Palindrome Partitioning DP Minimum Cuts in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C++

int n = s.size();
vector<int> cuts(n, [1]);

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An - 1

Bn

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing cuts with 0 instead of n - 1.

Using n instead of n - 1.

We initialize the cuts array with the maximum cuts possible, which is n - 1 for a string of length n.

2fill in blank

medium

Complete the code to check if the substring s[j..i] is a palindrome using previously computed results.

DSA C++

if (s[j] == s[i] && (i - j < 2 || [1][j + 1][i - 1])) {
    // substring s[j..i] is palindrome
}

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Acuts

Bpalindrome

Cdp

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong array name like cuts or dp.

Not checking the substring inside the current substring.

The palindrome 2D array stores whether the substring s[j+1..i-1] is a palindrome.

3fill in blank

hard

Fix the error in updating the minimum cuts when a palindrome substring is found.

DSA C++

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

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Acuts[j - 1]

Bcuts[i]

Ccuts[j]

Dcuts[i - 1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using cuts[j] instead of cuts[j - 1].

Using cuts[i] or cuts[i - 1] incorrectly.

We add 1 to the cuts needed before the palindrome substring, which is cuts[j - 1].

4fill in blank

hard

Fill both blanks to update the palindrome table and minimum cuts inside nested loops.

DSA C++

palindrome[j][i] = (s[j] == s[i]) && (i - j < 2 || [1][j + 1][i - 1]);
if (palindrome[j][i]) {
    cuts[i] = (j == 0) ? 0 : min(cuts[i], [2] + 1);
}

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Apalindrome

Bcuts

Cdp

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up palindrome and cuts arrays.

Using dp or visited incorrectly.

The palindrome table is updated using palindrome, and cuts are updated using cuts.

5fill in blank

hard

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

DSA C++

int minCut(string s) {
    int n = s.size();
    vector<int> cuts(n, [1]);
    vector<vector<bool>> palindrome(n, vector<bool>(n, false));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= i; j++) {
            if (s[j] == s[i] && (i - j < 2 || palindrome[j + 1][i - 1])) {
                palindrome[j][i] = true;
                cuts[i] = (j == 0) ? 0 : min(cuts[i], cuts[j - 1] + [2]);
            }
        }
    }
    return [3];
}

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An - 1

B1

Ccuts[n - 1]

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 or cuts[0] instead of cuts[n - 1].

Adding 0 instead of 1 when updating cuts.

Initializing cuts with 0 instead of n - 1.

Initialize cuts with n - 1, add 1 when updating cuts, and return cuts[n - 1] as the final answer.

Practice

1. What does the "Palindrome Partitioning DP Minimum Cuts" problem aim to find for a given string s?

easy

A. The longest palindrome that can be formed by rearranging s.

B. The maximum length of a palindrome substring in s.

C. The minimum number of cuts needed to partition s into palindromic substrings.

D. The total number of palindrome substrings in s.

Solution

1. Step 1: Understand the problem goal

   The problem asks for the fewest cuts needed to split the string into parts where each part is a palindrome.

2. Step 2: Clarify what minimum cuts mean

   Minimum cuts means the smallest number of splits between characters so that all resulting substrings are palindromes.

3. Final Answer:

   The minimum number of cuts needed to partition s into palindromic substrings. -> Option C

4. Quick Check:

   Minimum cuts = minimum splits for palindrome parts [OK]

Hint: Focus on splitting string into palindrome parts with fewest cuts [OK]

Common Mistakes:

• Confusing minimum cuts with maximum palindrome length
• Counting total palindrome substrings instead of cuts
• Thinking about rearranging characters

2. Which of the following C++ code snippets correctly initializes a 2D vector isPalindrome of size n x n with false values for palindrome checks?

easy

A. std::vector<bool> isPalindrome(n, std::vector<bool>(n, false));

B. std::vector isPalindrome(n)(n, false);

C. std::vector isPalindrome(n, n, false);

D. std::vector isPalindrome(n, std::vector<bool>(n, false));

Solution

1. Step 1: Recall correct 2D vector initialization syntax

   In C++, a 2D vector is initialized as vector of vectors: vector v(rows, vector<bool>(cols, value)).

