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

Palindrome Partitioning DP Minimum Cuts

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

🎯 Goal: You will build a program in C that uses dynamic programming to find the minimum number of cuts needed to partition a given string into palindromes.

📋 What You'll Learn

Create a string variable with the exact value "aab"

Create an integer variable to store the length of the string

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 find minimum cuts

Print the minimum cuts for the entire string

💡 Why This Matters

🌍 Real World

This technique helps in text processing tasks like DNA sequence analysis, data compression, and natural language processing where palindrome detection is important.

💼 Career

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

1

Create the input string and its length

Create a character array called s with the exact value "aab" and an integer variable called n that stores the length of s.

DSA C

char s[] = "aab";
int n = 0;  // Your code here

Hint

Use sizeof(s) / sizeof(s[0]) - 1 to get the length of the string excluding the null character.

2

Create the palindrome table and cuts array

Create a 2D boolean array called pal of size n by n to store palindrome checks, and an integer array called cuts of size n to store minimum cuts for substrings.

DSA C

char s[] = "aab";
int n = sizeof(s) / sizeof(s[0]) - 1;

int pal[n][n];
int cuts[n];  // Your code here

Hint

Use int pal[n][n]; and int cuts[n]; to declare the arrays.

3

Fill palindrome table and compute minimum cuts

Use nested for loops with variables j and i to fill the pal table for palindromes and compute minimum cuts in the cuts array using dynamic programming.

DSA C

char s[] = "aab";
int n = sizeof(s) / sizeof(s[0]) - 1;

int pal[n][n];
int cuts[n];

// Your code here

Hint

Use the condition s[i] == s[j] && (i - j < 2 || pal[j + 1][i - 1]) to check palindromes.

4

Print the minimum cuts for the entire string

Print the minimum cuts needed for the entire string by printing cuts[n - 1].

DSA C

char s[] = "aab";
int n = sizeof(s) / sizeof(s[0]) - 1;

int pal[n][n];
int cuts[n];

for (int i = 0; i < n; i++) {
    cuts[i] = i;  
    for (int j = 0; j <= i; j++) {
        if (s[i] == s[j] && (i - j < 2 || pal[j + 1][i - 1])) {
            pal[j][i] = 1;
            if (j == 0) {
                cuts[i] = 0;
            } else if (cuts[j - 1] + 1 < cuts[i]) {
                cuts[i] = cuts[j - 1] + 1;
            }
        } else {
            pal[j][i] = 0;
        }
    }
}

// Your code here

Hint

Use printf("%d\n", 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 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