Palindrome Partitioning II (Min Cuts)

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Steps

setup

Initialize dp array with maximum cuts

We initialize the dp array so that dp[i] = i, representing the worst case where each character is cut separately.

💡 This sets a baseline for minimum cuts, ensuring any found palindrome can only improve this value.

Line:

dp = [i for i in range(n)]

💡 Starting with maximum cuts allows the algorithm to minimize cuts by finding palindromes.

fill_cells

Start center expansion at index 0 (odd length palindrome)

Set left and right pointers to center=0 to check for odd length palindromes starting at 'a'.

💡 Odd length palindromes have a single center character; we expand around it.

Line:

left, right = center, center

💡 We begin palindrome checks at the first character.

compare

Check palindrome 'a' at indices [0..0]

Since s[left] == s[right] ('a' == 'a'), update dp[right] to 0 because left is 0 (no cuts needed).

💡 A single character is always a palindrome with zero cuts.

Line:

dp[right] = 0 if left == 0 else min(dp[right], dp[left - 1] + 1)

💡 Detecting a palindrome at the start means no cuts are needed for that substring.

expand

Expand palindrome pointers left= -1, right=1 (stop expansion)

Move left pointer left by 1 and right pointer right by 1 to check larger palindrome, but left < 0 stops expansion.

💡 Expansion stops when pointers go out of bounds or characters mismatch.

Line:

left -= 1
right += 1

💡 No larger palindrome centered at index 0.

fill_cells

Start center expansion at indices 0 and 1 (even length palindrome)

Set left=0 and right=1 to check for even length palindrome between 'a' and 'a'.

💡 Even length palindromes have two centers; we check pairs of characters.

Line:

left, right = center, center + 1

💡 We begin checking even length palindromes starting at indices 0 and 1.

compare

Check palindrome 'aa' at indices [0..1]

Characters s[left] and s[right] are equal ('a' == 'a'), update dp[1] to 0 since left is 0 (no cuts needed).

💡 Finding a palindrome substring reduces the cuts needed for that substring.

Line:

dp[right] = 0 if left == 0 else min(dp[right], dp[left - 1] + 1)

💡 The substring 'aa' is a palindrome with zero cuts needed.

expand

Expand palindrome pointers left= -1, right=2 (stop expansion)

Move left pointer left by 1 and right pointer right by 1, but left < 0 stops expansion.

💡 Expansion stops when pointers go out of bounds.

Line:

left -= 1
right += 1

💡 No larger even length palindrome starting at indices 0 and 1.

fill_cells

Start center expansion at index 1 (odd length palindrome)

Set left and right pointers to center=1 to check odd length palindromes starting at 'a'.

💡 Check palindromes centered at the second character.

Line:

left, right = center, center

💡 Begin palindrome checks at index 1.

compare

Check palindrome 'a' at indices [1..1]

Characters s[left] and s[right] are equal ('a' == 'a'), update dp[1] to 0 since left is 1 (dp[0] + 1 = 1), but dp[1] is already 0 so no change.

💡 Single character palindrome check updates dp if better, but here dp[1] is already minimal.

Line:

dp[right] = 0 if left == 0 else min(dp[right], dp[left - 1] + 1)

💡 No improvement in cuts for substring ending at index 1.

expand

Expand palindrome pointers left=0, right=2

Move left pointer left by 1 and right pointer right by 1 to check substring 'aab'.

💡 Check if larger palindrome exists by expanding pointers.

Line:

left -= 1
right += 1

💡 Characters differ, so no palindrome here.

fill_cells

Start center expansion at indices 1 and 2 (even length palindrome)

Set left=1 and right=2 to check for even length palindrome between 'a' and 'b'.

💡 Check even length palindromes starting at indices 1 and 2.

Line:

left, right = center, center + 1

💡 Prepare pointers for even length palindrome check.

compare

Check substring 'ab' at indices [1..2]

Characters s[left] and s[right] differ ('a' != 'b'), so no palindrome and stop expansion.

💡 Even length palindrome check fails immediately here.

Line:

while left >= 0 and right < n and s[left] == s[right]:

💡 No palindrome found between indices 1 and 2.

fill_cells

Start center expansion at index 2 (odd length palindrome)

Set left and right pointers to center=2 to check odd length palindrome at 'b'.

💡 Check palindrome centered at last character.

