After few months of gap posting an algo. The current post is pending from long time, and many readers (e.g. here, here, here may be few more, I am not keeping track of all) are posting requests for explanation of the below problem.

**Given an array of random numbers. Find longest increasing subsequence (LIS) in the array. I know many of you might have read recursive and dynamic programming (DP) solutions. There are few requests for O(N log N) algo in the forum posts.**

For the time being, forget about recursive and DP solutions. Let us take small samples and extend the solution to large instances. Even though it may look complex at first time, once if we understood the logic, coding is simple.

Consider an input array A = {2, 5, 3}. I will extend the array during explanation.

By observation we know that the LIS is either {2, 3} or {2, 5}. * Note that I am considering only strictly increasing sequences*.

Let us add two more elements, say 7, 11 to the array. These elements will extend the existing sequences. Now the increasing sequences are {2, 3, 7, 11} and {2, 5, 7, 11} for the input array {2, 5, 3, 7, 11}.

Further, we add one more element, say 8 to the array i.e. input array becomes {2, 5, 3, 7, 11, 8}. Note that the latest element 8 is greater than smallest element of any active sequence (*will discuss shortly about active sequences*). How can we extend the existing sequences with 8? First of all, can 8 be part of LIS? If yes, how? If we want to add 8, it should come after 7 (by replacing 11).

Since the approach is *offline (what we mean by offline?)*, we are not sure whether adding 8 will extend the series or not. Assume there is 9 in the input array, say {2, 5, 3, 7, 11, 8, 7, 9 …}. We can replace 11 with 8, as there is potentially *best* candidate (9) that can extend the new series {2, 3, 7, 8} or {2, 5, 7, 8}.

Our observation is, assume that the end element of largest sequence is E. We can add (replace) current element A[i] to the existing sequence if there is an element A[j] (j > i) such that E < A[i] < A[j] or (E > A[i] < A[j] – for replace). In the above example, E = 11, A[i] = 8 and A[j] = 9.

In case of our original array {2, 5, 3}, note that we face same situation when we are adding 3 to increasing sequence {2, 5}. I just created two increasing sequences to make explanation simple. Instead of two sequences, 3 can replace 5 in the sequence {2, 5}.

I know it will be confusing, I will clear it shortly!

*The question is, when will it be safe to add or replace an element in the existing sequence?*

Let us consider another sample A = {2, 5, 3}. Say, the next element is 1. How can it extend the current sequences {2,3} or {2, 5}. Obviously, it can’t extend either. Yet, there is a potential that the new smallest element can be start of an LIS. To make it clear, consider the array is {2, 5, 3, 1, 2, 3, 4, 5, 6}. Making 1 as new sequence will create new sequence which is largest.

*The observation is, when we encounter new smallest element in the array, it can be a potential candidate to start new sequence.*

From the observations, we need to maintain lists of increasing sequences.

In general, we have set of **active lists** of varying length. We are adding an element A[i] to these lists. We scan the lists (for end elements) in decreasing order of their length. We will verify the end elements of all the lists to find a list whose end element is smaller than A[i] (*floor* value).

Our strategy determined by the following conditions,

1. If A[i] is smallest among allendcandidates of active lists, we willstartnew active list of length 1.

2. If A[i] is largest among allendcandidates of active lists, we will clone thelargestactive list, and extend it by A[i].

3. If A[i] is in between, we will find a list withlargest end element that is smaller thanA[i]. Clone and extend this list by A[i]. We will discard all other lists of same length as that of this modified list.

Note that at any instance during our construction of active lists, the following condition is maintained.

*“end element of smaller list is smaller than end elements of larger lists”*.

It will be clear with an example, let us take example from wiki {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15}.

