Palindrome Partitioning DP Minimum Cuts in DSA Java - Build from Scratch

Palindrome Partitioning DP Minimum Cuts

📖 Scenario: Imagine you have a word, and you want to cut it into pieces so that each piece is a palindrome. A palindrome is a word that reads the same forwards and backwards, like "madam" or "racecar". You want to find the smallest number of cuts needed to split the word into palindromes.

🎯 Goal: Build a Java program that uses dynamic programming to find the minimum number of cuts needed to partition a given string into palindromic substrings.

📋 What You'll Learn

Create a string variable with the exact value "ababbbabbababa"

Create a 2D boolean array to store palindrome checks

Create an integer array to store minimum cuts for each substring

Use nested loops to fill the palindrome table

Use dynamic programming to calculate minimum cuts

Print the minimum cuts for the entire string

💡 Why This Matters

🌍 Real World

Palindrome partitioning is useful in text processing, DNA sequence analysis, and data compression where symmetrical patterns matter.

💼 Career

Understanding dynamic programming and string manipulation is important for software engineering roles, especially those involving algorithms and optimization.

1

Create the input string

Create a string variable called input and set it to the exact value "ababbbabbababa".

DSA Java

// Your code here

Hint

Use String input = "ababbbabbababa"; inside the main method.

2

Setup palindrome table and cuts array

Below the input string, create a 2D boolean array called palindrome with size input.length() by input.length(). Also create an integer array called cuts with size input.length().

DSA Java

public class PalindromePartition {
    public static void main(String[] args) {
        String input = "ababbbabbababa";
        // Your code here
    }
}

Hint

Use boolean[][] palindrome = new boolean[n][n]; and int[] cuts = new int[n]; where n = input.length().

3

Fill palindrome table and calculate minimum cuts

Use nested for loops with variables end from 0 to n-1 and start from 0 to end. Inside, set palindrome[start][end] to true if characters at start and end are equal and the substring between them is a palindrome or length less than 3. Then, update cuts[end] to the minimum cuts needed using the palindrome table.

DSA Java

public class PalindromePartition {
    public static void main(String[] args) {
        String input = "ababbbabbababa";
        int n = input.length();
        boolean[][] palindrome = new boolean[n][n];
        int[] cuts = new int[n];
        // Use nested loops to fill palindrome and cuts arrays
        // Your code here
    }
}

Hint

Use nested loops with end and start. Check palindrome condition and update cuts[end] accordingly.

4

Print the minimum cuts for the entire string

Add a print statement to display the minimum cuts needed for the entire string using cuts[n - 1].

DSA Java

public class PalindromePartition {
    public static void main(String[] args) {
        String input = "ababbbabbababa";
        int n = input.length();
        boolean[][] palindrome = new boolean[n][n];
        int[] cuts = new int[n];

        for (int end = 0; end < n; end++) {
            int minCuts = end; // maximum cuts
            for (int start = 0; start <= end; start++) {
                if (input.charAt(start) == input.charAt(end) && (end - start < 2 || palindrome[start + 1][end - 1])) {
                    palindrome[start][end] = true;
                    if (start == 0) {
                        minCuts = 0;
                    } else {
                        minCuts = Math.min(minCuts, cuts[start - 1] + 1);
                    }
                }
            }
            cuts[end] = minCuts;
        }
        // Your code here
    }
}

Hint

Use System.out.println(cuts[n - 1]); to print the minimum cuts.

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