palindrome-partitioning-iii.cpp

// Time:  O(k * n^2)
// Space: O(n^2)

class Solution {
public:
    int palindromePartition(string s, int k) {
        // dp1[i][j]: minimum number of changes to make s[i, j] palindrome
        vector<vector<int>> dp1(s.length(), vector<int>(s.length()));
        for (int l = 1; l <= s.length(); ++l) {
            for (int i = 0; i + l <= s.length(); ++i) {
                int j = i + l - 1;
                if (i == j - 1) {
                    dp1[i][j] = s[i] == s[j] ? 0 : 1;
                }  else if (i != j) {
                    dp1[i][j] = s[i] == s[j] ? dp1[i + 1][j - 1] : dp1[i + 1][j - 1] + 1;
                }
            }
        }
        // dp2[d][i]: minimum number of changes to divide s[0, i] into d palindromes
        vector<vector<int>> dp2(2, vector<int>(s.length(), numeric_limits<int>::max()));
        dp2[1] = dp1[0];
        for (int d = 2; d <= k; ++d) {
            dp2[d % 2] = vector<int>(s.length(), numeric_limits<int>::max());
            for (int i = d - 1; i < s.length(); ++i) {
                for (int j = d - 2; j < i; ++j) {
                    dp2[d % 2][i] = min(dp2[d % 2][i], dp2[(d - 1) % 2][j] + dp1[j + 1][i]);
                }
            }
        }
        return dp2[k % 2][s.length() - 1];
    }
};