Interleaving String

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Steps

setup

Initialize dp array with False

Create a dp array of length n+1 (6) filled with False, representing no prefixes matched yet.

💡 Initializing dp with False sets a baseline where no interleaving is assumed until proven otherwise.

Line:

dp = [False] * (n + 1)

💡 The dp array size depends on s2's length plus one, preparing to track all prefix combinations.

setup

Set dp[0] = True for empty prefixes

Set dp[0] to True because empty prefixes of s1 and s2 trivially interleave to form empty s3 prefix.

💡 This base case anchors the dp array, allowing subsequent prefix matches to build upon it.

Line:

dp[0] = True

💡 dp[0] = True means no characters taken from either string matches empty s3 prefix.

fill_row

Fill dp for first row based on s2 prefix matching s3

Iterate j from 1 to n, setting dp[j] = dp[j-1] and s2[j-1] == s3[j-1], checking if s2 prefix alone matches s3 prefix.

💡 This step checks if s3's prefix can be formed solely by s2's prefix, initializing the first row of the conceptual 2D dp.

Line:

for j in range(1, n + 1):
    dp[j] = dp[j - 1] and s2[j - 1] == s3[j - 1]

💡 This shows how the dp array accumulates valid prefix matches from s2 alone.

fill_cells

Check dp[1] to dp[5] for s2 prefixes

Check each dp[j] for j=1 to 5 by dp[j] = dp[j-1] and s2[j-1] == s3[j-1]. All remain False except dp[0].

💡 These checks confirm that s2 prefixes alone do not match s3 prefixes beyond dp[0].

Line:

dp[j] = dp[j - 1] and s2[j - 1] == s3[j - 1]  # for j=1 to 5

💡 No prefix of s2 alone matches s3 beyond empty prefix.

fill_row

Start iterating over s1: i=1, update dp[0]

Update dp[0] = dp[0] and s1[0] == s3[0], True and 'a' == 'a' → True.

💡 dp[0] tracks if s1 prefix alone matches s3 prefix at each i.

Line:

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

💡 Confirms s1 prefix 'a' matches s3 prefix 'a'.

fill_cells

Update dp[5] for i=1, j=5

Update dp[5] = (dp[5] and s1[0] == s3[5]) or (dp[4] and s2[4] == s3[5]) → (False and 'a'=='c') or (False and 'a'=='c') = False.

💡 Check if s3 prefix 'aadbbc' can be formed by interleaving s1[:1] and s2[:5].

Line:

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

💡 Shows how dp updates depend on previous dp values and character matches.

fill_cells

Update dp[4] for i=1, j=4

dp[4] = (dp[4] and s1[0] == s3[4]) or (dp[3] and s2[3] == s3[4]) → (False and 'a'=='b') or (False and 'c'=='b') = False.

💡 Check if s3 prefix 'aadbb' can be formed by s1[:1] and s2[:4].

Line:

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

💡 No valid interleaving at this prefix combination.

fill_cells

Update dp[3] for i=1, j=3

dp[3] = (dp[3] and s1[0] == s3[3]) or (dp[2] and s2[2] == s3[3]) → (False and 'a'=='b') or (False and 'b'=='b') = False.

💡 Check if s3 prefix 'aadb' can be formed by s1[:1] and s2[:3].

Line:

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

💡 No valid interleaving at this prefix combination.

fill_cells

Update dp[2] for i=1, j=2

dp[2] = (dp[2] and s1[0] == s3[2]) or (dp[1] and s2[1] == s3[2]) → (False and 'a'=='d') or (False and 'b'=='d') = False.

💡 Check if s3 prefix 'aad' can be formed by s1[:1] and s2[:2].

Line:

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

💡 No valid interleaving at this prefix combination.

fill_cells

Update dp[1] for i=1, j=1

dp[1] = (dp[1] and s1[0] == s3[1]) or (dp[0] and s2[0] == s3[1]) → (False and 'a'=='a') or (True and 'd'=='a') = False.

💡 Check if s3 prefix 'aa' can be formed by s1[:1] and s2[:1].

Line:

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

💡 No valid interleaving at this prefix combination.

fill_row

Start i=2, update dp[0]

dp[0] = dp[0] and s1[1] == s3[1], True and 'a' == 'a' → True.

💡 Check if s1 prefix 'aa' matches s3 prefix 'aa'.

Line:

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

💡 Confirms s1 prefix matches s3 prefix at i=2.

fill_cells

Update dp[1] for i=2, j=1

dp[1] = (dp[1] and s1[1] == s3[2]) or (dp[0] and s2[0] == s3[2]) → (False and 'a'=='d') or (True and 'd'=='d') = True.

💡 Check if s3 prefix 'aad' can be formed by s1[:2] and s2[:1].

Line:

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

💡 Shows how taking a character from s2 can enable interleaving.

fill_cells

Update dp[2] for i=2, j=2

dp[2] = (dp[2] and s1[1] == s3[3]) or (dp[1] and s2[1] == s3[3]) → (False and 'a'=='b') or (True and 'b'=='b') = True.

