Palindrome Partitioning DP Minimum Cuts in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a substring is a palindrome.

DSA Java

boolean isPalindrome(String s, int start, int end) {
    while (start < end) {
        if (s.charAt(start) != s.charAt([1])) {
            return false;
        }
        start++;
        end--;
    }
    return true;
}

Drag options to blanks, or click blank then click option'

Astart

Bend

Cstart + 1

Dend - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using start + 1 or end - 1 instead of end index.

Comparing wrong indices causing incorrect palindrome check.

To check palindrome, compare characters at start and end indices.

2fill in blank

medium

Complete the code to initialize the minimum cuts array with maximum cuts possible.

DSA Java

int n = s.length();
int[] cuts = new int[n];
for (int i = 0; i < n; i++) {
    cuts[i] = [1];
}

Drag options to blanks, or click blank then click option'

A0

Bn - 1

Ci

Di - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing cuts with 0 or n-1 causes wrong minimum cuts calculation.

Using i-1 causes negative index issues.

Initialize cuts[i] with i, meaning max cuts for substring s[0..i].

3fill in blank

hard

Fix the error in updating minimum cuts inside the nested loop.

DSA Java

for (int j = 0; j <= i; j++) {
    if (isPalindrome(s, j, i)) {
        cuts[i] = Math.min(cuts[i], [1]);
    }
}

Drag options to blanks, or click blank then click option'

Acuts[j - 1] + 1

Bcuts[j] + 1

Ccuts[i] + 1

Dcuts[i - 1] + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using cuts[j] + 1 causes off-by-one error.

Using cuts[i] + 1 or cuts[i-1] + 1 causes incorrect updates.

If substring s[j..i] is palindrome, cuts[i] can be updated by cuts[j-1] + 1.

4fill in blank

hard

Fill both blanks to handle the case when the palindrome starts at index 0.

DSA Java

for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++) {
        if (isPalindrome(s, j, i)) {
            if (j == 0) {
                cuts[i] = [1];
            } else {
                cuts[i] = Math.min(cuts[i], cuts[[2]] + 1);
            }
        }
    }
}

Drag options to blanks, or click blank then click option'

A0

Bj - 1

Ci - 1

Dj

Attempts:

3 left

💡 Hint

Common Mistakes

Setting cuts[i] to 1 when j=0 causes extra cut.

Using cuts[j] instead of cuts[j-1] causes wrong indexing.

If palindrome starts at 0, no cuts needed (0). Else use cuts[j-1] + 1.

5fill in blank

hard

Fill all three blanks to complete the palindrome partitioning minimum cuts function.

DSA Java

public int minCut(String s) {
    int n = s.length();
    int[] cuts = new int[n];
    for (int i = 0; i < n; i++) {
        cuts[i] = i;
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= i; j++) {
            if ([1](s, j, i)) {
                if (j == 0) {
                    cuts[i] = 0;
                } else {
                    cuts[i] = Math.min(cuts[i], cuts[[2]] + [3]);
                }
            }
        }
    }
    return cuts[n - 1];
}

Drag options to blanks, or click blank then click option'

AisPalindrome

Bj - 1

C1

Di - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong function name or wrong indices.

Adding 0 instead of 1 causes incorrect cut count.

Use isPalindrome to check substring, update cuts[i] with cuts[j-1] + 1.

Practice

1. What does the term minimum cuts mean in the context of palindrome partitioning of a string?

easy

A. The smallest number of cuts needed to split the string into palindromic substrings

B. The maximum number of palindromic substrings in the string

C. The length of the longest palindromic substring

D. The total number of substrings in the string

Solution

1. Step 1: Understand palindrome partitioning

   Palindrome partitioning means splitting a string into parts where each part is a palindrome.

2. Step 2: Define minimum cuts

   Minimum cuts is the least number of cuts needed so that every substring after cutting is a palindrome.

3. Final Answer:

   The smallest number of cuts needed to split the string into palindromic substrings -> Option A

4. Quick Check:

   Minimum cuts = smallest cuts to get palindromes [OK]

Hint: Minimum cuts = least splits for palindrome parts [OK]

Common Mistakes:

• Confusing minimum cuts with maximum palindromes
• Thinking minimum cuts means longest palindrome length
• Counting total substrings instead of cuts

2. Which of the following Java code snippets correctly initializes the DP array for minimum cuts in palindrome partitioning of a string s of length n?

easy

A. int[] cuts = new int[n]; Arrays.fill(cuts, -1);

B. int[] cuts = new int[n]; for (int i = 0; i <= n; i++) cuts[i] = 0;

C. int[] cuts = new int[n]; for (int i = 0; i < n; i++) cuts[i] = i;

D. int[] cuts = new int[n+1]; Arrays.fill(cuts, Integer.MAX_VALUE);

Solution

1. Step 1: Understand DP array initialization

   We initialize cuts array where cuts[i] = i means max cuts needed for substring s[0..i].

