Shortest Common Supersequence

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Problem

Imagine merging two versions of a document with minimal redundancy, preserving the order of characters from both versions.

Given two strings str1 and str2, return the shortest string that has both str1 and str2 as subsequences. If multiple answers exist, return any of them.

1 ≤ str1.length, str2.length ≤ 1000str1 and str2 consist of lowercase English letters

Edge cases: One string empty → output is the other stringBoth strings identical → output is the string itselfNo common characters → output is concatenation of both strings

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IDE

def scs(str1: str, str2: str) -> str:public String scs(String str1, String str2)string scs(string str1, string str2)function scs(str1, str2)

def scs(str1, str2):
    # Write your solution here
    pass

class Solution {
    public String scs(String str1, String str2) {
        // Write your solution here
        return "";
    }
}

#include <string>
using namespace std;

string scs(string str1, string str2) {
    // Write your solution here
    return "";
}

function scs(str1, str2) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: abcabcGreedy approach concatenating strings without considering LCS.✅ Use DP to find LCS and merge strings around it instead of simple concatenation.

Wrong: abacbIncorrect DP indexing causing wrong character order or missing characters.✅ Carefully implement DP with correct 1-based indexing and reconstruct supersequence accordingly.

Wrong: aabbccFailing to preserve subsequence order with repeated characters.✅ Track indices properly in DP and merge repeated characters in correct order.

Wrong: abcReturning shorter string when one input is empty instead of the other.✅ Return the non-empty string when the other is empty.

Wrong: abcdefConcatenating strings without checking for common subsequences.✅ Check for LCS and merge accordingly to minimize length.

✓

Test Cases

Focus on handling empty and identical strings correctly before moving on.

Beware of greedy traps and off-by-one errors in DP indexing.

Optimize your DP solution to run efficiently on large inputs.

t1_01basic

Input{"str1":"abac","str2":"cab"}

Expected5

⏱ Performance - must finish in 2000ms

One shortest common supersequence is "cabac" which contains both "abac" and "cab" as subsequences.

💡 Think about combining characters from both strings while preserving order.

💡 Use dynamic programming to find the longest common subsequence (LCS) first.

💡 Construct the shortest common supersequence by merging around the LCS.

Why it failed: Incorrect output length or characters - likely not merging subsequences properly. Fix by using LCS to guide construction of the supersequence.

✓ Correct shortest common supersequence found.

t1_02basic

Input{"str1":"abc","str2":"ac"}

Expected3

⏱ Performance - must finish in 2000ms

"abc" is a shortest common supersequence containing both "abc" and "ac" as subsequences.

💡 Check how common subsequences reduce the length of the supersequence.

💡 Longest common subsequence here is "ac", use it to merge strings.

💡 Include characters from both strings around the LCS to build the answer.

Why it failed: Output longer than necessary or missing characters - fix by correctly identifying and using the LCS to merge.

✓ Shortest common supersequence correctly constructed.

t2_01edge

Input{"str1":"","str2":"abc"}

Expected3

⏱ Performance - must finish in 2000ms

One string empty, so output is the other string.

💡 If one string is empty, the supersequence is the other string.

💡 Check base cases where one input is empty.

💡 Return the non-empty string directly when the other is empty.

Why it failed: Fails empty string base case - fix by returning the non-empty string when the other is empty.

✓ Correctly handles empty string input.

t2_02edge

Input{"str1":"aaa","str2":"aaa"}

Expected3

⏱ Performance - must finish in 2000ms

Both strings identical, so output is the string itself.

💡 If strings are identical, shortest supersequence is the string itself.

💡 Check if strings are equal before processing.

💡 Return one of the strings directly if they match.

Why it failed: Fails identical strings case - fix by returning the string directly if both inputs are equal.

✓ Correctly returns identical strings as supersequence.

t2_03edge

Input{"str1":"abc","str2":"def"}

Expected6

⏱ Performance - must finish in 2000ms

No common characters, so output is concatenation of both strings.

💡 If no common characters, supersequence is concatenation of both strings.

💡 Check if LCS length is zero.

💡 Return concatenation of str1 and str2 if no common subsequence.

Why it failed: Fails no common subsequence case - fix by concatenating both strings when LCS length is zero.

✓ Correctly concatenates strings with no common characters.

t3_01corner

Input{"str1":"abc","str2":"acb"}

Expected4

⏱ Performance - must finish in 2000ms

Greedy approach picking characters from one string first fails; correct SCS is "acbc".

💡 Greedy approach picking characters from one string first can fail.

💡 Check all subsequence orders, not just greedy merges.

💡 Use DP to find LCS and build SCS to avoid greedy traps.

Why it failed: Greedy approach fails by producing longer or incorrect supersequence. Fix by using DP-based LCS to guide construction.

✓ Correctly avoids greedy trap and constructs shortest supersequence.

t3_02corner

Input{"str1":"aabcc","str2":"dbbca"}

Expected8

⏱ Performance - must finish in 2000ms

Handles repeated characters and correct subsequence order; output "adbbccca" is shortest common supersequence.

💡 Repeated characters require careful subsequence order preservation.

💡 Check DP table construction for repeated characters.

💡 Build SCS by merging around LCS considering duplicates.

Why it failed: Fails repeated characters order preservation - fix by correctly tracking indices in DP and merging accordingly.

✓ Correctly handles repeated characters preserving subsequence order.

t3_03corner

Input{"str1":"abcde","str2":"ace"}

Expected5

⏱ Performance - must finish in 2000ms

Tests off-by-one errors in DP indexing; output "abcde" is shortest common supersequence.

