Palindrome Partitioning DP Minimum Cuts in DSA C

Mental Model

We want to split a string into the fewest pieces so that each piece reads the same forwards and backwards.

Analogy: Imagine cutting a ribbon into the smallest number of palindromic segments, like cutting a patterned ribbon where each piece shows a symmetrical design.

s: a b a c d c b a
cuts: 0 1 2 3 4 5 6 7 8
palindrome checks: true/false for each substring

Dry Run Walkthrough

Input: string: "aab"

Goal: Find the minimum cuts needed to split "aab" into palindromic substrings

Step 1: Check all substrings for palindrome

palindrome table:
[0][0]=true (a)
[0][1]=true (aa)
[0][2]=false (aab)
[1][1]=true (a)
[1][2]=false (ab)
[2][2]=true (b)

Why: Knowing which substrings are palindromes helps decide where to cut

Step 2: Initialize cuts array with max cuts (index)

cuts: [0, 1, 2]

Why: Start assuming worst case: cut before every character

Step 3: At index 1, substring "aa" is palindrome, update cuts[1] to 0

cuts: [0, 0, 2]

Why: No cut needed if substring from start is palindrome

Step 4: At index 2, check substrings ending at 2: "b" palindrome, cuts[2] = cuts[1] + 1 = 1 "ab" not palindrome "aab" not palindrome

cuts: [0, 0, 1]

Why: Minimum cuts updated based on palindrome substrings ending at current index

Result:

cuts array: [0, 0, 1]
Minimum cuts needed: 1

Annotated Code

DSA C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

// Function to check palindrome substrings using DP
void computePalindromeTable(char *s, bool palindrome[][100], int n) {
    for (int i = 0; i < n; i++) {
        palindrome[i][i] = true; // single char palindrome
    }
    for (int length = 2; length <= n; length++) {
        for (int start = 0; start <= n - length; start++) {
            int end = start + length - 1;
            if (s[start] == s[end]) {
                if (length == 2) {
                    palindrome[start][end] = true; // two chars same
                } else {
                    palindrome[start][end] = palindrome[start + 1][end - 1]; // depends on inner substring
                }
            } else {
                palindrome[start][end] = false;
            }
        }
    }
}

// Function to find minimum cuts for palindrome partitioning
int minCut(char *s) {
    int n = strlen(s);
    if (n == 0) return 0;
    bool palindrome[100][100] = {{false}};
    int cuts[100];

    computePalindromeTable(s, palindrome, n);

    for (int i = 0; i < n; i++) {
        cuts[i] = i; // max cuts
        for (int j = 0; j <= i; j++) {
            if (palindrome[j][i]) {
                if (j == 0) {
                    cuts[i] = 0; // whole substring is palindrome
                } else if (cuts[j - 1] + 1 < cuts[i]) {
                    cuts[i] = cuts[j - 1] + 1; // update min cuts
                }
            }
        }
    }
    return cuts[n - 1];
}

int main() {
    char s[] = "aab";
    int result = minCut(s);
    printf("Minimum cuts needed: %d\n", result);
    return 0;
}

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

mark single characters as palindrome

if (s[start] == s[end]) { ... palindrome[start][end] = palindrome[start + 1][end - 1]; }

check palindrome status of substring based on inner substring

cuts[i] = i; // max cuts

initialize cuts with worst case (cut before every character)

if (palindrome[j][i]) { if (j == 0) cuts[i] = 0; else cuts[i] = min(cuts[i], cuts[j - 1] + 1); }

update cuts if substring s[j..i] is palindrome

Output

Minimum cuts needed: 1

Complexity Analysis

Time: O(n^2) because we check all substrings for palindrome and compute cuts for each index

Space: O(n^2) for palindrome table and O(n) for cuts array

vs Alternative: Naive approach checks palindrome on the fly causing O(n^3) time; DP reduces it to O(n^2)

Edge Cases

empty string

returns 0 cuts as no partition needed

DSA C

int n = strlen(s); if n == 0 return 0;

string with all same characters, e.g. "aaaa"

returns 0 cuts since whole string is palindrome

DSA C

if (j == 0) cuts[i] = 0;

string with no palindrome longer than 1, e.g. "abc"

returns n-1 cuts, cutting before every character

DSA C

cuts[i] = i; // max cuts initialization

When to Use This Pattern

When asked to split a string into palindromic parts with minimum cuts, use DP with palindrome substring precomputation to optimize.

Common Mistakes

Mistake: Checking palindrome substrings repeatedly without DP causing slow O(n^3) time
Fix: Precompute palindrome table with DP to check in O(1) time

Mistake: Not updating cuts correctly when substring from start is palindrome
Fix: Set cuts[i] = 0 if substring s[0..i] is palindrome

Summary

Find the fewest cuts needed to split a string into palindromic substrings.

Use when you want to partition strings minimizing cuts with palindrome constraints.

Precompute palindrome substrings to quickly decide where to cut and update 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