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 find how the time needed grows when we find the minimum cuts to split a string into palindromes using dynamic programming.

How does the number of steps change as the string gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int minCut(string s) {
    int n = s.size();
    vector> palindrome(n, vector(n, false));
    vector cuts(n, 0);
    for (int end = 0; end < n; ++end) {
        int min_cuts = 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;
                min_cuts = start == 0 ? 0 : min(min_cuts, cuts[start - 1] + 1);
            }
        }
        cuts[end] = min_cuts;
    }
    return cuts[n - 1];
}
    

This code finds the minimum cuts needed to split the string into palindrome substrings 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 for each outer iteration.

How Execution Grows With Input

As the string length grows, the code checks more substrings, roughly all pairs of start and end positions.

Input Size (n)	Approx. Operations
10	About 100 checks
100	About 10,000 checks
1000	About 1,000,000 checks

Pattern observation: The number of operations grows roughly with the square of the input size.

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: "Since we use nested loops, the time must be O(n³) because of palindrome checks inside."

[OK] Correct: The palindrome check uses previously stored results, so it does not re-check substrings fully each time, keeping complexity at O(n²).

Interview Connect

Understanding this time complexity helps you explain how dynamic programming reduces repeated work, a key skill in problem solving.

Self-Check

"What if we removed the palindrome memoization and checked each substring from scratch? How would the time complexity change?"

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