Delete Operation for Two Strings

📋

Delete Operation for Two Strings

Edge cases:

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IDE

def minDistance(word1: str, word2: str) -> int:public int minDistance(String word1, String word2)int minDistance(string word1, string word2)function minDistance(word1, word2)

def minDistance(word1, word2):
    # Write your solution here
    pass

class Solution {
    public int minDistance(String word1, String word2) {
        // Write your solution here
        return 0;
    }
}

#include <string>
using namespace std;

int minDistance(string word1, string word2) {
    // Write your solution here
    return 0;
}

function minDistance(word1, word2) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: 0Returning zero deletions for all inputs, ignoring differences between strings.✅ Implement DP to compute LCS and calculate deletions as sum of lengths minus twice LCS length.

Wrong: len(word1) or len(word2) onlyOnly deleting characters from one string, ignoring the other string's deletions.✅ Calculate deletions as len(word1) + len(word2) - 2 * LCS(word1, word2).

Wrong: Incorrect value due to greedy matchingUsing greedy approach to match characters instead of DP for LCS.✅ Use DP table to find maximum LCS length instead of greedy matching.

Wrong: TLE or timeoutUsing pure recursion without memoization causing exponential time complexity.✅ Implement memoization or bottom-up DP to achieve O(m*n) time complexity.

Wrong: Off-by-one errors causing wrong LCS lengthIncorrect indexing in DP table or string character comparisons.✅ Carefully index DP table and strings; ensure loops and conditions are correct.

✓

Test Cases

Focus on handling empty and minimal input cases correctly.

Watch out for greedy traps and indexing errors in your DP.

Optimize your solution to run in O(m*n) time to handle large inputs.

t1_01basic

Input{"word1":"sea","word2":"eat"}

Expected2

⏱ Performance - must finish in 2000ms

Delete 's' from "sea" and 't' from "eat" to make both strings "ea".

💡 Think about the longest common subsequence (LCS) between the two strings.

💡 Use dynamic programming to find the LCS length, then calculate deletions as total length minus twice the LCS.

💡 Minimum deletions = len(word1) + len(word2) - 2 * LCS(word1, word2).

Why it failed: Incorrect output likely due to not correctly computing LCS or miscalculating deletions. Fix by implementing DP for LCS and then compute deletions as sum of lengths minus twice LCS length.

✓ Correctly computes minimum deletions using LCS-based DP.

t1_02basic

Input{"word1":"leetcode","word2":"etco"}

Expected4

⏱ Performance - must finish in 2000ms

LCS is "etco" with length 4; deletions = 8 + 4 - 2*4 = 4.

💡 Find the longest common subsequence between the two strings.

💡 Calculate deletions as total length minus twice the LCS length.

💡 Here, LCS length is 4, so deletions = 8 + 4 - 8 = 4.

Why it failed: Wrong output due to incorrect LCS calculation or deletion formula. Fix by ensuring DP computes LCS correctly and deletions = len(word1) + len(word2) - 2 * LCS.

✓ Correctly handles different length strings and computes minimum deletions.

t2_01edge

Input{"word1":"","word2":"abc"}

Expected3

⏱ Performance - must finish in 2000ms

One string empty, so all characters of the other must be deleted.

💡 If one string is empty, the answer is the length of the other string.

💡 Check base cases in your DP for empty strings.

💡 Return len(word1) if word2 is empty, else len(word2) if word1 is empty.

Why it failed: Fails empty string base case; fix by adding base case returns when either string is empty.

✓ Correctly handles empty string inputs.

t2_02edge

Input{"word1":"a","word2":"a"}

Expected0

⏱ Performance - must finish in 2000ms

Both strings identical; no deletions needed.

💡 If strings are identical, minimum deletions is zero.

💡 Check if your DP correctly identifies full matches.

💡 Ensure that when characters match, you move both pointers without adding deletions.

Why it failed: Fails identical strings case; fix by returning 0 when strings match completely or DP reaches end with no mismatches.

✓ Correctly returns zero deletions for identical strings.

t2_03edge

Input{"word1":"abc","word2":"def"}

Expected6

⏱ Performance - must finish in 2000ms

No common characters; must delete all characters from both strings.

