From 4e7045137cfed30e99201b1d52eafdf26ee382d2 Mon Sep 17 00:00:00 2001
From: Alon Firestein <57404551+alonfirestein@users.noreply.github.com>
Date: Wed, 24 Mar 2021 18:51:20 +0200
Subject: [PATCH] Update LongestIncreasingSubsequence.java

---
 .../LongestIncreasingSubsequence.java         | 44 +++++++++++++++++--
 1 file changed, 41 insertions(+), 3 deletions(-)

diff --git a/DynamicProgramming/LongestIncreasingSubsequence.java b/DynamicProgramming/LongestIncreasingSubsequence.java
index 06885fbdd..28f54afdc 100644
--- a/DynamicProgramming/LongestIncreasingSubsequence.java
+++ b/DynamicProgramming/LongestIncreasingSubsequence.java
@@ -9,13 +9,15 @@ public class LongestIncreasingSubsequence {
     Scanner sc = new Scanner(System.in);
     int n = sc.nextInt();
 
-    int ar[] = new int[n];
+    int arr[] = new int[n];
     for (int i = 0; i < n; i++) {
-      ar[i] = sc.nextInt();
+      arr[i] = sc.nextInt();
     }
 
-    System.out.println(LIS(ar));
+    System.out.println(LIS(arr));
+    System.out.println(findLISLen(arr));
     sc.close();
+    
   }
 
   private static int upperBound(int[] ar, int l, int r, int key) {
@@ -56,4 +58,40 @@ public class LongestIncreasingSubsequence {
 
     return length;
   }
+  
+    /** @author Alon Firestein (https://github.com/alonfirestein) */
+  
+    // A function for finding the length of the LIS algorithm in O(nlogn) complexity.  
+  	public static int findLISLen(int a[]) { 
+		int size = a.length;
+		int arr[] = new int[size];
+		arr[0] = a[0];
+		int lis = 1;
+		for (int i = 1; i < size; i++) {
+			int index = binarySearchBetween(arr, lis, a[i]);
+			arr[index] = a[i];
+			if (index > lis)
+				lis++;
+		}
+		return lis;
+	}
+  // O(logn)
+	private static int binarySearchBetween(int[] t, int end, int key) { 
+		int left = 0;
+		int right = end;
+		if (key < t[0])
+			return 0;
+		if (key > t[end])
+			return end + 1;
+		while (left < right - 1) {
+			int middle = (left + right) / 2;
+			if (t[middle] < key)
+				left = middle;
+			else
+				right = middle;
+		}
+		return right;
+	}
+  
+  
 }