Palindrome Partitioning II (Min Cuts)

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Palindrome Partitioning II (Min Cuts)

Edge cases:

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IDE

def minCut(s: str) -> int:public int minCut(String s)int minCut(string s)function minCut(s)

def minCut(s: str) -> int:
    # Write your solution here
    pass

class Solution {
    public int minCut(String s) {
        // Write your solution here
        return 0;
    }
}

#include <string>
using namespace std;

int minCut(string s) {
    // Write your solution here
    return 0;
}

function minCut(s) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: 0Returning zero cuts for all inputs, ignoring actual partitions.✅ Implement dp array and update minimum cuts based on palindrome substrings.

Wrong: n-1Assuming no palindromes longer than 1 without checking substrings.✅ Add palindrome checks for substrings and update dp accordingly.

Wrong: TLEUsing brute force recursion without memoization or DP.✅ Use DP with palindrome preprocessing or center expansion to achieve O(n^2).

Wrong: Incorrect cuts for all identical charactersNot recognizing full string palindrome for identical chars.✅ Ensure palindrome table or checks mark entire string as palindrome.

✓

Test Cases

Focus on handling base cases like empty and single character strings.

Beware of greedy traps and off-by-one errors in palindrome detection.

Optimize your solution to O(n^2) time complexity to pass large inputs.

t1_01basic

Input{"s":"aab"}

Expected1

⏱ Performance - must finish in 2000ms

The palindrome partitioning ['aa','b'] requires 1 cut.

💡 Think about checking all substrings for palindrome property.

💡 Use dynamic programming to store minimum cuts up to each index.

💡 dp[i] = min(dp[j] + 1) for all j < i where s[j+1:i] is palindrome.

Why it failed: Returned value incorrect - likely missed counting cuts properly or palindrome checks are wrong. Fix by ensuring dp[i] updates only when substring s[j+1:i] is palindrome.

✓ Correct minimum cuts computed for canonical example.

t1_02basic

Input{"s":"abccba"}

Expected0

⏱ Performance - must finish in 2000ms

The entire string 'abccba' is a palindrome, so no cuts needed.

💡 Check if the whole string is palindrome first.

💡 If yes, minimum cuts is zero.

💡 dp array initialization should handle this base case.

Why it failed: Incorrect output - likely not checking full string palindrome or dp initialization error. Fix by adding base case dp[0] = -1 or dp[i] = 0 if s[0:i] palindrome.

✓ Correctly identified no cuts needed for full palindrome string.

t2_01edge

Input{"s":""}

Expected0

⏱ Performance - must finish in 2000ms

Empty string requires zero cuts by definition.

💡 Consider empty input as a base case.

💡 Return 0 cuts when string length is zero.

💡 Add a condition to handle empty string input explicitly.

Why it failed: Fails on empty string - likely no base case for empty input. Fix by returning 0 immediately if s is empty.

✓ Handles empty string input correctly with zero cuts.

t2_02edge

Input{"s":"z"}

Expected0

⏱ Performance - must finish in 2000ms

Single character string is always palindrome, so zero cuts needed.

💡 Single character strings are palindromes by definition.

💡 Ensure dp initialization accounts for single character substrings.

💡 Minimum cuts for length 1 string should be zero.

Why it failed: Incorrect output for single character - likely dp initialization or base case missing. Fix by setting dp[0] = -1 or dp[i] = 0 for single chars.

✓ Correctly returns zero cuts for single character string.

t2_03edge

Input{"s":"abcde"}

Expected4

⏱ Performance - must finish in 2000ms

No palindromic substrings longer than 1, so cuts = length-1 = 4.

💡 Check substrings longer than 1 for palindrome property.

💡 If none found, minimum cuts equals string length minus one.

💡 dp array should reflect maximum cuts when no palindromes longer than 1.

Why it failed: Output less than expected - likely missing max cuts case when no palindromes longer than 1. Fix by initializing dp[i] = i-1 or similar max cuts.

✓ Correctly computes maximum cuts when no palindromes longer than 1.

t2_04edge

Input{"s":"aaaaa"}

Expected0

⏱ Performance - must finish in 2000ms

All identical characters form a palindrome, zero cuts needed.

💡 All identical characters form a palindrome.

💡 DP should recognize entire string as palindrome.

💡 Initialize dp properly to avoid unnecessary cuts.

