Palindrome Partitioning DP Minimum Cuts in DSA C - 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 changes as the input string gets longer.

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int minCut(char* s) {
    int n = strlen(s);
    int cuts[n+1];
    bool palindrome[n][n];
    memset(palindrome, 0, sizeof(palindrome));
    for (int i = 0; i <= n; i++) cuts[i] = i - 1;

    for (int end = 0; end < n; end++) {
        for (int start = 0; start <= end; start++) {
            if (s[start] == s[end] && (end - start < 2 || palindrome[start+1][end-1])) {
                palindrome[start][end] = true;
                cuts[end+1] = fmin(cuts[end+1], cuts[start] + 1);
            }
        }
    }
    return cuts[n];
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops checking all substrings for palindrome property.
• How many times: Outer loop runs n times, inner loop runs up to n times, so roughly n x n = n² times.

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: Doubling the input size roughly quadruples the work because of the nested loops.

Final Time Complexity

Time Complexity: O(n²)

This means the time needed grows roughly with the square of the string length.

Common Mistake

[X] Wrong: "The algorithm runs in linear time because it just scans the string twice."

[OK] Correct: The nested loops check all possible substrings, which is much more work than just scanning once or twice.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming improves over brute force and shows your skill in analyzing nested loops.

Self-Check

"What if we used a different method to check palindromes in constant time? How would the time complexity change?"

Practice

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

easy

A. The total number of substrings in the string

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

C. The length of the longest palindromic substring

D. The maximum number of palindromic 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 to divide the string so that every piece is a palindrome.

3. Final Answer:

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

4. Quick Check:

   Minimum cuts = smallest cuts for palindrome parts [OK]

Hint: Minimum cuts split string into 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 is the correct way to declare a 2D boolean array isPalindrome of size n x n in C for palindrome checks?

easy

A. bool isPalindrome[n][n];

B. bool isPalindrome = new bool[n][n];

C. bool* isPalindrome = malloc(n * n * sizeof(bool));

D. bool isPalindrome[n];

Solution

1. Step 1: Understand 2D array declaration in C

   In C, a 2D array of booleans with size n x n is declared as bool isPalindrome[n][n];

2. Step 2: Check other options

   bool isPalindrome = new bool[n][n]; uses Java-like syntax, bool* isPalindrome = malloc(n * n * sizeof(bool)); is a pointer but not a 2D array, bool isPalindrome[n]; is 1D array.

3. Final Answer:

   bool isPalindrome[n][n]; -> Option A

4. Quick Check:

   2D array in C = bool isPalindrome[n][n]; [OK]

Hint: Use bool isPalindrome[n][n]; for 2D boolean arrays in C [OK]

Common Mistakes:

• Using Java or C++ syntax in C
• Declaring 1D array instead of 2D
• Using malloc without proper pointer handling

3. Consider the following C code snippet for minimum cuts palindrome partitioning. What is the output for input string "aab"?

int minCut(char* s) {
    int n = strlen(s);
    int cuts[n+1];
    bool isPal[n][n];
    for (int i = 0; i < n; i++) {
        cuts[i] = i - 1;
        for (int j = 0; j <= i; j++) {
            if (s[j] == s[i] && (i - j < 2 || isPal[j+1][i-1])) {
                isPal[j][i] = true;
                cuts[i] = j == 0 ? 0 : fmin(cuts[i], cuts[j-1] + 1);
            } else {
                isPal[j][i] = false;
            }
        }
    }
    return cuts[n-1];
}
printf("%d", minCut("aab"));

medium

A. 2

B. 0

C. 1

D. 3

Solution

1. Step 1: Trace palindrome checks and cuts for "aab"

   For i=0 ('a'): cuts[0]= -1 initially, palindrome is 'a', cuts[0]=0.
   For i=1 ('a'): substring 'aa' is palindrome, cuts[1]=0.
   For i=2 ('b'): 'b' is palindrome, cuts[2] = min(cuts[2], cuts[1]+1) = 1.

2. Step 2: Final minimum cuts calculation

   Minimum cuts needed to partition "aab" into palindromes is 1 (split as "aa" | "b").

3. Final Answer:

   1 -> Option C

4. Quick Check:

   Minimum cuts for "aab" = 1 [OK]

Hint: Trace cuts array updates for each palindrome end [OK]

Common Mistakes:

• Assuming no cuts needed for non-palindromic parts
• Misunderstanding substring palindrome checks
• Incorrectly initializing cuts array

4. The following code snippet for palindrome partitioning minimum cuts has a bug. What is the error?

int minCut(char* s) {
    int n = strlen(s);
    int cuts[n];
    bool isPal[n][n];
    for (int i = 0; i < n; i++) {
        cuts[i] = i;
        for (int j = 0; j <= i; j++) {
            if (s[j] == s[i] && (i - j < 2 || isPal[j+1][i-1])) {
                isPal[j][i] = true;
                cuts[i] = j == 0 ? 0 : fmin(cuts[i], cuts[j-1] + 1);
            } else {
                isPal[j][i] = false;
            }
        }
    }
    return cuts[n-1];
}

medium

A. The condition (i - j < 2) is incorrect

B. isPal array is not initialized before use

C. The function returns cuts[0] instead of cuts[n-1]

D. cuts array should be of size n+1 and initialized with i-1

Solution

1. Step 1: Check cuts array size and initialization

   cuts array should be size n+1 to handle base case and initialized with i-1 to represent max cuts.

2. Step 2: Identify impact of incorrect cuts initialization

   Initializing cuts[i] = i causes wrong minimum cuts calculation because it doesn't consider zero-based indexing properly.

3. Final Answer:

   cuts array should be of size n+1 and initialized with i-1 -> Option D

4. Quick Check:

   Correct cuts size and init = n+1 and i-1 [OK]

Hint: Initialize cuts with i-1 and size n+1 [OK]

Common Mistakes:

• Using cuts size n instead of n+1
• Not initializing cuts properly
• Ignoring base case for cuts

5. Given the string "abcbm", what is the minimum number of cuts needed to partition it into palindromic substrings using dynamic programming?

hard

A. 2

B. 1

C. 3

D. 0

Solution

1. Step 1: Identify palindromic substrings in "abcbm"

   "a" is palindrome, "bcb" is palindrome, "m" is palindrome.

2. Step 2: Calculate minimum cuts

   Partition as "a" | "bcb" | "m" requires 2 cuts.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Minimum cuts for "abcbm" = 2 [OK]

Hint: Look for longest palindromes to reduce cuts [OK]

Common Mistakes:

• Counting cuts as number of palindromes
• Missing middle palindrome "bcb"
• Assuming zero cuts for non-palindromic string