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This approach involves using two DP tables. One to check if a substring is a palindrome and another to compute the minimum cuts required.
We maintain a 2D boolean table where palindrome[i][j] is true if the substring s[i...j] is a palindrome. Using this table, we calculate the minimum number of palindrome partitions.
Time Complexity: O(n^2) due to filling the palindrome table and computing cuts.
Space Complexity: O(n^2) as well for the DP tables storage.
1var minCut = function(s) {
2 const n = s.length;
3 const palindrome = Array.from({ length: n }, () => Array(n).fill(false));
4 const cuts = Array(n).fill(0);
5
6 for (let i = 0; i < n; i++) {
7 let minCut = i;
8 for (let j = 0; j <= i; j++) {
9 if (s[j] == s[i] && (i - j <= 2 || palindrome[j + 1][i - 1])) {
10 palindrome[j][i] = true;
11 minCut = j == 0 ? 0 : Math.min(minCut, cuts[j - 1] + 1);
12 }
13 }
14 cuts[i] = minCut;
15 }
16
17 return cuts[n - 1];
18};
19
20console.log("Minimum cuts needed for Palindrome Partitioning: " + minCut("aab"));
21The JavaScript solution also initializes a palindrome table using nested arrays and determines cuts through checks of palindrome properties of substrings. It closely adheres to the same logic as described in other languages within this approach.
This approach utilizes the idea of expanding around potential palindrome centers, combined with a memoization strategy to store minimum cuts. It significantly reduces redundant calculations by only considering centers and keeping track of the best solutions observed so far.
Time Complexity: O(n^2) due to potentially expanding and updating cuts for each center.
Space Complexity: O(n) focused on the cuts array.
1
Python implementation embraces center expansion combined with a dynamic programming array to track minimal cuts, aiming to tighten the calculation around palindrome potentials.