A[0] = 0. Case 1. There are no active lists, create one. 0. ----------------------------------------------------------------------------- A[1] = 8. Case 2. Clone and extend. 0. 0, 8. ----------------------------------------------------------------------------- A[2] = 4. Case 3. Clone, extend and discard. 0. 0, 4. 0, 8. Discarded ----------------------------------------------------------------------------- A[3] = 12. Case 2. Clone and extend. 0. 0, 4. 0, 4, 12. ----------------------------------------------------------------------------- A[4] = 2. Case 3. Clone, extend and discard. 0. 0, 2. 0, 4. Discarded. 0, 4, 12. ----------------------------------------------------------------------------- A[5] = 10. Case 3. Clone, extend and discard. 0. 0, 2. 0, 2, 10. 0, 4, 12. Discarded. ----------------------------------------------------------------------------- A[6] = 6. Case 3. Clone, extend and discard. 0. 0, 2. 0, 2, 6. 0, 2, 10. Discarded. ----------------------------------------------------------------------------- A[7] = 14. Case 2. Clone and extend. 0. 0, 2. 0, 2, 6. 0, 2, 6, 14. ----------------------------------------------------------------------------- A[8] = 1. Case 3. Clone, extend and discard. 0. 0, 1. 0, 2. Discarded. 0, 2, 6. 0, 2, 6, 14. ----------------------------------------------------------------------------- A[9] = 9. Case 3. Clone, extend and discard. 0. 0, 1. 0, 2, 6. 0, 2, 6, 9. 0, 2, 6, 14. Discarded. ----------------------------------------------------------------------------- A[10] = 5. Case 3. Clone, extend and discard. 0. 0, 1. 0, 1, 5. 0, 2, 6. Discarded. 0, 2, 6, 9. ----------------------------------------------------------------------------- A[11] = 13. Case 2. Clone and extend. 0. 0, 1. 0, 1, 5. 0, 2, 6, 9. 0, 2, 6, 9, 13. ----------------------------------------------------------------------------- A[12] = 3. Case 3. Clone, extend and discard. 0. 0, 1. 0, 1, 3. 0, 1, 5. Discarded. 0, 2, 6, 9. 0, 2, 6, 9, 13. ----------------------------------------------------------------------------- A[13] = 11. Case 3. Clone, extend and discard. 0. 0, 1. 0, 1, 3. 0, 2, 6, 9. 0, 2, 6, 9, 11. 0, 2, 6, 9, 13. Discarded. ----------------------------------------------------------------------------- A[14] = 7. Case 3. Clone, extend and discard. 0. 0, 1. 0, 1, 3. 0, 1, 3, 7. 0, 2, 6, 9. Discarded. 0, 2, 6, 9, 11. ---------------------------------------------------------------------------- A[15] = 15. Case 2. Clone and extend. 0. 0, 1. 0, 1, 3. 0, 1, 3, 7. 0, 2, 6, 9, 11.0, 2, 6, 9, 11, 15. <-- LIS List----------------------------------------------------------------------------

It is required to understand above strategy to devise an algorithm. Also, ensure we have maintained the condition, “*end element of smaller list is smaller than end elements of larger lists*“. Try with few other examples, before reading further. It is important to understand what happening to end elements.

**Algorithm:**

Querying length of longest is fairly easy. Note that we are dealing with end elements only. We need not to maintain all the lists. We can store the end elements in an array. Discarding operation can be simulated with replacement, and extending a list is analogous to adding more elements to array.

We will use an auxiliary array to keep end elements. The maximum length of this array is that of input. In the worst case the array divided into N lists of size one (*note that it does’t lead to worst case complexity*). To discard an element, we will trace ceil value of A[i] in auxiliary array (again observe the end elements in your rough work), and replace ceil value with A[i]. We extend a list by adding element to auxiliary array. We also maintain a counter to keep track of auxiliary array length.

**Bonus:** You have learnt Patience Sorting technique partially 🙂

Here is a proverb, “*Tell me and I will forget. Show me and I will remember. Involve me and I will understand*.” So, pick a suit from deck of cards. Find the longest increasing sub-sequence of cards from the shuffled suit. You will never forget the approach. 🙂

**Update – 17 July, 2016**: Quite impressive reponses from the readers and few sites referring the post, feeling happy as my hardwork helping others. It looks like readers are not doing any homework prior to posting comments. Requesting to run through some examples after reading the article, and please do your work on paper (don’t use editor/compiler). The request is to help yourself. Profess to ‘know’ is different from real understanding (no disrespect). Given below was my personal experience.

