Palindrome Partitioning DP Minimum Cuts in DSA C++

Mental Model

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

Analogy: Imagine cutting a ribbon into the smallest number of palindromic segments, like cutting a necklace into symmetrical beads.

s = "a b a c"
Indices: 0 1 2 3
Partitions: a | b | a | c or aba | c
Goal: minimize cuts between characters

String: a -> b -> a -> c -> null
Cuts: ^    ^    ^    ^
       0    1    2    3

Dry Run Walkthrough

Input: string: "aab"

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

Step 1: Check all substrings if they are palindrome

Palindrome table:
[true, true, false]
[ , true, false]
[ ,  , true]

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

Step 2: Initialize cuts array with max cuts (worst case: cut before every char)

cuts = [0, 1, 2]

Why: Start with maximum cuts, then try to reduce

Step 3: For i=1 (substring "aa"), since palindrome, cuts[1] = 0

cuts = [0, 0, 2]

Why: No cut needed if substring from start is palindrome

Step 4: For i=2 (substring "aab"), check partitions: - cut after 0: if s[1..2] palindrome? no - cut after 1: if s[2..2] palindrome? yes, cuts[2] = cuts[1] + 1 = 1

cuts = [0, 0, 1]

Why: Choose minimum cuts by checking palindrome partitions

Result:

cuts array: [0, 0, 1]
Minimum cuts needed: 1
Partitions: "aa" | "b"

Annotated Code

DSA C++

#include <iostream>
#include <vector>
#include <string>
using namespace std;

class Solution {
public:
    int minCut(string s) {
        int n = s.size();
        if (n == 0) return 0;
        vector<vector<bool>> palindrome(n, vector<bool>(n, false));
        vector<int> cuts(n, 0);

        // Precompute palindrome substrings
        for (int end = 0; end < n; ++end) {
            cuts[end] = end; // max cuts
            for (int start = 0; start <= end; ++start) {
                if (s[start] == s[end] && (end - start < 2 || palindrome[start + 1][end - 1])) {
                    palindrome[start][end] = true;
                    if (start == 0) {
                        cuts[end] = 0; // no cut if palindrome from start
                    } else {
                        cuts[end] = min(cuts[end], cuts[start - 1] + 1); // cut before start
                    }
                }
            }
        }
        return cuts[n - 1];
    }
};

int main() {
    Solution sol;
    string input = "aab";
    cout << sol.minCut(input) << endl;
    return 0;
}

for (int end = 0; end < n; ++end) {

iterate over substring end indices to build palindrome info and cuts

cuts[end] = end; // max cuts

initialize cuts with worst case: cut before every character

for (int start = 0; start <= end; ++start) {

check all substring starts for palindrome ending at end

if (s[start] == s[end] && (end - start < 2 || palindrome[start + 1][end - 1])) {

check palindrome condition: chars match and inner substring palindrome

palindrome[start][end] = true;

mark substring as palindrome

if (start == 0) { cuts[end] = 0; } else { cuts[end] = min(cuts[end], cuts[start - 1] + 1); }

update cuts: no cut if palindrome from start, else cut before start

Output

1

Complexity Analysis

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

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

vs Alternative: Naive approach tries all partitions recursively with exponential time; DP reduces it to polynomial time

Edge Cases

empty string

returns 0 cuts since no partitions needed

DSA C++

int n = s.size(); if n == 0, cuts array is empty and function returns 0

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

returns 0 cuts since whole string is palindrome

DSA C++

if (start == 0) cuts[end] = 0;

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

returns n-1 cuts, cutting between every character

DSA C++

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

When to Use This Pattern

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

Common Mistakes

Mistake: Not precomputing palindrome substrings and checking palindrome repeatedly causing TLE
Fix: Use DP table to store palindrome results and reuse them

Mistake: Incorrectly updating cuts array without considering no-cut case when palindrome starts at index 0
Fix: Set cuts[end] = 0 when palindrome substring starts at 0

Summary

Finds minimum cuts to partition a string into palindromic substrings.

Use when you need to split strings into palindromes with fewest cuts.

Key insight: precompute palindrome substrings and use DP to minimize cuts efficiently.

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