Minimum ASCII Delete Sum

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Steps

setup

Initialize dp array with zeros

Create a dp array of length n+1 (4) initialized to zero, representing costs when s1 is empty.

💡 This sets the base case: no characters from s1 means cost depends only on s2 deletions.

Line:

dp = [0] * (n + 1)

💡 dp array length matches s2 prefix lengths; initially zero since no deletions yet.

fill_row

Fill dp for empty s1 and s2 prefixes

Fill dp[j] for j=1 to n with cumulative ASCII sums of s2 prefixes, representing cost to delete s2 chars when s1 is empty.

💡 This shows the cost of deleting all characters from s2 up to j when s1 is empty.

Line:

for j in range(1, n + 1):
    dp[j] = dp[j-1] + ord(s2[j-1])

💡 dp[j] accumulates ASCII values of s2 characters, reflecting deletion cost.

fill_cells

Start processing s1 character 's' (i=1)

Begin first iteration over s1 with character 's'. Update dp[0] by adding ASCII of 's' to represent deleting 's' when s2 is empty.

💡 dp[0] tracks cost when s2 prefix is empty, so deleting s1 chars accumulates here.

Line:

prev = dp[0]
dp[0] += ord(s1[i-1])

💡 Deleting 's' costs 115, so dp[0] updates from 0 to 115.

compare

Compare s1='s' and s2='e' (j=1)

Compare characters 's' and 'e'. They differ, so choose minimum cost between deleting 's' or 'e'.

💡 This decision shows how the algorithm picks cheaper deletion to minimize total cost.

Line:

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

💡 Both deletion options cost 202, so dp[1] updates to 202.

compare

Compare s1='s' and s2='a' (j=2)

Compare 's' and 'a'. They differ, so compute minimal deletion cost similarly.

💡 Each character mismatch requires evaluating deletion costs from either string.

Line:

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

💡 Deleting 'a' from s2 is cheaper here, so dp[2] updates to 299.

compare

Compare s1='s' and s2='t' (j=3)

Compare 's' and 't'. They differ, compute minimal deletion cost.

💡 This step continues evaluating deletion costs for all s2 prefixes.

Line:

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

💡 Deleting 't' from s2 is cheaper, dp[3] updates to 415.

fill_cells

Start processing s1 character 'e' (i=2)

Update dp[0] by adding ASCII of 'e' to represent deleting 'e' when s2 is empty.

💡 dp[0] accumulates deletion cost for s1 prefix when s2 is empty.

Line:

prev = dp[0]
dp[0] += ord(s1[i-1])

💡 dp[0] updates from 115 to 115 + 101 = 216.

compare

Compare s1='e' and s2='e' (j=1)

Characters match ('e' == 'e'), so copy previous diagonal value to dp[1].

💡 Matching characters mean no deletion cost added here.

Line:

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

💡 Matching chars allow skipping deletion, reducing cost.

compare

Compare s1='e' and s2='a' (j=2)

Characters differ ('e' != 'a'), choose minimal deletion cost.

💡 Evaluating whether to delete from s1 or s2 to minimize cost.

Line:

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

💡 Deleting 'a' from s2 is cheaper, dp[2] updates to 212.

compare

Compare s1='e' and s2='t' (j=3)

Characters differ ('e' != 't'), compute minimal deletion cost.

💡 Continuing to evaluate deletion costs for all s2 prefixes.

Line:

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

💡 Deleting 't' from s2 is cheaper, dp[3] updates to 328.

fill_cells

Start processing s1 character 'a' (i=3)

Update dp[0] by adding ASCII of 'a' to represent deleting 'a' when s2 is empty.

💡 dp[0] accumulates deletion cost for s1 prefix when s2 is empty.

Line:

prev = dp[0]
dp[0] += ord(s1[i-1])

💡 dp[0] updates from 216 to 216 + 97 = 313.

compare

Compare s1='a' and s2='e' (j=1)

Characters differ ('a' != 'e'), choose minimal deletion cost.

💡 Evaluating deletion cost choices for prefixes ending with 'a' and 'e'.

Line:

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

💡 Deleting 'a' from s1 is cheaper, dp[1] updates to 212.

compare

Compare s1='a' and s2='a' (j=2)

Characters match ('a' == 'a'), copy previous diagonal value to dp[2].

💡 Matching characters mean no deletion cost added here.

Line:

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

💡 Matching chars allow skipping deletion, reducing cost.

compare

Compare s1='a' and s2='t' (j=3)

Characters differ ('a' != 't'), compute minimal deletion cost.

💡 Final comparison for this s1 character iteration.

Line:

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

💡 Deleting 't' from s2 is cheaper, dp[3] updates to 231.

reconstruct

Final answer read from dp[n]

After processing all characters, dp[n] holds the minimum ASCII delete sum to make s1 and s2 equal.

