 2: Evaluate algorithm suitability

   Greedy approaches fail because they can miss longer subsequences by making local choices. Sorting changes order and breaks subsequence constraints. Divide and conquer without overlapping subproblems cannot efficiently combine partial results. Dynamic programming systematically explores all prefixes and stores intermediate results, ensuring optimality.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   DP is the standard approach for LCS problems [OK]

Hint: LCS requires DP, greedy fails on order constraints [OK]

Common Mistakes:

• Assuming greedy matching always works
• Confusing subsequence with substring
• Trying to sort sequences to simplify

3. The following code attempts to implement the space-optimized bottom-up DP for the Delete Operation for Two Strings problem. Which line contains a subtle bug that can cause incorrect results or index errors?

medium

A. Line 7: for i in range(m - 1, -1, -1):

B. Line 12: dp[j] = 1 + min(dp[j], dp[j + 1])

C. Line 10: if word1[j] == word2[j]:

D. Line 14: prev = temp

Solution

1. Step 1: Identify indices used in inner loop

   Inner loop iterates j over word2 indices; word1 index should be i, not j.

2. Step 2: Spot incorrect character comparison

   Line 10 compares word1[j] instead of word1[i], mixing indices causes wrong subproblem computations.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Correct index usage is critical for DP correctness [OK]

Hint: Check index variables carefully in nested loops [OK]

Common Mistakes:

• Mixing i and j indices
• Incorrect base case initialization
• Wrong iteration order

4. The following code attempts to find the longest palindromic substring using DP. Identify the line containing the subtle bug that causes incorrect results for some inputs.

medium

A. Line: if length < 3:

B. Line: dp[i][i] = True

C. Line: dp[i][j] = dp[i + 1][j - 1]

D. Line: for i in range(n - length + 1):

Solution

1. Step 1: Compare condition with correct code

   The correct condition is length <= 3, but here it is length < 3, so length=3 substrings are not handled correctly.

2. Step 2: Understand impact

   For length=3, dp[i][j] should be True if s[i]==s[j], but this code skips that, causing missed palindromes of length 3.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Changing < to <= fixes the bug [OK]

Hint: Check boundary conditions for length 3 palindromes [OK]

Common Mistakes:

• Off-by-one in length condition
• Confusing substring indices
• Incorrect DP table initialization

5. Suppose you want to find the longest palindromic subsequence in a string where characters can be reused unlimited times. Which modification to the DP approach correctly handles this variant?

hard

A. Modify DP to consider infinite reuse by allowing i to stay fixed while j moves, and vice versa

B. Use a greedy approach selecting matching characters repeatedly from both ends

C. Use the same DP but allow indices i and j to move independently without restriction

D. This problem reduces to longest palindromic substring, so use Manacher's algorithm

Solution

1. Step 1: Understand reuse means characters can be counted multiple times

   Standard LPS counts each character once; reuse requires relaxing index movement constraints.

2. Step 2: Modify DP transitions to allow staying at i or j to reuse characters

   DP must consider cases where i or j does not advance to reuse characters, effectively allowing repeated use.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Allowing indices to stay or move independently models reuse correctly [OK]

Hint: Reuse means relaxing index movement in DP transitions [OK]

Common Mistakes:

• Using greedy approach
• Confusing substring with subsequence
• Applying substring algorithms incorrectly