Why it failed: Incorrect cuts for all identical chars - likely missing full palindrome detection. Fix by ensuring palindrome table or checks cover this case.

✓ Correctly returns zero cuts for all identical characters.

t3_01corner

Input{"s":"ababbbabbababa"}

Expected3

⏱ Performance - must finish in 2000ms

Complex palindrome partitions require careful DP, greedy approach fails here.

💡 Greedy approach picking largest palindrome first may fail.

💡 Use DP to consider all partitions and choose minimum cuts.

💡 Check all substrings and update dp with minimum cuts accordingly.

Why it failed: Greedy approach fails - returns more cuts than minimum. Fix by implementing full DP with palindrome table instead of greedy.

✓ DP correctly finds minimum cuts avoiding greedy trap.

t3_02corner

Input{"s":"aaaaa"}

Expected0

⏱ Performance - must finish in 2000ms

All identical characters form a palindrome, zero cuts needed.

💡 Check for all identical characters as a special case.

💡 DP should recognize entire string as palindrome.

💡 Initialize dp properly to avoid unnecessary cuts.

Why it failed: Incorrect cuts for all identical chars - likely missing full palindrome detection. Fix by ensuring palindrome table or checks cover this case.

✓ Correctly returns zero cuts for all identical characters.

t3_03corner

Input{"s":"abcbaefgfe"}

Expected1

⏱ Performance - must finish in 2000ms

Multiple palindromic substrings with overlapping centers require careful DP.

💡 Check palindrome centers carefully, including odd and even lengths.

💡 Use palindrome table or center expansion to mark palindromes.

💡 Update dp with minimum cuts considering all palindrome partitions.

Why it failed: Off-by-one or incorrect palindrome detection causes wrong cuts. Fix by correctly marking palindromes and updating dp accordingly.

✓ DP correctly handles overlapping palindromes and returns minimum cuts.

t4_01performance

Input{"s":"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"}

Expectednull

⏱ Performance - must finish in 2000ms

Input length n=100, O(n^2) DP must complete within 2 seconds.

💡 Brute force exponential solutions will time out here.

💡 Use DP with palindrome preprocessing for O(n^2) time.

💡 Optimize palindrome checks with center expansion or DP table.

Why it failed: TLE due to exponential or inefficient palindrome checks. Fix by implementing O(n^2) DP with palindrome table or center expansion.

✓ Solution runs within time limit using O(n^2) DP approach.

Practice

1. 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

2. You are given two strings and need to find the shortest string that contains both as subsequences. Which algorithmic approach guarantees finding this shortest common supersequence efficiently?

easy

A. Greedy approach that merges characters from both strings whenever they differ

B. Dynamic programming based on longest common subsequence to build the shortest supersequence

C. Pure brute force recursion enumerating all supersequences and selecting the shortest

D. Sorting characters of both strings and merging them in lexicographical order

Solution

1. Step 1: Understand the problem requires minimal length supersequence containing both strings as subsequences

   Greedy or sorting approaches fail because they do not consider subsequence order constraints.

2. Step 2: Recognize that longest common subsequence (LCS) helps identify overlapping parts to minimize duplication

   Dynamic programming using LCS length allows building the shortest common supersequence efficiently.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   LCS-based DP is the classic and optimal approach for shortest common supersequence [OK]

Hint: Shortest common supersequence uses LCS-based DP [OK]

Common Mistakes:

• Thinking greedy merging always works
• Confusing substring with subsequence
• Assuming sorting helps

3. 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

4. The following code attempts to implement the space-optimized DP for maximum uncrossed lines. Which line contains a subtle bug that can cause incorrect results?

medium

A. Line 7: temp = dp[j]

B. Line 12: prev = temp

C. Line 11: dp[j] = max(dp[j], dp[j - 1])

D. Line 9: dp[j] = dp[j - 1] + 1

Solution

1. Step 1: Understand role of prev variable

   In space-optimized DP, dp[j] depends on prev (dp[j-1] from previous iteration), not dp[j-1] of current iteration.

2. Step 2: Identify incorrect dp update

   Line 9 uses dp[j - 1] from current iteration, which may have been updated already, causing incorrect results. It should use prev instead.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Using dp[j-1] instead of prev breaks correctness in space-optimized DP [OK]

Hint: Use prev to avoid overwriting dp[j-1] before use [OK]

Common Mistakes:

• Using dp[j-1] instead of prev in update
• Not preserving previous dp[j] value before update

5. 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