Minimum ASCII Delete Sum

📋

Minimum ASCII Delete Sum

Edge cases:

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IDE

def minimumDeleteSum(s1: str, s2: str) -> int:public int minimumDeleteSum(String s1, String s2)int minimumDeleteSum(string s1, string s2)function minimumDeleteSum(s1, s2)

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

class Solution {
    public int minimumDeleteSum(String s1, String s2) {
        // Write your solution here
        return "0";
    }
}

#include <string>
using namespace std;

string minimumDeleteSum(string s1, string s2) {
    // Write your solution here
    return "0";
}

function minimumDeleteSum(s1, s2) {
    // Write your solution here
    return "0";
}

Common Bugs to Avoid

Wrong: 0Ignoring delete costs and returning zero for all inputs.✅ Implement dp recurrence that adds ASCII delete costs when characters differ.

Wrong: sum of ASCII of only one stringOnly deleting characters from one string, ignoring the other.✅ In dp recurrence, consider deleting from either s1 or s2 and take minimum.

Wrong: incorrect sums due to off-by-one indexingMisaligned dp indices causing wrong character access.✅ Use dp[i][j] to represent s1[:i] and s2[:j], carefully index s1[i-1], s2[j-1].

Wrong: too large sums from greedy deletion of all duplicatesGreedy approach deletes all duplicates instead of minimal needed.✅ Use DP to consider all prefixes and minimal deletions, not greedy.

Wrong: TLE or timeoutExponential brute force recursion without memoization.✅ Implement memoization or bottom-up DP to achieve O(m*n) time.

✓

Test Cases

Focus on handling empty strings and identical strings correctly in your DP base cases.

Test your solution against repeated characters and subtle indexing bugs.

Optimize your solution to O(m*n) DP to handle large inputs efficiently.

t1_01basic

Input{"s1":"sea","s2":"eat"}

Expected231

⏱ Performance - must finish in 2000ms

Delete 's' from s1 (ASCII 115) and 't' from s2 (ASCII 116), total 231.

💡 Think about aligning characters and deleting mismatches with minimum ASCII cost.

💡 Use dynamic programming to track minimum delete sums for prefixes of s1 and s2.

💡 Define dp[i][j] as minimum delete sum for s1[:i] and s2[:j], then build up.

Why it failed: Output incorrect - likely missing correct handling of character matches or delete costs. Fix by ensuring dp recurrence considers both delete options and matches correctly.

✓ Correctly computes minimum ASCII delete sum using DP.

t1_02basic

Input{"s1":"delete","s2":"leet"}

Expected403

⏱ Performance - must finish in 2000ms

Delete 'd' (100) and 'e' (101) from s1 and 'e' (101) from s2 to match 'leet', total 403.

💡 Consider how to handle partial overlaps and deletions in both strings.

💡 Check dp transitions for deleting from s1 or s2 when characters differ.

💡 Ensure dp[i][j] = min(dp[i-1][j] + ASCII(s1[i-1]), dp[i][j-1] + ASCII(s2[j-1])) when chars differ.

Why it failed: Incorrect sum - likely ignoring delete cost from one string or miscomputing ASCII sums. Fix by adding both delete options in dp recurrence.

✓ DP correctly accounts for all delete costs and matches.

t2_01edge

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

Expected294

⏱ Performance - must finish in 2000ms

s1 empty, must delete all chars in s2: 'a'(97)+'b'(98)+'c'(99)=294.

💡 Check behavior when one string is empty.

💡 Minimum delete sum should be sum of ASCII of all characters in the non-empty string.

💡 Initialize dp base cases to sum of ASCII values for empty prefixes.

Why it failed: Fails empty string base case - dp base row/column not initialized properly. Fix by setting dp[i][0] and dp[0][j] to cumulative ASCII sums.

✓ Handles empty string cases correctly.

t2_02edge

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

Expected0

⏱ Performance - must finish in 2000ms

Both strings identical, no deletions needed, cost 0.

