longest-palindromic-subsequence Algorithm
Therefore, it provides a linear time solution to the longest palindromic substring problem. However, as detected e.g., by Apostolico, Breslauer & Galil (1995), the same algorithm can also be used to find all maximal palindromic substrings anywhere within the input string, again in linear time.
// longest palindromic subsequence
// http://www.geeksforgeeks.org/dynamic-programming-set-12-longest-palindromic-subsequence/
package longestPalindromicSubsequence
// package main
import "fmt"
func max(a, b int) int {
if a > b {
return a
}
return b
}
func lpsRec(word string, i, j int) int {
if i == j {
return 1
}
if i > j {
return 0
}
if word[i] == word[j] {
return 2 + lpsRec(word, i+1, j-1)
}
return max(lpsRec(word, i, j-1), lpsRec(word, i+1, j))
}
func lpsDp(word string) int {
N := len(word)
dp := make([][]int, N)
for i := 0; i < N; i++ {
dp[i] = make([]int, N)
dp[i][i] = 1
}
for l := 2; l <= N; l++ {
// for length l
for i := 0; i < N-l+1; i++ {
j := i + l - 1
if word[i] == word[j] {
if l == 2 {
dp[i][j] = 2
} else {
dp[i][j] = 2 + dp[i+1][j-1]
}
} else {
dp[i][j] = max(dp[i+1][j], dp[i][j-1])
}
}
}
return dp[1][N-1]
}
/*
func main() {
// word := "aaabbbbababbabbabbabf"
word := "aaaabbbba"
fmt.Printf("%d\n", lpsRec(word, 0, len(word)-1))
fmt.Printf("%d\n", lpsDp(word))
}
*/