💡 Off-by-one errors in DP indices cause incorrect results.

💡 Verify base cases and indexing in DP table.

💡 Ensure DP uses 1-based indexing consistently for strings.

Why it failed: Off-by-one indexing error in DP causes wrong output length or characters. Fix by carefully handling indices in DP and reconstruction.

✓ Correctly handles DP indexing and produces correct supersequence.

t4_01performance

Input{"_description":"n=1000 at constraint boundary - executor generates this input"}

Expectednull

⏱ Performance - must finish in 2000ms

Input size n=1000, O(m*n) DP solution must complete within 2 seconds.

💡 Use bottom-up DP with O(m*n) time complexity.

💡 Avoid exponential brute force recursion.

💡 Optimize space if needed but focus on time complexity.

Why it failed: TLE due to exponential or inefficient solution. Fix by implementing O(m*n) DP approach.

✓ Efficient DP solution confirmed by passing performance test.

Practice

1. You need to transform one string into another using the minimum number of insertions, deletions, and substitutions. Which algorithmic approach guarantees finding the optimal solution efficiently?

easy

A. Dynamic Programming that builds solutions for all prefixes of both strings

B. Greedy algorithm that replaces characters whenever they differ

C. Divide and conquer by splitting strings at midpoint and solving recursively

D. Backtracking all possible sequences of edits without memoization

Solution

1. Step 1: Understand problem requires minimal edits including insertions, deletions, and substitutions

   These operations depend on overlapping subproblems, making DP suitable.

2. Step 2: Identify approach that solves all subproblems efficiently

   Dynamic Programming computes minimal edits for all prefixes, ensuring optimality and polynomial time complexity.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   DP covers all prefix pairs systematically [OK]

Hint: Optimal edit distance requires DP over all prefix pairs [OK]

Common Mistakes:

• Assuming greedy always works
• Thinking divide and conquer alone suffices
• Ignoring overlapping subproblems

2. You are given a string and need to partition it into substrings such that each substring is a palindrome. The goal is to minimize the number of cuts needed to achieve this. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Pure recursion checking all partitions without memoization

B. Greedy approach that cuts whenever a palindrome substring is found from the start

C. Dynamic Programming using a 1D dp array with palindrome center expansion to minimize cuts

D. Sorting substrings by length and choosing the longest palindrome first

Solution

1. Step 1: Understand problem constraints

   The problem requires minimizing cuts so that every substring is a palindrome, which is a classic optimization problem.

2. Step 2: Identify algorithmic approach

   Greedy or sorting approaches fail because palindrome partitions depend on global structure, not local choices. Pure recursion is correct but inefficient. The optimal approach uses DP with palindrome center expansion to efficiently find minimal cuts.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   DP with palindrome center expansion is known optimal [OK]

Hint: Optimal solution uses DP with palindrome center expansion [OK]

Common Mistakes:

• Assuming greedy cuts yield minimal partitions
• Ignoring palindrome checks in DP
• Using pure recursion without memoization

3. Suppose the problem is modified so that characters can be deleted multiple times (i.e., you can delete the same character repeatedly if it appears multiple times), and you want to find the minimum number of deletions to make the two strings equal. Which modification to the DP approach is necessary to handle this variant correctly?

hard

A. Use the same DP but add a third dimension to track deletion counts

B. Switch to a greedy approach deleting all duplicates first

C. No change needed; the original DP already accounts for multiple deletions of the same character

D. Modify the DP to allow reuse of characters by changing the recurrence to consider repeated deletions

Solution

1. Step 1: Understand repeated deletions mean characters can be deleted multiple times

   This breaks the assumption that each character is deleted at most once.

2. Step 2: Modify DP recurrence to consider repeated deletions

   DP must allow transitions that delete characters multiple times, possibly by adjusting state or recurrence to reflect reuse.

3. Step 3: Recognize original DP does not handle repeated deletions explicitly

   Original DP assumes each character is deleted once; greedy or no change fail to capture repeated deletions correctly.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Repeated deletions require recurrence changes, not just dimensions or greedy [OK]

Hint: Repeated deletions -> recurrence must allow reuse [OK]

Common Mistakes:

• Assuming original DP suffices
• Adding unnecessary DP dimensions
• Using greedy incorrectly

4. Suppose the problem is modified so that characters from s1 and s2 can be reused any number of times to form s3. Which of the following algorithmic changes correctly adapts the DP solution to this new requirement?

hard

A. Use the same DP but allow transitions to stay at the same indices in s1 or s2 when characters match, enabling reuse

B. No change needed; the original DP already handles reuse implicitly

C. Change the DP to a shortest path problem in a graph where edges represent character matches with reuse allowed

D. Modify the DP to consider infinite loops by adding a visited set to avoid cycles in recursion

Solution

1. Step 1: Understand reuse impact

   Allowing reuse means indices in s1 and s2 can stay the same or move forward, potentially causing cycles.

2. Step 2: Adapt algorithm

   To avoid infinite recursion or loops, the DP or recursion must track visited states to prevent cycles.

3. Step 3: Evaluate options

   Modify the DP to consider infinite loops by adding a visited set to avoid cycles in recursion correctly adds a visited set to recursion to handle cycles; others either do not prevent cycles or incorrectly claim no change is needed.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Cycle detection is essential when reuse is allowed to ensure termination [OK]

Hint: Reuse allows cycles; track visited states to avoid infinite loops [OK]

Common Mistakes:

• Assuming original DP handles reuse without modification
• Trying to model reuse with simple DP transitions without cycle checks

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