Line:

left, right = center, center

💡 Begin palindrome check at index 2.

compare

Check palindrome 'b' at indices [2..2]

Characters s[left] and s[right] are equal ('b' == 'b'), update dp[2] to min(dp[2], dp[1] + 1) = min(2, 0 + 1) = 1.

💡 Single character palindrome reduces cuts if better than current dp value.

Line:

dp[right] = 0 if left == 0 else min(dp[right], dp[left - 1] + 1)

💡 Partitioning 'aab' as ['aa','b'] requires only 1 cut.

expand

Expand palindrome pointers left=1, right=3 (stop expansion)

Move left pointer left by 1 and right pointer right by 1, but right >= n stops expansion.

💡 Expansion stops when pointers go out of bounds.

Line:

left -= 1
right += 1

💡 No larger palindrome centered at index 2.

fill_cells

Start center expansion at indices 2 and 3 (even length palindrome)

Set left=2 and right=3 to check even length palindrome, but right >= n stops immediately.

💡 No even length palindrome possible at the last character.

Line:

left, right = center, center + 1

💡 No palindrome found for even length at last center.

reconstruct

Return final minimum cuts from dp array

Return dp[n-1] which holds the minimum cuts needed for the entire string.

💡 The dp array now contains the minimum cuts for all substrings ending at each index.

Line:

return dp[-1]

💡 The minimum cuts for 'aab' is 1, partitioning as ['aa','b'].

def minCut(s: str) -> int:
    n = len(s)
    dp = [i for i in range(n)]  # STEP 1: Initialize dp with max cuts

    for center in range(n):
        # Odd length palindromes
        left, right = center, center  # STEP 2: Set pointers for odd palindrome
        while left >= 0 and right < n and s[left] == s[right]:  # STEP 3: Check palindrome
            dp[right] = 0 if left == 0 else min(dp[right], dp[left - 1] + 1)  # STEP 3: Update dp
            left -= 1  # STEP 4: Expand left
            right += 1  # STEP 4: Expand right

        # Even length palindromes
        left, right = center, center + 1  # STEP 5: Set pointers for even palindrome
        while left >= 0 and right < n and s[left] == s[right]:  # STEP 6: Check palindrome
            dp[right] = 0 if left == 0 else min(dp[right], dp[left - 1] + 1)  # STEP 6: Update dp
            left -= 1  # STEP 7: Expand left
            right += 1  # STEP 7: Expand right

    return dp[-1]  # STEP 8: Return result

📊

Palindrome Partitioning II (Min Cuts) - Watch the Algorithm Execute, Step by Step

Watching the algorithm step through each palindrome expansion and dp update reveals how the minimum cuts are computed efficiently without redundant checks.

Item 0 - wt:0 val:0

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

dp[0]=0

Item 0 - wt:0 val:0

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

center=0

Item 0 - wt:0 val:0

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

dp[0] updated to 0

Item 0 - wt:0 val:0

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

center=0

Item 0 - wt:0 val:0

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

left=0

Item 0 - wt:0 val:0

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

dp[1] updated to 0

Item 0 - wt:0 val:0

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

right=2

Item 1 - wt:0 val:0

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

center=1

Item 1 - wt:0 val:0

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

dp[1] unchanged

Item 1 - wt:0 val:0

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

left=0

Item 1 - wt:0 val:0

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

left=1

Item 1 - wt:0 val:0

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

left=1

Item 2 - wt:0 val:0

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

center=2

Item 2 - wt:0 val:0

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

dp[2] updated to 1

Item 2 - wt:0 val:0

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

left=1

Item 2 - wt:0 val:0

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

left=2

Item 2 - wt:0 val:1

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

Answer: dp[2] = 1

Key Takeaways

✓ The dp array starts with maximum cuts and is improved by expanding palindromes around centers.

This insight is hard to see from code alone because the dp updates happen inside nested loops with pointer expansions.

✓ Odd and even length palindrome expansions cover all palindrome substrings efficiently without extra space.

Visualizing pointer movements clarifies why both expansions are necessary and how they complement each other.

✓ Updating dp[right] with dp[left-1] + 1 captures the minimal cuts needed before the palindrome plus one cut.

This decision is subtle and seeing it in action with actual indices helps understand the partitioning logic.

