Palindrome Partitioning DP Minimum Cuts in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a substring is a palindrome.

DSA C

int isPalindrome(char* s, int start, int end) {
    while (start < end) {
        if (s[start] != s[end]) {
            return [1];
        }
        start++;
        end--;
    }
    return 1;
}

Drag options to blanks, or click blank then click option'

A2

B1

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 1 instead of 0 on mismatch

Not returning immediately on mismatch

If characters at start and end don't match, the substring is not a palindrome, so return 0 (false).

2fill in blank

medium

Complete the code to initialize the minimum cuts array with maximum possible cuts.

DSA C

for (int i = 0; i < n; i++) {
    minCuts[i] = [1];
}

Drag options to blanks, or click blank then click option'

A0

Bi

Cn - 1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing with 0 which is incorrect

Using n instead of i

Initialize minCuts[i] with i, the maximum cuts needed if every character is a separate palindrome.

3fill in blank

hard

Fix the error in the code that updates minimum cuts when a palindrome substring is found.

DSA C

if (isPalindrome(s, j, i)) {
    if (j == 0) {
        minCuts[i] = 0;
    } else {
        minCuts[i] = [1];
    }
}

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AminCuts[i] + 1

BminCuts[j] + 1

CminCuts[j - 1] + 1

DminCuts[i - 1] + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using minCuts[i] + 1 causes infinite increase

Using minCuts[j] instead of minCuts[j-1]

If substring s[j..i] is palindrome, minimum cuts at i can be updated as minCuts[j-1] + 1.

4fill in blank

hard

Fill both blanks to complete the nested loops for DP minimum cuts calculation.

DSA C

for (int i = 1; i < n; i++) {
    for (int j = 0; j <= [1]; j++) {
        if (isPalindrome(s, j, i)) {
            if (j == 0) {
                minCuts[i] = 0;
            } else {
                minCuts[i] = min(minCuts[i], [2]);
            }
        }
    }
}

Drag options to blanks, or click blank then click option'

Ai

Bj - 1

Ci - 1

DminCuts[j - 1] + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Looping j beyond i

Using wrong expression for minCuts update

Inner loop runs j from 0 to i. Update minCuts[i] with minimum of current and minCuts[j-1]+1.

5fill in blank

hard

Fill all three blanks to complete the function that returns minimum cuts for palindrome partitioning.

DSA C

int minCut(char* s) {
    int n = strlen(s);
    int minCuts[n];
    for (int i = 0; i < n; i++) {
        minCuts[i] = [1];
    }
    for (int i = 1; i < n; i++) {
        for (int j = 0; j <= [2]; j++) {
            if (isPalindrome(s, j, i)) {
                if (j == 0) {
                    minCuts[i] = 0;
                } else {
                    minCuts[i] = min(minCuts[i], [3]);
                }
            }
        }
    }
    return minCuts[n - 1];
}

Drag options to blanks, or click blank then click option'

Ai

Bn - 1

CminCuts[j - 1] + 1

DminCuts[i - 1] + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing minCuts with wrong value

Looping j incorrectly

Updating minCuts[i] with wrong expression

Initialize minCuts[i] = i; inner loop j runs to i; update minCuts[i] with minCuts[j-1]+1.

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