💡 If no common characters, LCS length is zero.

💡 Minimum deletions equals sum of lengths of both strings.

💡 Check that your DP returns zero for LCS when no matches found.

Why it failed: Fails no common characters case; fix by ensuring DP returns 0 LCS when no matches and deletions = sum of lengths.

✓ Correctly handles strings with no common characters.

t2_04edge

Input{"word1":"aaa","word2":"aa"}

Expected1

⏱ Performance - must finish in 2000ms

LCS is "aa" length 2; deletions = 3 + 2 - 2*2 = 1.

💡 Repeated characters require careful DP to count LCS correctly.

💡 Avoid greedy matching; use full DP table to find max LCS.

💡 Check your DP transitions to handle repeated characters properly.

Why it failed: Fails repeated characters case; fix by using full DP table for LCS instead of greedy matching.

✓ Correctly handles repeated characters with DP.

t3_01corner

Input{"word1":"aabbcc","word2":"abcabc"}

Expected4

⏱ Performance - must finish in 2000ms

LCS is "abcc" or "aabc" with length 4; deletions = 6 + 6 - 2*4 = 4.

💡 Repeated characters require careful DP to count LCS correctly.

💡 Avoid greedy matching; use full DP table to find max LCS.

💡 Check your DP transitions to handle repeated characters properly.

Why it failed: Greedy approach fails repeated characters; fix by using full DP table for LCS instead of greedy matching.

✓ Correctly handles repeated characters with DP.

t3_02corner

Input{"word1":"abc","word2":"abcabc"}

Expected3

⏱ Performance - must finish in 2000ms

LCS is "abc" length 3; deletions = 3 + 6 - 2*3 = 3.

💡 Check that your solution treats this as 0/1 deletion per character, not unbounded.

💡 Do not count multiple deletions of the same character beyond string length.

💡 Ensure DP indices and transitions respect string lengths.

Why it failed: Confuses 0/1 deletion with unbounded; fix by limiting deletions to characters present and using DP indices correctly.

✓ Correctly handles 0/1 deletion constraint.

t3_03corner

Input{"word1":"abcde","word2":"abfde"}

Expected2

⏱ Performance - must finish in 2000ms

LCS is "abde" length 4; deletions = 5 + 5 - 2*4 = 2.

💡 Watch for off-by-one errors in DP indexing.

💡 Ensure you compare characters at correct indices before deciding deletion.

💡 Verify DP table initialization and boundary conditions.

Why it failed: Off-by-one indexing error causes wrong LCS; fix by carefully indexing characters and DP table.

✓ Correctly handles off-by-one indexing in DP.

t4_01performance

Input{"word1":"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa","word2":"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"}

Expectednull

⏱ Performance - must finish in 2000ms

n=500, O(m*n) DP must complete within 2s.

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

💡 Avoid exponential recursion without memoization.

💡 Optimize space if needed but focus on time complexity.

Why it failed: TLE due to exponential brute force recursion; fix by implementing DP memoization or bottom-up approach.

✓ Efficient O(m*n) DP solution confirmed.

Practice

1. Consider the following Python function implementing the space-optimized DP for Minimum ASCII Delete Sum:

def minimumDeleteSum(s1: str, s2: str) -> int:
    m, n = len(s1), len(s2)
    dp = [0] * (n + 1)
    for j in range(1, n + 1):
        dp[j] = dp[j-1] + ord(s2[j-1])
    for i in range(1, m + 1):
        prev = dp[0]
        dp[0] += ord(s1[i-1])
        for j in range(1, n + 1):
            temp = dp[j]
            if s1[i-1] == s2[j-1]:
                dp[j] = prev
            else:
                dp[j] = min(dp[j] + ord(s1[i-1]), dp[j-1] + ord(s2[j-1]))
            prev = temp
    return dp[n]

print(minimumDeleteSum("sea", "eat"))

What is the final returned value for input s1 = "sea" and s2 = "eat"?

easy

A. 231

B. 197

C. 314

D. 259

Solution

1. Step 1: Initialize dp array for s2="eat"

   dp after first loop: dp=[0, 101('e'), 101+97('a')=198, 198+116('t')=314]