💡 Check if algorithm correctly identifies matching characters.

💡 When characters match, no deletion cost should be added.

💡 Ensure dp[i][j] = dp[i-1][j-1] if s1[i-1] == s2[j-1].

Why it failed: Incorrectly adds deletion cost on matching chars. Fix by skipping cost when characters match in dp recurrence.

✓ Correctly returns zero cost for identical strings.

t2_03edge

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

Expected597

⏱ Performance - must finish in 2000ms

No common chars, delete all: 'a'(97)+'b'(98)+'c'(99)+'d'(100)+'e'(101)+'f'(102)=597.

💡 Consider when no characters match between strings.

💡 Minimum delete sum is sum of ASCII of all characters in both strings.

💡 Check dp initialization and transitions handle no matches correctly.

Why it failed: Fails no-common-character case - dp may incorrectly match chars or skip deletes. Fix by ensuring dp considers all delete options.

✓ Correctly sums ASCII values when no matches.

t2_04edge

Input{"s1":"aaa","s2":"aa"}

Expected97

⏱ Performance - must finish in 2000ms

Delete one 'a'(97) from s1 to match s2.

💡 Check handling of repeated characters with different counts.

💡 DP must consider minimal deletes, not greedy removal of all duplicates.

💡 Ensure dp recurrence correctly accounts for extra characters.

Why it failed: Greedy deletion of all duplicates causes overcounting. Fix by using DP to consider all prefixes and minimal deletes.

✓ Correctly handles repeated characters with minimal deletions.

t3_01corner

Input{"s1":"aabbcc","s2":"abc"}

Expected294

⏱ Performance - must finish in 2000ms

Delete one 'a'(97), one 'b'(98), one 'c'(99) from s1 to match s2, total 294.

💡 Check handling of repeated characters and partial matches.

💡 DP must correctly handle multiple occurrences and choose minimal deletes.

💡 Ensure dp recurrence does not greedily delete all duplicates but considers minimal cost.

Why it failed: Greedy deletion of all duplicates causes overcounting. Fix by using DP to consider all prefixes and minimal deletes.

✓ Correctly handles repeated characters with minimal deletions.

t3_02corner

Input{"s1":"abc","s2":"abcc"}

Expected99

⏱ Performance - must finish in 2000ms

Delete extra 'c'(99) from s2 to match s1, cost 99.

💡 Beware of confusing 0/1 deletion with unbounded deletion.

💡 Each character can be deleted once; extra chars must be deleted explicitly.

💡 DP must consider all prefixes, not assume unlimited deletions.

Why it failed: Confuses 0/1 deletion with unbounded - deletes fewer chars than needed. Fix by ensuring DP considers exact prefixes and deletions.

✓ Correctly handles 0/1 deletion constraints.

t3_03corner

Input{"s1":"abcd","s2":"abdc"}

Expected198

⏱ Performance - must finish in 2000ms

Must delete 'c'(99) from s1 and 'd'(100) from s2 or vice versa, total 198.

💡 Watch for off-by-one errors in indexing dp arrays.

💡 Ensure dp indices align with string characters correctly.

💡 Check dp recurrence uses i-1 and j-1 indices properly.

Why it failed: Off-by-one indexing causes incorrect dp values. Fix by carefully indexing dp and strings consistently.

✓ Correctly computes minimal deletes with proper indexing.

t4_01performance

Input{"s1":"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa","s2":"bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"}

Expectednull

⏱ Performance - must finish in 2000ms

n=100, O(m*n) DP must complete in 2s.

💡 DP solution runs in O(m*n) time, suitable for n=100.

💡 Avoid exponential brute force recursion which TLEs here.

💡 Use bottom-up or memoized DP to pass performance test.

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

✓ Efficient DP solution confirmed by passing performance test.

