Minimum Insertion Steps to Make a String Palindrome
Leetcode Daily Challenge (22nd April, 2023)
Problem Statement:-
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.
Link: https://leetcode.com/problems/minimum-insertion-steps-to-make-a-string-palindrome/description/
Problem Explanation with examples:-
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 <= 500sconsists of lowercase English letters.
Intuition:-
We are allowed to insert any number of characters at any position.
So, if we find the longest palindromic subsequence, we can be sure about how many characters are already making up the palindrome part.
The remaining characters need to be inserted to make the whole string a palindrome.
The number of characters to be inserted is the difference between the length of the string and the length of the longest palindromic subsequence.
To find the longest palindromic subsequence, we can use the longest common subsequence algorithm with the string and its reverse.
Solution:-
Define a function lcs(s1, s2, idx1, idx2) to find the longest common subsequence of s1 and s2.
If idx1 < 0 or idx2 < 0, return 0.
If s1[idx1] == s2[idx2], return 1 + lcs(s1, s2, idx1 - 1, idx2 - 1).
Else, return max(lcs(s1, s2, idx1 - 1, idx2), lcs(s1, s2, idx1, idx2 - 1)).
Define a function longestPalindromeSubseq(s) to find the longest palindromic subsequence of s.
Return lcs(s, s[::-1], len(s) - 1, len(s) - 1) where s[::-1] is the reverse of s.
Finally our answer is len(s) - longestPalindromeSubseq(s) where s is the input string.
Code:-
JAVA Solution
class Solution {
public int minInsertions(String s) {
return s.length() - longestPalindromeSubseq(s);
}
public int longestPalindromeSubseq(String s) {
Integer[][][] dp = new Integer[s.length()][s.length()][2];
return lcs(s, new StringBuilder(s).reverse().toString(), s.length() - 1, s.length() - 1, dp);
}
public int lcs(String s1, String s2, int idx1, int idx2, Integer[][][] dp) {
if (idx1 < 0 || idx2 < 0) {
return 0;
}
if (dp[idx1][idx2][0] != null) {
return dp[idx1][idx2][0];
}
if (s1.charAt(idx1) == s2.charAt(idx2)) {
dp[idx1][idx2][0] = 1 + lcs(s1, s2, idx1 - 1, idx2 - 1, dp);
} else {
dp[idx1][idx2][0] = Math.max(lcs(s1, s2, idx1 - 1, idx2, dp), lcs(s1, s2, idx1, idx2 - 1, dp));
}
return dp[idx1][idx2][0];
}
}
Python Solution
class Solution:
def minInsertions(self, s: str) -> int:
@cache
def lcs(s1, s2, idx1, idx2):
if idx1<0 or idx2<0:
return 0
if s1[idx1] == s2[idx2]:
return 1 + lcs(s1, s2, idx1-1, idx2-1)
return max(lcs(s1, s2, idx1-1, idx2), lcs(s1, s2, idx1, idx2-1))
def longestPalindromeSubseq(s: str) -> int:
return lcs(s, s[::-1], len(s)-1, len(s)-1)
return len(s) - longestPalindromeSubseq(s)
Complexity Analysis:-
TIME:-
The time complexity of the code is O(n^2) where n is the length of the input string as we are using the longest common subsequence algorithm to find the longest palindromic subsequence.
SPACE:-
The space complexity of the code is O(n^2) where n is the length of the input string as we are using a 2D dp array to store the results of the longest common subsequence algorithm.