132. Palindrome Partitioning II

Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.
 

Example 1:

Input: s = “aab”
Output: 1
Explanation: The palindrome partitioning [“aa”,“b”] could be produced using 1 cut.

Example 2:

Input: s = “a”
Output: 0

Example 3:

Input: s = “ab”
Output: 1

Constraints:

• 1 <= s.length <= 2000
• s consists of lowercase English letters only.

From: LeetCode
Link: 132. Palindrome Partitioning II

---

Solution:

Ideas:

1. Helper Function isPalindrome:

• This function checks if a substring s[start:end] is a palindrome by comparing characters from the start and end moving towards the center.

2. Dynamic Programming Array dp:

• dp[i] represents the minimum cuts needed for the substring s[0:i-1].
• Initialize dp[i] to i-1 since the worst case requires i-1 cuts (each character is a palindrome).

3. Nested Loops:

• The outer loop iterates over the end index of the substring.
• The inner loop iterates over the start index.
• If the substring s[start:end-1] is a palindrome, update dp[end] to the minimum of its current value or dp[start] + 1.

4. Return the Result:

• The final result is stored in dp[n] which represents the minimum cuts for the entire string s.

Code:

bool isPalindrome(char* s, int start, int end) {while (start < end) {if (s[start] != s[end]) {return false;}start++;end--;}return true;
}int minCut(char* s) {int n = strlen(s);if (n == 0) {return 0;}// Array to store the minimum cuts for each prefix of the stringint* dp = (int*)malloc((n + 1) * sizeof(int));for (int i = 0; i <= n; i++) {dp[i] = i - 1;}// Iterate over the end of the substringfor (int end = 1; end <= n; end++) {// Iterate over the start of the substringfor (int start = 0; start < end; start++) {if (isPalindrome(s, start, end - 1)) {dp[end] = (dp[end] < dp[start] + 1) ? dp[end] : dp[start] + 1;}}}int result = dp[n];free(dp);return result;
}