*Initial content preparation took roughly 6 hours to me. But, it was a good lesson. I finished initial code in an hour. When I start writing content to explain the reader, I realized I didn’t understand the cases. Took my note book (I have habit of maintaining binded note book to keep track of my rough work), and after few hours I filled nearly 15 pages of rough work. Whatever the content you are seeing in the gray colored example is from these pages. All the thought process for the solution triggered by a note in the book ‘Introduction to Algorithms by Udi Manber’, I strongly recommend to practice the book.*

I suspect, many readers might not get the logic behind CeilIndex (binary search). I leave it as an exercise to the reader to understand how it works. Run through few examples on paper. I realized I have already covered the algorithm in another post.

**Update – 5th August, 2016**:

The following link worth referring after you do your work. I got to know the link via my recently created **Disqus** profile. The link has explanation of approach mentioned in the Wiki.

Given below is code to find length of LIS (*updated to C++11 code, no C-style arrays*),

## C++

#include <iostream> #include <vector> // Binary search (note boundaries in the caller) int CeilIndex(std::vector<int> &v, int l, int r, int key) { while (r-l > 1) { int m = l + (r-l)/2; if (v[m] >= key) r = m; else l = m; } return r; } int LongestIncreasingSubsequenceLength(std::vector<int> &v) { if (v.size() == 0) return 0; std::vector<int> tail(v.size(), 0); int length = 1; // always points empty slot in tail tail[0] = v[0]; for (size_t i = 1; i < v.size(); i++) { if (v[i] < tail[0]) // new smallest value tail[0] = v[i]; else if (v[i] > tail[length-1]) // v[i] extends largest subsequence tail[length++] = v[i]; else // v[i] will become end candidate of an existing subsequence or // Throw away larger elements in all LIS, to make room for upcoming grater elements than v[i] // (and also, v[i] would have already appeared in one of LIS, identify the location and replace it) tail[CeilIndex(tail, -1, length-1, v[i])] = v[i]; } return length; } int main() { std::vector<int> v{ 2, 5, 3, 7, 11, 8, 10, 13, 6 }; std::cout << "Length of Longest Increasing Subsequence is " << LongestIncreasingSubsequenceLength(v) << '\n'; return 0; }

## Java

// Java program to find length of longest increasing subsequence // in O(n Log n) time import java.io.*; import java.util.*; import java.lang.Math; class LIS { // Binary search (note boundaries in the caller) // A[] is ceilIndex in the caller static int CeilIndex(int A[], int l, int r, int key) { while (r - l > 1) { int m = l + (r - l)/2; if (A[m]>=key) r = m; else l = m; } return r; } static int LongestIncreasingSubsequenceLength(int A[], int size) { // Add boundary case, when array size is one int[] tailTable = new int[size]; int len; // always points empty slot tailTable[0] = A[0]; len = 1; for (int i = 1; i < size; i++) { if (A[i] < tailTable[0]) // new smallest value tailTable[0] = A[i]; else if (A[i] > tailTable[len-1]) // A[i] wants to extend largest subsequence tailTable[len++] = A[i]; else // A[i] wants to be current end candidate of an existing // subsequence. It will replace ceil value in tailTable tailTable[CeilIndex(tailTable, -1, len-1, A[i])] = A[i]; } return len; } // Driver program to test above function public static void main(String[] args) { int A[] = { 2, 5, 3, 7, 11, 8, 10, 13, 6 }; int n = A.length; System.out.println("Length of Longest Increasing Subsequence is "+ LongestIncreasingSubsequenceLength(A, n)); } } /* This code is contributed by Devesh Agrawal*/

Output:

Length of Longest Increasing Subsequence is 6

**Complexity:**

The loop runs for N elements. In the worst case (what is worst case input?), we may end up querying ceil value using binary search (log *i*) for many A[i].

Therefore, T(n) < O( log N! ) = O(N log N). Analyse to ensure that the upper and lower bounds are also O( N log N ). The complexity is THETA (N log N).

**Exercises:**

1. Design an algorithm to construct the longest increasing list. Also, model your solution using DAGs.

2. Design an algorithm to construct **all** increasing list**s of equal longest size**.

3. Is the above algorithm an *online* algorithm?

4. Design an algorithm to construct the longest *decreasing* list..

— Venki. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.