Palindrome Partitioning DP Minimum Cuts in DSA Java - Time & Space Complexity

Time Complexity: Palindrome Partitioning DP Minimum Cuts

O(n²)

Understanding Time Complexity

We want to know how the time needed to find the minimum cuts for palindrome partitioning grows as the input string gets longer.

How does the number of steps change when the string size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code finds the minimum number of cuts needed to split a string into palindromes using dynamic programming.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops checking substrings and palindrome status.
• How many times: Outer loop runs n times; inner loop runs up to n times for each outer iteration.

How Execution Grows With Input

As the string length grows, the number of substring checks grows roughly with the square of the length.

Input Size (n)	Approx. Operations
10	About 100 checks
100	About 10,000 checks
1000	About 1,000,000 checks

Pattern observation: The operations grow roughly by the square of the input size.

Final Time Complexity

Time Complexity: O(n²)

This means if the string doubles in length, the time to compute minimum cuts roughly quadruples.

Common Mistake

[X] Wrong: "The time grows linearly because we only loop through the string once."

[OK] Correct: There is a nested loop inside the main loop, so many substring checks happen, making the time grow faster than just once through the string.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming improves efficiency when working with substrings and palindromes, a common pattern in coding challenges.

Self-Check

"What if we precomputed all palindrome substrings first and then calculated minimum cuts? How would that affect the time complexity?"

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