[Coding Made Simple] Longest Palindromic Subsequence
Given a string, find longest palindromic subsequence in this string.
lps[start, end] = lps[start + 1, end - 1] + 2, if s.charAt(start) == s.charAt(end);
lps[start, end] = Math.max(lps[start + 1][end], lps[start][end - 1]), if s.charAt(start) != s.charAt(end).
Base case: start > end, lps = 0; start == end, lps = 1.
Based on the above formula, we can use either recursion or dynamic programming to solve this problem.
Recursive solution.
1 public class LongestPalSubseq { 2 public int getLenOfLPSRecursion(String s) { 3 if(s == null) { 4 return 0; 5 } 6 return recursiveHelper(s, 0, s.length() - 1); 7 } 8 private int recursiveHelper(String s, int start, int end) { 9 if(start > end) { 10 return 0; 11 } 12 else if(start == end) { 13 return 1; 14 } 15 if(s.charAt(start) == s.charAt(end)) { 16 return 2 + recursiveHelper(s, start + 1, end - 1); 17 } 18 return Math.max(recursiveHelper(s, start + 1, end), recursiveHelper(s, start, end - 1)); 19 } 20 public static void main(String[] args) { 21 String s1 = "agbdba", s2 = "", s3 = "a", s4 = "abcefegdba"; 22 LongestPalSubseq test = new LongestPalSubseq(); 23 System.out.println(test.getLenOfLPSRecursion(s1)); 24 System.out.println(test.getLenOfLPSRecursion(s2)); 25 System.out.println(test.getLenOfLPSRecursion(s3)); 26 System.out.println(test.getLenOfLPSRecursion(s4)); 27 } 28 }
Dynamic programming solution. T[i][j]: the length of longest palindromic subsequence in substring s[i....j].
1 import java.util.ArrayList; 2 3 public class LongestPalSubseq { 4 private ArrayList<Character> palinSeq; 5 public int getLenOfLpsDp(String s) { 6 palinSeq = new ArrayList<Character>(); 7 if(s == null|| s.length() == 0) { 8 return 0; 9 } 10 int[][] T = new int[s.length()][s.length()]; 11 for(int i = 0; i < T.length; i++) { 12 T[i][i] = 1; 13 } 14 for(int len = 2; len <= s.length(); len++) { 15 for(int i = 0; i <= s.length() - len; i++) { 16 if(s.charAt(i) == s.charAt(i + len - 1)) { 17 T[i][i + len - 1] = len >= 3 ? T[i + 1][i + len - 2] + 2 : 2; 18 } 19 else { 20 T[i][i + len - 1] = Math.max(T[i + 1][i + len - 1], T[i][i + len - 2]); 21 } 22 } 23 } 24 int start = 0, end = s.length() - 1; 25 while(start <= end) { 26 if(s.charAt(start) == s.charAt(end)) { 27 palinSeq.add(s.charAt(start)); 28 start++; 29 end--; 30 } 31 else{ 32 if(T[start + 1][end] > T[start][end - 1]) { 33 start++; 34 } 35 else { 36 end--; 37 } 38 } 39 } 40 for(int i = T[0][s.length() - 1] / 2 - 1; i >= 0; i--) { 41 palinSeq.add(palinSeq.get(i)); 42 } 43 return T[0][s.length() - 1]; 44 } 45 }
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