💡 This final value is the solution to the problem.

Line:

return dp[n]

💡 dp[3] = 231 is the minimal ASCII delete sum for s1='sea' and s2='eat'.

def minimumDeleteSum(s1: str, s2: str) -> int:
    m, n = len(s1), len(s2)
    dp = [0] * (n + 1)  # STEP 1: Initialize dp array
    for j in range(1, n + 1):  # STEP 2: Fill base case for empty s1
        dp[j] = dp[j-1] + ord(s2[j-1])
    for i in range(1, m + 1):  # STEP 3: Iterate over s1
        prev = dp[0]
        dp[0] += ord(s1[i-1])  # STEP 3: Update dp[0] for s1 prefix
        for j in range(1, n + 1):  # STEP 4-14: Iterate over s2
            temp = dp[j]
            if s1[i-1] == s2[j-1]:
                dp[j] = prev  # STEP when chars match
            else:
                dp[j] = min(dp[j] + ord(s1[i-1]), dp[j-1] + ord(s2[j-1]))  # STEP when chars differ
            prev = temp
    return dp[n]  # STEP 15: Return final answer

if __name__ == '__main__':
    print(minimumDeleteSum("sea", "eat"))  # Output: 231

📊

Minimum ASCII Delete Sum - Watch the Algorithm Execute, Step by Step

Watching the dp array update step-by-step reveals how the algorithm balances deleting characters from either string to minimize total ASCII cost.

Initializing dp array

i\w	0	1	2	3
i=0	0	0	0	0

dp array initialized

Initializing dp array

i\w	0	1	2	3
i=0	0	101	202	318

dp[j] updated

Item 0 - wt:115 val:115

i\w	0	1	2	3
i=0	115	101	202	318

dp[0] updated

Item 0 - wt:115 val:202

i\w	0	1	2	3
i=0	115	202	202	318

dp[1] updated

Item 0 - wt:115 val:299

i\w	0	1	2	3
i=0	115	202	299	318

dp[2] updated

Item 0 - wt:115 val:415

i\w	0	1	2	3
i=0	115	202	299	415

dp[3] updated

Item 1 - wt:101 val:216

i\w	0	1	2	3
i=0	216	202	299	415

dp[0] updated

Item 1 - wt:101 val:115

i\w	0	1	2	3
i=0	216	115	299	415

chars match, dp[1] copied

Item 1 - wt:101 val:212

i\w	0	1	2	3
i=0	216	115	212	415

dp[2] updated

Item 1 - wt:101 val:328

i\w	0	1	2	3
i=0	216	115	212	328

dp[3] updated

Item 2 - wt:97 val:313

i\w	0	1	2	3
i=0	313	115	212	328

dp[0] updated

Item 2 - wt:97 val:212

i\w	0	1	2	3
i=0	313	212	212	328

dp[1] updated

Item 2 - wt:97 val:115

i\w	0	1	2	3
i=0	313	212	115	328

chars match, dp[2] copied

Item 2 - wt:97 val:231

i\w	0	1	2	3
i=0	313	212	115	231

dp[3] updated

Item 2 - wt:97 val:231

i\w	0	1	2	3
i=0	313	212	115	231

final answer

Key Takeaways

✓ The dp array represents the minimal ASCII delete sum for prefixes of s1 and s2, updated iteratively.

This insight is hard to see from code alone because the 1D dp array reuses space and depends on previous iteration values.

✓ Matching characters allow copying the diagonal dp value, meaning no deletion cost is added at that step.

This shows how the algorithm leverages character matches to reduce cost, a key dynamic programming optimization.

✓ When characters differ, the algorithm chooses the cheaper deletion between deleting from s1 or s2, reflected in dp updates.

Seeing the cost comparisons step-by-step clarifies how the minimal cost path is constructed.

Practice

1. Consider the following code snippet for finding the longest palindromic substring using dynamic programming. What is the value of max_len after processing the input s = "abba"?

easy

A. 1

B. 3

C. 2

D. 4

Solution

1. Step 1: Trace dp for input "abba"

   Single chars dp[i][i] = True, max_len=1. For length=2, "ab" dp[0][1]=False, "bb" dp[1][2]=True, max_len=2. For length=3, "abb" dp[0][2]=False, "bba" dp[1][3]=False. For length=4, "abba" dp[0][3]=True (since s[0]==s[3] and dp[1][2]=True), max_len=4.

2. Step 2: Confirm max_len update

   max_len updated to 2 at "bb", then to 4 at "abba".

