feat: Add Longest Palindromic Subsequence algorithm in R

string_manipulation/longest.palindromic.subsequence.R

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+# Longest Palindromic Subsequence in R
+# Author: sgindeed
+# Description: Finds and prints the longest palindromic subsequence and its length
+
+# Ask for user input
+input.string <- readline(prompt = "Enter a string: ")
+
+# Convert string to lowercase for consistency
+clean.string <- tolower(input.string)
+
+# Get length of string
+n <- nchar(clean.string)
+
+# Split string into characters
+chars <- strsplit(clean.string, "")[[1]]
+
+# Initialize DP table for lengths
+dp <- matrix(0, nrow = n, ncol = n)
+
+# Each single character is a palindrome of length 1
+for (i in seq_len(n)) {
+  dp[i, i] <- 1
+}
+
+# Fill the DP table
+for (cl in 2:n) {
+  for (i in 1:(n - cl + 1)) {
+    j <- i + cl - 1
+    if (chars[i] == chars[j] && cl == 2) {
+      dp[i, j] <- 2
+    } else if (chars[i] == chars[j]) {
+      dp[i, j] <- dp[i + 1, j - 1] + 2
+    } else {
+      dp[i, j] <- max(dp[i + 1, j], dp[i, j - 1])
+    }
+  }
+}
+
+# Function to reconstruct the subsequence
+reconstructLPS <- function(chars, dp, i, j) {
+  if (i > j) {
+    return("")
+  }
+  if (i == j) {
+    return(chars[i])
+  }
+  if (chars[i] == chars[j]) {
+    return(paste0(chars[i], reconstructLPS(chars, dp, i + 1, j - 1), chars[j]))
+  }
+  if (dp[i + 1, j] > dp[i, j - 1]) {
+    return(reconstructLPS(chars, dp, i + 1, j))
+  } else {
+    return(reconstructLPS(chars, dp, i, j - 1))
+  }
+}
+
+# Get the longest palindromic subsequence
+lps <- reconstructLPS(chars, dp, 1, n)
+
+# Display the result
+cat("Longest Palindromic Subsequence:", lps, "\n")
+cat("Length:", nchar(lps), "\n")