Palindrome Partitioning DP Minimum Cuts in DSA Java

Mental Model

We want to split a string into parts where each part is a palindrome, using the fewest cuts possible. We remember which parts are palindromes and build up the answer step by step.

Analogy: Imagine cutting a ribbon into pieces so that each piece is a perfect mirror image of itself. You want to make as few cuts as possible to get all mirror pieces.

s = "abac"
Indices: 0 1 2 3
Cuts: between indices
Initial: a|b|a|c
Possible: abac (not pal) -> a|b|a|c (3 cuts) or aba|c (1 cut)

Dry Run Walkthrough

Input: string: "abac"

Goal: Find the minimum number of cuts to split "abac" into palindromes

Step 1: Check all substrings for palindrome property

Palindrome table (P):
[[T,F,T,F],
 [F,T,F,F],
 [F,F,T,F],
 [F,F,F,T]]

Why: Knowing which substrings are palindromes helps avoid repeated checks later

Step 2: Initialize cuts implicitly to max (i) in loop

cuts starts as [0,0,0,0] but set per i

Why: Worst case: cut before every char after first

Step 3: For i=0 ('a'): P[0][0]=T -> cuts[0]=0

cuts = [0, ?, ?, ?]

Why: Single char always palindrome

Step 4: For i=1 ('ab'): !P[0][1], set cuts[1]=1, check j=1: P[1][1]=T, cuts[0]+1=1

cuts = [0, 1, ?, ?]

Why: 'ab' not pal from 0, but split after 0

Step 5: For i=2 ('aba'): P[0][2]=T -> cuts[2]=0

cuts = [0, 1, 0, ?]

Why: Whole 'aba' is palindrome, no cuts needed

Step 6: For i=3 ('abac'): !P[0][3], set cuts[3]=3, check splits: j=1 P[1][3]=F j=2 P[2][3]=F j=3 P[3][3]=T -> cuts[2]+1=1 <3

cuts = [0, 1, 0, 1]

Why: Best: cut after 'aba', then 'c'

Result:

Minimum cuts: 1 ("aba" | "c")

Annotated Code

DSA Java

public class PalindromePartitioning {

    // Check if substring s[left..right] is palindrome
    private static boolean[][] buildPalindromeTable(String s) {
        int n = s.length();
        boolean[][] P = new boolean[n][n];
        for (int i = 0; i < n; i++) {
            P[i][i] = true; // single char palindrome
        }
        for (int length = 2; length <= n; length++) {
            for (int left = 0; left <= n - length; left++) {
                int right = left + length - 1;
                if (s.charAt(left) == s.charAt(right) && (length == 2 || P[left + 1][right - 1])) {
                    P[left][right] = true;
                }
            }
        }
        return P;
    }

    // Compute minimum cuts needed for palindrome partitioning
    public static int minCut(String s) {
        int n = s.length();
        if (n == 0) return 0;
        boolean[][] P = buildPalindromeTable(s);
        int[] cuts = new int[n];
        for (int i = 0; i < n; i++) {
            if (P[0][i]) {
                cuts[i] = 0;
            } else {
                cuts[i] = i;
                for (int j = 1; j <= i; j++) {
                    if (P[j][i] && cuts[j - 1] + 1 < cuts[i]) {
                        cuts[i] = cuts[j - 1] + 1;
                    }
                }
            }
        }
        return cuts[n - 1];
    }

    public static void main(String[] args) {
        String s = "abac";
        System.out.println("Minimum cuts needed: " + minCut(s));
    }
}

P[i][i] = true; // single char palindrome

Base case: every single character is a palindrome

if (s.charAt(left) == s.charAt(right) && (length == 2 || P[left + 1][right - 1]))

Optimized palindrome check: ends match and inner is palindrome

if (P[0][i]) { cuts[i] = 0; }

If entire prefix s[0..i] is palindrome, 0 cuts needed

for (int j = 1; j <= i; j++) { if (P[j][i] && cuts[j - 1] + 1 < cuts[i]) ... }

Try all possible last palindrome endings at i, take min cuts

Output

Minimum cuts needed: 1

Complexity Analysis

Time: O(n^2): O(n^2) for palindrome table + O(n^2) for cuts DP

Space: O(n^2): mainly for P table

vs Alternative: Brute force recursion: exponential; naive DP without table: O(n^3)

Edge Cases

Empty string ""

n==0, return 0

DSA Java

if (n == 0) return 0;

All palindromic "aaa"

P[0][2]=true, cuts[2]=0

DSA Java

if (P[0][i]) cuts[i]=0

No long palindromes "abc"

cuts[2]=2 (a|b|c)

DSA Java

cuts[i]=i initially, only single char splits

When to Use This Pattern

Minimum cuts/partitions into special substrings (palindromes) -> Precompute substring property table + 1D DP on prefixes.

Common Mistakes

Mistake: Missing && (length == 2 || P[left+1][right-1]) in pal table
Fix: Always check inner palindrome unless length=2

Mistake: cuts[i] = cuts[i-1] + 1 without checking if s[i] can attach
Fix: Only add cut if previous + new pal; try all split points

Summary

Minimum cuts to partition string into palindromes using DP.

Key: Palindrome table O(n^2) + cuts DP where cuts[i] = min over j where s[j..i] pal of (cuts[j-1] + 1)

Classic: property table + prefix minimization DP.

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