2. Step 2: Iterate over s1="sea" updating dp

   After processing 's': dp=[115, 101, 198, 314] After 'e': dp=[216, 115, 198, 314] After 'a': dp final dp[n]=231

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Matches known example output 231 [OK]

Hint: Trace dp array updates carefully for each character [OK]

Common Mistakes:

• Off-by-one errors in indexing dp or strings

2. The following code attempts to implement the space-optimized LCS algorithm. Which line contains a subtle bug that can cause incorrect results on some inputs?

medium

A. Line 7: for j in range(1, n + 1): # iterating forwards

B. Line 5: for i in range(1, m + 1):

C. Line 9: if text1[i - 1] == text2[j - 1]:

D. Line 12: prev = temp

Solution

1. Step 1: Identify iteration order in inner loop

   The inner loop iterates forwards (j=1 to n), which causes dp[j-1] to be overwritten before it is used in the max calculation.

2. Step 2: Understand impact of forward iteration

   Because dp[j-1] is updated before dp[j], the algorithm uses updated values instead of previous iteration's values, leading to incorrect LCS length.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Inner loop must iterate backwards to preserve dp[j-1] values [OK]

Hint: Inner loop must run backwards to avoid overwriting dp array [OK]

Common Mistakes:

• Iterating inner loop forwards in space optimized DP
• Mixing up indices in character comparison
• Incorrect initialization of prev variable

3. Suppose you want to find the length of the longest common subsequence between two strings, but now characters from the second string can be reused multiple times. Which modification to the standard LCS algorithm correctly handles this scenario?

hard

A. Use the same DP but allow dp[j] to depend on dp[j] itself without decrementing j in inner loop

B. Change the inner loop to iterate forwards and update dp[j] using dp[j] and dp[j-1] to allow reuse

C. Use a greedy approach matching characters from left to right allowing reuse

D. Use a bottom-up DP with a 2D table but ignore previous matches to allow reuse

Solution

1. Step 1: Understand reuse impact

   Allowing reuse of characters in the second string means dp[j] can be updated based on dp[j] itself (not only dp[j-1]) and the inner loop must run forwards to propagate reuse correctly.

2. Step 2: Identify correct DP iteration

   Iterating forwards in the inner loop allows dp[j] to accumulate matches with reuse, unlike standard LCS which iterates backwards to avoid reuse.

3. Step 3: Evaluate other options

   Greedy fails to find optimal subsequence. Ignoring previous matches breaks DP correctness. Using 2D DP without reuse logic is insufficient.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Forward iteration in inner loop enables reuse of characters [OK]

Hint: Forward inner loop iteration enables reuse in DP [OK]

Common Mistakes:

• Using backward iteration preventing reuse
• Trying greedy approach
• Ignoring DP state dependencies

4. Suppose you want to find the longest common substring between two strings, but now substrings can be formed by reusing characters multiple times (i.e., characters can be repeated any number of times). Which modification to the DP approach correctly handles this variant?

hard

A. Use the same DP but allow dp[j] to accumulate without resetting to zero on mismatch

B. Modify DP to allow transitions from dp[j] and dp[j-1] to handle reuse, iterating forward over j

C. Use a suffix automaton or suffix trie data structure instead of DP

D. Apply longest common subsequence DP since reuse allows gaps and repeats

Solution

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

   Longest common substring with reuse breaks contiguity and requires allowing repeated matches.

2. Step 2: Modify DP to allow reuse by iterating forward and allowing dp[j] to build on itself

   Iterating forward over j and allowing dp[j] to depend on dp[j] and dp[j-1] enables reuse of characters multiple times.

3. Step 3: Recognize other options fail

   Resetting dp on mismatch breaks reuse, suffix automaton is complex but not DP, subsequence DP allows gaps but not reuse as described.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Forward iteration with reuse transitions correctly models repeated characters [OK]

Hint: Reuse requires forward iteration and dp[j] depending on itself [OK]

Common Mistakes:

• Not resetting dp breaks reuse
• Using subsequence DP incorrectly
• Ignoring iteration order for reuse

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