implemented Longest Palindromic Subsequence DP

Author: JakubSchwenkbeck

src/main/scala/algorithms/dynamic/Palindrome.scala

@@ -0,0 +1,53 @@
+package algorithms.dynamic
+
+/// Compute the length of the longest palindromic subsequence for a given string.
+/// A subsequence is a sequence derived from the original string by deleting some or no characters
+/// without changing the order of the remaining characters.
+///
+///   - dp(i)(j) represents the length of the longest palindromic subsequence in s(i..j).
+/// The recurrence relation is:
+///   - If s(i) == s(j) then:
+///         dp(i)(j) = dp(i+1)(j-1) + 2
+///         (if the substring length is 2, then dp(i)(j) = 2)
+///   - Otherwise:
+///         dp(i)(j) = max(dp(i+1)(j), dp(i)(j-1))
+/// Base case:
+///   - dp(i)(i) = 1 (each individual character is a palindrome of length 1)
+///
+/// Time Complexity: O(n²) – Each cell of the DP table is computed once.
+
+def longestPalindromicSubsequence(s: String): (Int, Array[Array[Int]]) = {
+  val n: Int = s.length
+  // Initialize DP table with zeros.
+  val dp: Array[Array[Int]] = Array.fill(n, n)(0)
+
+  // Base Case: every single character is a palindrome of length 1.
+  for (i <- 0 until n) {
+    dp(i)(i) = 1
+  }
+
+  // Build the DP table for substrings of length 2 to n.
+  for (len <- 2 to n) {
+    for (i <- 0 to n - len) {
+      val j = i + len - 1
+      if (s(i) == s(j)) {
+        dp(i)(j) = if (len == 2) 2 else dp(i + 1)(j - 1) + 2
+      } else {
+        dp(i)(j) = Math.max(dp(i + 1)(j), dp(i)(j - 1))
+      }
+    }
+  }
+
+  (dp(0)(n - 1), dp)
+}
+
+@main
+def mainLongestPalindromicSubsequence(): Unit = {
+  val inputString: String = "character"
+  val (lpsLength, dpMatrix) = longestPalindromicSubsequence(inputString)
+  println(s"Longest Palindromic Subsequence length for '$inputString' is: $lpsLength")
+  println("DP Matrix:")
+  dpMatrix.foreach { row =>
+    println(row.mkString("  "))
+  }
+}