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1312. Minimum Insertion Steps to Make a String Palindrome

Description

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 do not 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".

 

Constraints:

  • 1 <= s.length <= 500
  • s consists of lowercase English letters.

Solutions

  • class Solution {
    private Integer[][] f;
    private String s;

    public int minInsertions(String s) {
        this.s = s;
        int n = s.length();
        f = new Integer[n][n];
        return dfs(0, n - 1);
    }

    private int dfs(int i, int j) {
        if (i >= j) {
            return 0;
        }
        if (f[i][j] != null) {
            return f[i][j];
        }
        int ans = 1 << 30;
        if (s.charAt(i) == s.charAt(j)) {
            ans = dfs(i + 1, j - 1);
        } else {
            ans = Math.min(dfs(i + 1, j), dfs(i, j - 1)) + 1;
        }
        return f[i][j] = ans;
    }
}

  • class Solution {
public:
    int minInsertions(string s) {
        int n = s.size();
        int f[n][n];
        memset(f, -1, sizeof(f));
        function<int(int, int)> dfs = [&](int i, int j) -> int {
            if (i >= j) {
                return 0;
            }
            if (f[i][j] != -1) {
                return f[i][j];
            }
            int ans = 1 << 30;
            if (s[i] == s[j]) {
                ans = dfs(i + 1, j - 1);
            } else {
                ans = min(dfs(i + 1, j), dfs(i, j - 1)) + 1;
            }
            return f[i][j] = ans;
        };
        return dfs(0, n - 1);
    }
};

  • class Solution:
    def minInsertions(self, s: str) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if i >= j:
                return 0
            if s[i] == s[j]:
                return dfs(i + 1, j - 1)
            return 1 + min(dfs(i + 1, j), dfs(i, j - 1))

        return dfs(0, len(s) - 1)


  • func minInsertions(s string) int {
	n := len(s)
	f := make([][]int, n)
	for i := range f {
		f[i] = make([]int, n)
		for j := range f[i] {
			f[i][j] = -1
		}
	}
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		if i >= j {
			return 0
		}
		if f[i][j] != -1 {
			return f[i][j]
		}
		ans := 1 << 30
		if s[i] == s[j] {
			ans = dfs(i+1, j-1)
		} else {
			ans = min(dfs(i+1, j), dfs(i, j-1)) + 1
		}
		f[i][j] = ans
		return ans
	}
	return dfs(0, n-1)
}

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