131. Palindrome Partitioning - cocoder39/coco39_LC GitHub Wiki

131. Palindrome Partitioning

the helper() takes O(n) to check if a substring is a palindrome, which can be optimized through preprocessed by dp. we can record the partition point that can produce palindrome. dp[i][j] = true iff s[i .. j] is a palindrome. Anyway, the time complexity is O(n * 2^n) for the worst case where s is "aaaaa...."

class Solution {
public:
    vector<vector<string>> partition(string s) {
        vector<vector<string>> res;
        vector<string> cur;
        helper(res, cur, s, 0);
        return res;
    }
private:
    void helper(vector<vector<string>>& res, vector<string>& cur, string& s, int start) {
        if (start == s.length()) {
            res.push_back(cur);
            return;
        }
        
        for (int i = start; i < s.length(); i++) {
            if (isPalindrome(s, start, i)) {
                string str = s.substr(start, i - start + 1);
                cur.push_back(str);
                helper(res, cur, s, i + 1);
                cur.pop_back();
            }
        }
    }
    
    bool isPalindrome(string& s, int start, int end) {
        while (start < end) {
            if (s[start] != s[end]) {
                return false;
            }
            start++;
            end--;
        }
        return true;
    }
};

dp records the range that is a palindrome, thus saving the O(n) palindrome check. time is O(n ^ 2 + 2 ^ n) = O(2 ^ n)

class Solution {
public:
    vector<vector<string>> partition(string s) {
        int len = s.length();
        vector<vector<bool>> isPal(len, vector<bool>(len)); 
        for (int i = 0; i < len; i++) {
            isPal[i][i] = true;
            if (i > 0 && s[i - 1] == s[i]) {
                isPal[i - 1][i] = true;
            }
        }
        for (int l = 3; l <= len; l++) {
            for (int i = 0; i - 1 + l < len; i++) {
                int j = i - 1 + l;
                //i + 1 <= j - 1 => j - i + 1 >= 3 => l = 1, 2 are basecase
                isPal[i][j] = (isPal[i + 1][j - 1] && s[i] == s[j]);
            }
        }
        
        vector<vector<string>> res;
        vector<string> cur;
        helper(res, cur, s, 0, isPal);
        return res;
    }
private:
    void helper(vector<vector<string>>& res, vector<string>& cur, string& s, int start, vector<vector<bool>>& isPal) {
        if (start == s.length()) {
            res.push_back(cur);
            return;
        }
        
        for (int i = start; i < s.length(); i++) {
            if (isPal[start][i]) {
                cur.push_back(s.substr(start, i - start + 1));
                helper(res, cur, s, i + 1, isPal);
                cur.pop_back();
            }
        }
    }
};

unlike word break, here vector cutable does not help since each position is cutable (each input has at least one solution which contains only single char)

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