Practice

1. Consider the following Python code implementing the space-optimized DP solution for maximum uncrossed lines. What is the value of dp after the outer loop completes for i=2 when nums1 = [1,4,2] and nums2 = [1,2,4]?

easy

A. [0, 1, 2, 2]

B. [0, 1, 1, 1]

C. [0, 1, 1, 2]

D. [0, 0, 1, 1]

Solution

1. Step 1: Trace dp after i=1

   For i=1 (nums1[0]=1), dp updates to [0,1,1,1] because nums2 has 1 at j=1, matches increment dp[1].

2. Step 2: Trace dp after i=2

   For i=2 (nums1[1]=4), dp updates as follows: - j=1: no match, dp[1]=max(dp[1], dp[0])=1 - j=2: no match, dp[2]=max(dp[2], dp[1])=1 - j=3: match (4==4), dp[3]=prev+1=1+1=2 where prev is dp[2] before update which was 1, so dp after i=2 is [0,1,1,2].

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Matches at positions (1,1) and (2,3) give dp[3]=2 after i=2 [OK]

Hint: Track prev and dp[j] carefully to avoid off-by-one errors [OK]

Common Mistakes:

• Confusing prev and dp[j] updates
• Off-by-one indexing in dp array

2. What is the time complexity of the space-optimized bottom-up DP solution for the Delete Operation for Two Strings problem, where the input strings have lengths m and n respectively?

medium

A. O(m * n)

B. O(m + n)

C. O(m * n * log(m + n))

D. O(m^2 + n^2)

Solution

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

   The outer loop runs m times, inner loop runs n times, leading to m*n iterations.

2. Step 2: Confirm no extra hidden factors

   Each iteration does O(1) work; no sorting or recursion overhead.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   DP table size m*n and each cell computed once [OK]

Hint: Nested loops over m and n -> O(m*n) [OK]

Common Mistakes:

• Confusing linear with quadratic
• Assuming sorting or recursion adds log factor
• Ignoring nested loops

3. What is the time and space complexity of the space-optimized bottom-up dynamic programming solution for the Longest Common Subsequence problem, where the input strings have lengths m and n respectively?

medium

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

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

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

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

Solution

1. Step 1: Identify time complexity

   The algorithm uses two nested loops: outer loop runs m times, inner loop runs n times, resulting in O(m*n) time.

2. Step 2: Identify space complexity

   Space optimized DP uses a single array of size n+1, so space complexity is O(n). The recursion stack or full DP table is not used.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Nested loops give O(m*n) time; single dp array gives O(n) space [OK]

Hint: Nested loops multiply time; single dp array reduces space [OK]

Common Mistakes:

• Confusing full DP table space with optimized space
• Assuming linear time due to single dp array
• Forgetting recursion stack space in brute force

4. 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

5. The following code attempts to implement space-optimized DP for regex matching. Which line contains a subtle bug that causes incorrect results on some inputs?

def isMatch(s: str, p: str) -> bool:
    m, n = len(s), len(p)
    dp = [False] * (n + 1)
    dp[0] = True
    for j in range(2, n + 1):
        if p[j - 1] == '*':
            dp[j] = dp[j - 2]
    for i in range(1, m + 1):
        prev = dp[0]
        dp[0] = False
        for j in range(1, n + 1):
            temp = dp[j]
            if p[j - 1] == '*':
                dp[j] = dp[j] or dp[j - 2]
                if p[j - 2] == '.' or p[j - 2] == s[i - 1]:
                    dp[j] = dp[j] or prev
            else:
                dp[j] = prev and (p[j - 1] == '.' or p[j - 1] == s[i - 1])
            prev = temp
    return dp[n]

medium

A. Line: dp[0] = False inside the outer loop

B. Line: for j in range(1, n + 1): iterating forwards instead of backwards

C. Line: prev = dp[0] before inner loop starts

D. Line: dp[j] = dp[j] or dp[j - 2] inside the '*' handling block

Solution

1. Step 1: Understand dp array update dependencies

   When updating dp[j], dp[j - 2] must not be overwritten before use. Iterating forwards overwrites dp[j - 2] too early.

2. Step 2: Identify iteration direction bug

   Iterating forwards causes dp[j - 2] to be stale or overwritten, leading to incorrect matches. Iterating backwards avoids this.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Known bug in space-optimized regex DP is forward iteration [OK]

Hint: DP updates with '*' require backward iteration to preserve dp[j-2] [OK]

Common Mistakes:

• Assuming forward iteration is safe in space-optimized DP