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LongestPalindromicSubsequence.cpp
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LongestPalindromicSubsequence.cpp
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// Source : https://leetcode.com/problems/longest-palindromic-subsequence/
// Author : Hao Chen
// Date : 2021-03-27
/*****************************************************************************************************
*
* Given a string s, find the longest palindromic subsequence's length in s.
*
* A subsequence is a sequence that can be derived from another sequence by deleting some or no
* elements without changing the order of the remaining elements.
*
* Example 1:
*
* Input: s = "bbbab"
* Output: 4
* Explanation: One possible longest palindromic subsequence is "bbbb".
*
* Example 2:
*
* Input: s = "cbbd"
* Output: 2
* Explanation: One possible longest palindromic subsequence is "bb".
*
* Constraints:
*
* 1 <= s.length <= 1000
* s consists only of lowercase English letters.
******************************************************************************************************/
/*
supposed s = "abbcba"
we can have a matrix,
- dp[start, end] is the longest from s[start] to s[end]
- if (start == end) dp[statr, end] = 1, it means every char can be palindromic
a b b c b a
a 1 0 0 0 0 0
b 0 1 0 0 0 0
b 0 0 1 0 0 0
c 0 0 0 1 0 0
b 0 0 0 0 1 0
a 0 0 0 0 0 1
calculating from the bottom to up. (Note: only care about the top-right trangle)
a b b c b a
a 1 1 2 2 3 [5] <-- a == a , so "abbcba" comes from "bbcb" + 2
b 0 1 [2] 2 3 3 <-- b == b , so "bb" comes from "" + 2
b 0 0 1 1 [3] 3 <-- b == b , so "bcb" comes from "c" + 2
c 0 0 0 1 1 [1] <-- c != a , so "cba" comes from max("cb", "a")
b 0 0 0 0 1 [1] <-- b != a , so "ba" comes from max ("b", "a")
a 0 0 0 0 0 1
So, we can have the following formular:
s[start] != s[end] ==> dp[start, end] = max (dp[start+1, end], dp[start, end-1]);
s[start] == s[end] ==> dp[start, end] = dp[start+1, end-1] + 2;
*/
class Solution {
public:
int longestPalindromeSubseq(string s) {
int n = s.size();
vector<vector<int>> dp(n, vector<int>(n, 0));
for (int start = n-1; start>=0; start--) {
for (int end = start ; end < n ; end++) {
if (start == end) {
dp[start][end] = 1;
continue;
}
if (s[start] == s[end]) {
dp[start][end] = dp[start+1][end-1] + 2;
}else{
dp[start][end] = max (dp[start+1][end], dp[start][end-1]);
}
}
}
return dp[0][n-1];
}
};