2. Step 2: Check each option

   std::vector isPalindrome(n, std::vector<bool>(n, false)); matches correct syntax. Options A, C, D have syntax errors or wrong types.

3. Final Answer:

   std::vector isPalindrome(n, std::vector<bool>(n, false)); -> Option D

4. Quick Check:

   Correct 2D vector init = std::vector isPalindrome(n, std::vector<bool>(n, false)); [OK]

Hint: Use vector of vectors with inner vector size and default value [OK]

Common Mistakes:

• Using wrong vector types or missing inner vector
• Trying to initialize 2D vector with wrong constructor
• Confusing parentheses and commas in initialization

3. What is the output of the following C++ code snippet that calculates minimum cuts for s = "aab"?

std::string s = "aab";
int n = s.size();
std::vector isPalindrome(n, std::vector<bool>(n, false));
for (int i = 0; i < n; i++) {
  isPalindrome[i][i] = true;
}
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) isPalindrome[start][end] = true;
      else isPalindrome[start][end] = isPalindrome[start + 1][end - 1];
    }
  }
}
std::vector<int> cuts(n, 0);
for (int i = 0; i < n; i++) {
  int minCut = i;
  for (int j = 0; j <= i; j++) {
    if (isPalindrome[j][i]) {
      minCut = j == 0 ? 0 : std::min(minCut, cuts[j - 1] + 1);
    }
  }
  cuts[i] = minCut;
}
std::cout << cuts[n - 1] << std::endl;

medium

A. 1

B. 0

C. 2

D. 3

Solution

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

   "a", "a", "b" are palindromes; "aa" is palindrome; "aab" is not.

2. Step 2: Calculate minimum cuts using DP

   Minimum cuts: "aa" | "b" needs 1 cut. So cuts[n-1] = 1.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   Minimum cuts for "aab" = 1 [OK]

Hint: Check palindrome substrings and count cuts between them [OK]

Common Mistakes:

• Assuming no cuts needed for non-palindrome whole string
• Misindexing palindrome checks
• Forgetting to update minimum cuts properly

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

for (int i = 0; i < n; i++) {
  int minCut = i;
  for (int j = 0; j <= i; j++) {
    if (isPalindrome[j][i]) {
      minCut = std::min(minCut, cuts[j] + 1);
    }
  }
  cuts[i] = minCut;
}

medium

A. Using cuts[j] instead of cuts[j - 1] causes incorrect indexing.

B. The outer loop should start from 1, not 0.

C. The inner loop should run from i to n, not 0 to i.

D. The isPalindrome check should be reversed to isPalindrome[i][j].

Solution

1. Step 1: Analyze the indexing in cuts array

   When j == 0, cuts[j - 1] is invalid, so we must handle j == 0 separately.

2. Step 2: Identify the bug in minCut update

   Using cuts[j] + 1 is wrong; it should be cuts[j - 1] + 1 except when j == 0.

3. Final Answer:

   Using cuts[j] instead of cuts[j - 1] causes incorrect indexing. -> Option A

4. Quick Check:

   Indexing cuts[j - 1] correctly fixes bug [OK]

Hint: Remember cuts[j - 1] for previous partition, handle j=0 separately [OK]

Common Mistakes:

• Ignoring boundary condition j=0
• Mixing up indices in cuts array
• Incorrect loop ranges

5. Given the string s = "ababbbabbababa", what is the minimum number of cuts needed to partition it into palindromic substrings using the DP approach?

hard

A. 5

B. 3

C. 4

D. 6

Solution

1. Step 1: Understand the string and palindrome partitions

   The string "ababbbabbababa" can be partitioned into palindromes with minimum cuts by checking all substrings using DP.

2. Step 2: Use DP to find minimum cuts

   Known solution for this string is 3 cuts, for example: "a" | "babbbab" | "b" | "ababa".

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Minimum cuts for given string = 3 [OK]

Hint: Use DP palindrome checks and track minimum cuts carefully [OK]

Common Mistakes:

• Underestimating cuts by missing longer palindrome substrings
• Overcounting cuts by splitting too often
• Not using DP to optimize palindrome checks