💡 Check if s3 prefix 'aadb' can be formed by s1[:2] and s2[:2].

Line:

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

💡 Interleaving can continue by taking s2's character.

fill_cells

Update dp[3] for i=2, j=3

dp[3] = (dp[3] and s1[1] == s3[4]) or (dp[2] and s2[2] == s3[4]) → (False and 'a'=='b') or (True and 'b'=='b') = True.

💡 Check if s3 prefix 'aadbb' can be formed by s1[:2] and s2[:3].

Line:

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

💡 Interleaving continues successfully.

fill_cells

Update dp[4] for i=2, j=4

dp[4] = (dp[4] and s1[1] == s3[5]) or (dp[3] and s2[3] == s3[5]) → (False and 'a'=='c') or (True and 'c'=='c') = True.

💡 Check if s3 prefix 'aadbbc' can be formed by s1[:2] and s2[:4].

Line:

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

💡 Interleaving remains valid.

fill_cells

Update dp[5] for i=2, j=5

dp[5] = (dp[5] and s1[1] == s3[6]) or (dp[4] and s2[4] == s3[6]) → (False and 'a'=='b') or (True and 'a'=='b') = False.

💡 Check if s3 prefix 'aadbbcb' can be formed by s1[:2] and s2[:5].

Line:

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

💡 No match at this prefix combination.

reconstruct

Final step: Return dp[n] after all iterations

After processing all characters of s1 and s2, return dp[n] which indicates if s3 is an interleaving.

💡 The final dp[n] value tells if s3 can be formed by interleaving s1 and s2.

Line:

return dp[n]

💡 The algorithm confirms s3 is a valid interleaving.

def isInterleave(s1, s2, s3):
    m, n = len(s1), len(s2)
    if m + n != len(s3):
        return False
    dp = [False] * (n + 1)  # STEP 1
    dp[0] = True  # STEP 2
    for j in range(1, n + 1):  # STEP 3
        dp[j] = dp[j - 1] and s2[j - 1] == s3[j - 1]  # STEPS 4-8
    for i in range(1, m + 1):
        dp[0] = dp[0] and s1[i - 1] == s3[i - 1]  # STEPS 9,15
        for j in range(n, 0, -1):
            dp[j] = (dp[j] and s1[i - 1] == s3[i + j - 1]) or (dp[j - 1] and s2[j - 1] == s3[i + j - 1])  # STEPS 10-20
    return dp[n]  # STEP 21

📊

Interleaving String - Watch the Algorithm Execute, Step by Step

Watching each update to the dp array reveals how the algorithm builds up the solution from smaller subproblems, making the logic clear without needing to mentally simulate the code.

Item 0 - wt:0 val:0

i\w	0	1	2	3	4	5
i=0

False initialization

Item 0 - wt:0 val:0

i\w	0	1	2	3	4	5
i=0

dp[0] = True

Item 0 - wt:0 val:0

i\w	0	1	2	3	4	5
i=0

dp array after first row initialization

Item 0 - wt:0 val:0

i\w	0	1	2	3	4	5
i=0

dp[1] = False

Item 0 - wt:0 val:0

i\w	0	1	2	3	4	5
i=0

dp[0] updated

Item 0 - wt:0 val:0

i\w	0	1	2	3	4	5
i=0

dp[5] = False

Item 0 - wt:0 val:0

i\w	0	1	2	3	4	5
i=0

dp[4] = False

Item 0 - wt:0 val:0

i\w	0	1	2	3	4	5
i=0

dp[3] = False

Item 0 - wt:0 val:0

i\w	0	1	2	3	4	5
i=0

dp[2] = False

Item 0 - wt:0 val:0

i\w	0	1	2	3	4	5
i=0

dp[1] = False

Item 1 - wt:0 val:0

i\w	0	1	2	3	4	5
i=0

dp[0] updated

Item 1 - wt:0 val:0

i\w	0	1	2	3	4	5
i=0

dp[1] = True

Item 1 - wt:0 val:0

i\w	0	1	2	3	4	5
i=0

dp[2] = True

Item 1 - wt:0 val:0

i\w	0	1	2	3	4	5
i=0

dp[3] = True

Item 1 - wt:0 val:0

i\w	0	1	2	3	4	5
i=0

dp[4] = True

Item 1 - wt:0 val:0

i\w	0	1	2	3	4	5
i=0

dp[5] = False

Item 4 - wt:0 val:0

i\w	0	1	2	3	4	5
i=0

Answer dp[n] = True

Key Takeaways

✓ The dp array tracks whether prefixes of s1 and s2 can interleave to form prefixes of s3.

This is hard to see from code alone because the dp array is 1D and updated in a tricky order.

✓ Updating dp from right to left preserves previous states needed for correct computation.

