1312

1312. Minimum Insertion Steps to Make a String Palindrome (Hard)

Given a string s. In one step you can insert any character at any index of the string.

Return the minimum number of steps to make s palindrome.

A Palindrome String is one that reads the same backward as well as forward.

 

Example 1:

Input: s = "zzazz"
Output: 0
Explanation: The string "zzazz" is already palindrome we don't need any insertions.

Example 2:

Input: s = "mbadm"
Output: 2
Explanation: String can be "mbdadbm" or "mdbabdm".

Example 3:

Input: s = "leetcode"
Output: 5
Explanation: Inserting 5 characters the string becomes "leetcodocteel".

Example 4:

Input: s = "g"
Output: 0

Example 5:

Input: s = "no"
Output: 1

 

Constraints:

• 1 <= s.length <= 500
• All characters of s are lower case English letters.

Related Topics:
Dynamic Programming

Solution 1. DP

This problem is similar to "given two strings s and t, count how many insertions are needed to make them the same".

In this problem, t is the reversed s.

Let dp[i][j] be the minimum insertions needed to make s[0..(i-1)] and t[0..(j-1)] the same.

dp[i][j] = dp[i-1][j-1]                     If s[i-1] == t[j-1]
         = 1 + min(dp[i-1][j], dp[i][j-1])  If s[i-1] != t[j-1]
dp[0][i] = dp[i][0] = i

The answer of this problem is dp[N][N] / 2.

// OJ: https://leetcode.com/problems/minimum-insertion-steps-to-make-a-string-palindrome/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N^2)
class Solution {
public:
    int minInsertions(string s) {
        int N = s.size();
        string t(s.rbegin(), s.rend());
        vector<vector<int>> dp(N + 1, vector<int>(N + 1));
        for (int i = 0; i <= N; ++i) {
            for (int j = 0; j <= N; ++j) {
                if (i == 0 || j == 0) dp[i][j] = i + j;
                else if (s[i - 1] == t[j - 1]) dp[i][j] = dp[i - 1][j - 1];
                else dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        return dp[N][N] / 2;
    }
};

Solution 2. DP + Space Optimization

Since dp[i][j] is only dependent on dp[i-1][j-1], dp[i-1][j] and dp[i][j-1], we can reduce the space of dp array from N * N to 1 * N with a temporary variable storing dp[i-1][j-1].

// OJ: https://leetcode.com/problems/minimum-insertion-steps-to-make-a-string-palindrome/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int minInsertions(string s) {
        int N = s.size();
        vector<int> dp(N + 1);
        for (int i = 0; i <= N; ++i) {
            int prev;
            for (int j = 0; j <= N; ++j) {
                int cur = dp[j];
                if (i == 0 || j == 0) dp[j] = i + j;
                else if (s[i - 1] == s[N - j]) dp[j] = prev;
                else dp[j] = 1 + min(dp[j], dp[j - 1]);
                prev = cur;
            }
        }
        return dp[N] / 2;
    }
};;

Solution 3. Longest Palindrome Subsequence

If we know the length of the longest palindrome subsequence of s is M, then the answer to this problem would be s.size() - M.

So we can reuse the solution to 516. Longest Palindromic Subsequence (Medium).

// OJ: https://leetcode.com/problems/minimum-insertion-steps-to-make-a-string-palindrome/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N)
class Solution {
    int longestPalindromeSubseq(string &s) {
        int N = s.size();
        vector<int> dp(N + 1);
        for (int i = 1; i <= N; ++i) {
            int prev = 0;
            for (int j = 1; j <= N; ++j) {
                int cur = dp[j];
                if (s[i - 1] == s[N - j]) dp[j] = 1 + prev;
                else dp[j] = max(dp[j], dp[j - 1]);
                prev = cur;
            }
        }
        return dp[N];
    }
public:
    int minInsertions(string s) {
        return s.size() - longestPalindromeSubseq(s);
    }
};