2. Step 2: Check code correctness

   int[] cuts = new int[n]; for (int i = 0; i < n; i++) cuts[i] = i; initializes cuts array of size n and sets cuts[i] = i correctly. Other options have errors like wrong size or wrong initialization.

3. Final Answer:

   int[] cuts = new int[n]; for (int i = 0; i < n; i++) cuts[i] = i; -> Option C

4. Quick Check:

   Initialize cuts with max cuts = index [OK]

Hint: Initialize cuts[i] = i for max cuts before optimization [OK]

Common Mistakes:

• Using wrong array size causing IndexOutOfBounds
• Filling cuts with zero instead of max cuts
• Using negative values which are invalid

3. What is the output of the following Java code snippet for string s = "aab"?

public class Solution {
  public static int minCut(String s) {
    int n = s.length();
    int[] cuts = new int[n];
    boolean[][] palindrome = new boolean[n][n];
    for (int i = 0; i < n; i++) cuts[i] = i;
    for (int end = 0; end < n; end++) {
      for (int start = 0; start <= end; start++) {
        if (s.charAt(start) == s.charAt(end) && (end - start < 2 || palindrome[start + 1][end - 1])) {
          palindrome[start][end] = true;
          cuts[end] = start == 0 ? 0 : Math.min(cuts[end], cuts[start - 1] + 1);
        }
      }
    }
    return cuts[n - 1];
  }
  public static void main(String[] args) {
    System.out.println(minCut("aab"));
  }
}

medium

A. 0

B. 1

C. 2

D. 3

Solution

1. Step 1: Analyze palindrome substrings in "aab"

   "a", "a", "b" are palindromes. "aa" is palindrome. Minimum cuts needed is 1 (cut between "aa" and "b").

2. Step 2: Trace DP updates

   cuts array updates to [0,0,1]. Final cuts[n-1] = 1.

3. Final Answer:

   1 -> Option B

4. Quick Check:

   Minimum cuts for "aab" = 1 [OK]

Hint: Check palindromes and update cuts with min previous cuts +1 [OK]

Common Mistakes:

• Assuming no cuts needed for "aab"
• Counting cuts as number of palindromes
• Ignoring palindrome "aa" substring

4. Identify the bug in the following Java code snippet for palindrome partitioning minimum cuts:

public int minCut(String s) {
  int n = s.length();
  int[] cuts = new int[n];
  boolean[][] palindrome = new boolean[n][n];
  for (int i = 0; i <= n; i++) cuts[i] = i;
  for (int end = 0; end < n; end++) {
    for (int start = 0; start <= end; start++) {
      if (s.charAt(start) == s.charAt(end) && (end - start < 2 || palindrome[start + 1][end - 1])) {
        palindrome[start][end] = true;
        cuts[end] = start == 0 ? 0 : Math.min(cuts[end], cuts[start - 1] + 1);
      }
    }
  }
  return cuts[n - 1];
}

medium

A. The return statement should be 'cuts[n]' instead of 'cuts[n - 1]'

B. The palindrome check condition is incorrect

C. The cuts array is not initialized properly

D. The loop 'for (int i = 0; i <= n; i++)' causes IndexOutOfBoundsException

Solution

1. Step 1: Check array initialization loop

   cuts array size is n, but loop runs i from 0 to n inclusive, causing IndexOutOfBoundsException at i = n.

2. Step 2: Confirm other parts are correct

   Palindrome check and return statement are correct. Initialization loop must be 'i < n'.

3. Final Answer:

   The loop 'for (int i = 0; i <= n; i++)' causes IndexOutOfBoundsException -> Option D

4. Quick Check:

   Array size n but loop runs to n inclusive causes error [OK]

Hint: Loop to n-1 for array size n to avoid out-of-bounds [OK]

Common Mistakes:

• Using <= n instead of < n in loops
• Misunderstanding palindrome condition
• Returning wrong index from cuts array

5. Given the string s = "ababbbabbababa", what is the minimum number of cuts needed to partition it into palindromic substrings using the DP minimum cuts approach?

hard

A. 4

B. 5

C. 3

D. 6

Solution

1. Step 1: Understand the problem and DP approach

   We find minimum cuts so that each substring is palindrome. The string "ababbbabbababa" has palindromic partitions with 4 cuts.

2. Step 2: Trace or recall known result

   Known minimum cuts for this string is 4, splitting into palindromes like "a|babbbab|b|ababa" or similar.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Minimum cuts for "ababbbabbababa" = 4 [OK]

Hint: Use DP to find min cuts; known result for this string is 4 [OK]

Common Mistakes:

• Guessing cuts without checking palindrome substrings
• Counting number of palindromes instead of cuts
• Confusing cuts with substring length