LeetCode 132 Palindrome Partitioning II (C++, Python)

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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.
For example, given s = "aab",
Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.

Solution C++


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class Solution {
public:
    int minCut(string s) {
        bool palin[s.length()][s.length()];
        for (int i = s.length() - 1; i >= 0; --i) {
            for (int j = i; j < s.length(); ++j) {
                if ((s[i] == s[j]) && ((j - i < 3) || palin[i + 1][j - 1])) {
                    palin[i][j] = true;
                } else {
                    palin[i][j] = false;
                }
            }
        }
        vector<int> mincuts (s.length(), 0);
        for (int i = 1; i < s.length(); ++i) {
            if (palin[0][i]) {
                mincuts[i] = 0;
            } else {
                int tmp = i;
                for (int k = 0; k < i; ++k) {
                    if (palin[k + 1][i]) tmp = min(tmp, 1 + mincuts[k]);        
                }
                mincuts[i] = tmp;
            }
        }
        return mincuts.back();
    }
};
Solution Python


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class Solution:
    # @param s, a string
    # @return an integer
    def minCut(self, s):
        row = [False] * len(s)
        palin = []
        for _ in range(len(s)):
            palin.append(row[:])
        for i in range(len(s) - 1, -1, -1):
            for j in range(i, len(s)):
                if s[i] == s[j] and ((j - i < 3) or palin[i + 1][j - 1]):
                    palin[i][j] = True
        
        mincuts = [0] * len(s)
        for i in range(1, len(s)):
            if palin[0][i]:
                mincuts[i] = 0
            else:
                tmp = i
                for k in range(0, i):
                    if palin[k + 1][i]:
                        tmp = min(tmp, 1 + mincuts[k])
                mincuts[i] = tmp
        return mincuts[-1]

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