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Longest palindrome is "abba" length 4 [OK]

Hint: Check dp updates for length 4 substring "abba" [OK]

Common Mistakes:

• Off-by-one errors in length calculation
• Missing even length palindrome updates
• Confusing max_len update condition

2. Consider the following buggy code for the space-optimized edit distance. Which line contains the subtle bug that causes incorrect results for empty strings or boundary cases?

def minDistance(word1: str, word2: str) -> int:
    m, n = len(word1), len(word2)
    dp = list(range(n + 1))
    for i in range(1, m + 1):
        prev = dp[0]
        dp[0] = 0  # Bug here
        for j in range(1, n + 1):
            temp = dp[j]
            if word1[i - 1] == word2[j - 1]:
                dp[j] = prev
            else:
                dp[j] = min(dp[j] + 1, dp[j - 1] + 1, prev + 1)
            prev = temp
    return dp[n]

medium

A. Line 4: dp = list(range(n + 1))

B. Line 6: prev = dp[0]

C. Line 7: dp[0] = 0 # Bug here

D. Line 11: dp[j] = min(dp[j] + 1, dp[j - 1] + 1, prev + 1)

Solution

1. Step 1: Understand dp[0] initialization role

   dp[0] represents cost of deleting all characters from word1 prefix to match empty word2 prefix, so it must be i, not 0.

2. Step 2: Identify incorrect dp[0] reset

   Setting dp[0] = 0 inside loop resets deletion cost incorrectly, breaking base case for empty prefixes.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Correct dp[0] update is dp[0] = i [OK]

Hint: dp[0] must track deletions, so dp[0] = i each iteration [OK]

Common Mistakes:

• Resetting dp[0] to zero
• Confusing prev and temp
• Misindexing dp array

3. What is the time and space complexity of the space-optimized bottom-up DP solution for the interleaving string problem, where m and n are the lengths of s1 and s2 respectively?

medium

A. Time: O(m*n), Space: O(n)

B. Time: O(m*n), Space: O(m*n)

C. Time: O(2^(m+n)), Space: O(m+n)

D. Time: O(m+n), Space: O(m+n)

Solution

1. Step 1: Analyze time complexity

   The DP fills a table of size m*n (or simulates it with 1D array), so time is O(m*n).

2. Step 2: Analyze space complexity

   Space optimized approach uses a 1D dp array of size n+1, so space is O(n).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   DP avoids exponential recursion and uses linear space in smaller dimension [OK]

Hint: DP table size dictates time; 1D dp array dictates space [OK]

Common Mistakes:

• Confusing recursion stack space with DP space
• Assuming full 2D dp array space is used in optimized version

4. Examine the following buggy code for shortest common supersequence length. Which line contains the subtle bug that causes incorrect results when one input string is empty?

medium

A. Line 7: Condition comparing str1[i-1] and str2[j-1]

B. Line 5: Initialization of curr with [i] + [0]*n

C. Line 10: Updating prev = curr

D. Line 3: Initialization of prev with zeros instead of range

Solution

1. Step 1: Check base case initialization of prev array

   Prev should be initialized as range(n+1) to represent base case when str1 is empty; zeros cause incorrect base lengths.

2. Step 2: Confirm other lines correctly implement DP transitions

   Lines initializing curr, comparing characters, and updating prev are correct.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Incorrect base initialization leads to wrong results when one string is empty [OK]

Hint: Base case initialization must reflect empty string lengths [OK]

Common Mistakes:

• Initializing dp arrays with zeros instead of incremental ranges
• Mixing up indices in conditions
• Forgetting to update prev array

5. Suppose you want to compute the edit distance where characters can be replaced multiple times (reuse allowed) and the cost of replacement depends on character pairs. Which modification to the classic DP approach correctly handles this variant?

hard

A. Use bottom-up DP with a cost matrix for replacements and update dp accordingly

B. Modify DP to allow revisiting states multiple times with memoization

C. Switch to a greedy approach that picks cheapest replacement at each mismatch

D. Use the same DP but add a third dimension for replacement cost per character pair

Solution

1. Step 1: Understand replacement cost depends on character pairs

   Replacement cost varies, so DP must incorporate a cost matrix for each character pair.

2. Step 2: Adapt bottom-up DP to use cost matrix in dp updates

   When characters differ, dp[j] updates using the specific replacement cost from the matrix, ensuring correct minimal cost.

3. Step 3: Confirm approach handles reuse and varying costs

   Bottom-up DP with cost matrix correctly accumulates minimal costs allowing multiple replacements.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   DP with cost matrix handles variable replacement costs [OK]

Hint: Incorporate cost matrix into DP transitions for variable replacements [OK]

Common Mistakes:

• Adding extra DP dimension unnecessarily
• Using memoization without cost matrix
• Trying greedy which fails globally