Visualizing the backward update order clarifies why the algorithm avoids overwriting needed values.

✓ Each dp[j] update chooses between taking a character from s1 or s2, showing the branching decision.

Seeing the dp updates side-by-side reveals how the algorithm explores both possibilities at each step.

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. Consider the following code snippet for counting distinct subsequences of t in s. What is the final returned value when s = "abc" and t = "ac"?

easy

A. 0

B. 1

C. 3

D. 2

Solution

1. Step 1: Trace dp array updates for s="abc", t="ac"

   Initialize dp=[1,0,0]. Iteration i=1 (s[0]='a'): j=2 no match, j=1 matches t[0]='a', dp[1]+=dp[0]=1 -> dp=[1,1,0].

2. Step 2: Continue tracing for i=2 (s[1]='b') and i=3 (s[2]='c')

   i=2 no matches, dp unchanged. i=3 matches t[1]='c', dp[2]+=dp[1]=0+1=1 -> dp=[1,1,1]. Final dp[n]=dp[2]=1.

3. Step 3: Count all subsequences matching "ac"

   Subsequences: "ac" only, total 1.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   dp array correctly counts 1 subsequence [OK]

Hint: Trace dp array backward updates carefully [OK]

Common Mistakes:

• Off-by-one errors in indices
• Forgetting to iterate j backward

3. Consider the following Python function implementing a space-optimized DP for longest common substring. What is the returned value when called with s1 = 'abc' and s2 = 'xbc'?

def longest_common_substring_space_optimized(s1, s2):
    n, m = len(s1), len(s2)
    dp = [0] * (m + 1)
    max_len = 0
    for i in range(1, n + 1):
        for j in range(m, 0, -1):
            if s1[i - 1] == s2[j - 1]:
                dp[j] = dp[j - 1] + 1
                if dp[j] > max_len:
                    max_len = dp[j]
            else:
                dp[j] = 0
    return max_len

easy

A. 2

B. 1

C. 3

D. 0

Solution

1. Step 1: Trace dp array updates for i=1 (s1[0]='a')

   Compare 'a' with 'c','b','x' in s2 backwards: no matches, dp remains zeros.

2. Step 2: Trace dp updates for i=2 (s1[1]='b') and i=3 (s1[2]='c')

   At i=2, j=2 matches 'b' == 'b', dp[2] = dp[1]+1=1; max_len=1. At i=3, j=3 matches 'c' == 'c', dp[3] = dp[2]+1=2; max_len=2.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Longest common substring is 'bc' length 2 [OK]

Hint: Trace dp updates backward to avoid overwriting [OK]

Common Mistakes:

• Iterating dp forward overwrites needed values
• Off-by-one indexing errors
• Confusing max_len update timing

4. The following code attempts to count palindromic substrings by expanding around centers. Identify the line containing the subtle bug that causes incorrect counts for even-length palindromes.

def countSubstrings(s: str) -> int:
    n = len(s)
    count = 0

    def expand(left, right):
        nonlocal count
        while left >= 0 and right < n and s[left] == s[right]:
            count += 1
            left -= 1
            right += 1

    for i in range(n):
        expand(i, i)      # Odd length palindromes
        expand(i, i)      # Even length palindromes (bug here)

    return count

medium

A. Line calling expand(i, i) for even length palindromes

B. Line calling expand(i, i) for odd length palindromes

C. Line decrementing left inside expand function

D. Line incrementing right inside expand function

Solution

1. Step 1: Understand even length palindrome expansion

   Even length palindromes require expanding around two adjacent indices, i and i+1.

2. Step 2: Identify incorrect call

   Calling expand(i, i) twice means even length expansions are not checked; should be expand(i, i+1).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Even length palindromes like "aa" missed due to wrong center [OK]

Hint: Even length palindromes need expand(i, i+1) call [OK]

Common Mistakes:

• Using same indices for odd and even expansions

5. Examine the following buggy code snippet for wildcard matching using space-optimized DP. Which line contains the subtle bug that causes incorrect results on empty strings or patterns?

medium

A. Line 14: curr[j] = curr[j - 1] or prev[j]

B. Line 10: curr = [False] * (n + 1) initialization

C. Line 6: prev[j] = True instead of prev[j] = prev[j - 1]

D. Line 17: curr[j] = prev[j - 1]

Solution

1. Step 1: Analyze initialization of prev array for '*' characters

   Line 6 sets prev[j] = True unconditionally when p[j-1] == '*', which incorrectly assumes all prefixes match.

2. Step 2: Correct initialization logic

   It should propagate previous state: prev[j] = prev[j - 1], ensuring '*' can match empty sequence only if previous prefix matched.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Incorrect initialization causes false positives on empty strings [OK]

Hint: Initialization must propagate previous True states for '*' [OK]

Common Mistakes:

• Setting prev[j] = True blindly
• Ignoring empty string edge cases
• Confusing '*' handling in initialization