Practice

1. What is the time complexity of the space-optimized bottom-up DP solution for the longest common substring problem, where n and m are the lengths of the two input strings?

medium

A. O(n * m) due to nested loops comparing all character pairs

B. O(n * m) but with additional O(n + m) recursion stack space

C. O(n * m * min(n, m)) due to checking all substrings explicitly

D. O(n + m) due to single pass over both strings

Solution

1. Step 1: Identify nested loops over n and m

   The algorithm uses two nested loops: outer over n and inner over m, leading to O(n*m) iterations.

2. Step 2: Confirm no recursion or substring extraction

   Space optimization uses iterative DP with O(m) space, no recursion stack or substring extraction, so no extra multiplicative factors.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Nested loops over both strings dominate time complexity [OK]

Hint: Nested loops over both strings imply O(n*m) time [OK]

Common Mistakes:

• Confusing with linear O(n+m)
• Assuming recursion adds stack space
• Thinking substring extraction adds extra factor

2. Suppose you want to count palindromic substrings in a string where characters can be replaced any number of times (reusable characters). Which modification to the expand around center approach correctly handles this variant?

hard

A. No modification needed; the original expand around center approach already counts palindromes with replacements.

B. Modify expand to allow skipping mismatched characters to find palindromes with replacements.

C. Precompute all possible character replacements and run the original expand around center for each variant.

D. Use dynamic programming with memoization to store counts for substrings allowing character replacements.

Solution

1. Step 1: Understand the variant

   Allowing character replacements means palindrome checks must consider substitutions, not exact matches.

2. Step 2: Evaluate approaches

   Expand around center relies on exact matches; skipping mismatches breaks palindrome definition. Precomputing all variants is exponential. DP with memoization can track allowed replacements efficiently.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   DP can handle flexible palindrome definitions with substitutions [OK]

Hint: Exact match expansion fails when replacements allowed [OK]

Common Mistakes:

• Assuming original approach works without changes

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 you want to find the shortest common supersequence when characters from the input strings can be reused unlimited times. Which modification to the DP approach correctly handles this variant?

hard

A. Change the recurrence to allow staying at the same index in one string while advancing in the other

B. Modify the DP to consider infinite loops by adding memoization to avoid repeated states

C. Use the same DP but allow indices i and j to reset to zero after reaching the end, enabling reuse

D. No change needed; the original DP already accounts for unlimited reuse of characters

Solution

1. Step 1: Understand that unlimited reuse means characters can be used multiple times, so indices may not always advance

   Original DP always advances both indices or one index forward, disallowing reuse.

2. Step 2: Modify recurrence to allow staying at the same index in one string while advancing in the other

   This models reuse by allowing repeated characters from one string to be matched multiple times.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Allowing index to stay while advancing the other correctly models reuse [OK]

Hint: Reuse means indices may not always advance simultaneously [OK]

Common Mistakes:

• Assuming original DP handles reuse
• Trying to reset indices naively
• Adding memoization without changing recurrence

5. Suppose the problem is modified so that each integer in nums1 and nums2 can be used multiple times to form uncrossed lines (i.e., reuse allowed). Which of the following algorithmic changes correctly adapts the solution?

hard

A. Modify the DP to consider unlimited reuse by allowing transitions that do not increment both indices simultaneously after a match

B. Change the DP to allow staying on the same index in nums1 or nums2 after a match, effectively converting to a longest common substring problem

C. Use a greedy approach to connect all matching pairs regardless of crossing

D. Use the same DP but allow matching the same indices multiple times by removing the increment of i and j after a match

Solution

1. Step 1: Understand reuse implications

   Allowing reuse means after matching nums1[i] and nums2[j], we can match nums1[i] or nums2[j] again, so indices may not both increment.

2. Step 2: Adapt DP transitions

   The DP must allow transitions that increment only one index or none after a match, unlike classic LCS where both indices increment. This requires modifying the recurrence to consider reuse.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Allowing reuse requires DP transitions that do not always increment both indices [OK]

Hint: Reuse means indices may not both advance after a match [OK]

Common Mistakes:

• Removing increments without adjusting DP
• Confusing reuse with substring problem
• Using greedy approach ignoring crossing