Given an array of random numbers. Find longest monotonically increasingsubsequence (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 monotone 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 all end candidates of active lists, we will start new active list of length 1.
2. If A[i] is largest among all end candidates of active lists, we will clone the largestactive list, and extend it by A[i].
3. If A[i] is in between, we will find a list with largest end element that is smaller than A[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. 
Given below is code to find length of LIS,
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
#define ARRAY_SIZE(A) sizeof(A)/sizeof(A[0])
int CeilIndex(int A[], int l, int r, int key) {
int m;
while( r - l > 1 ) {
m = l + (r - l)/2;
(A[m] >= key ? r : l) = m;
}
return r;
}
int LongestIncreasingSubsequenceLength(int A[], int size) {
int *tailTable = new int[size];
int len;
memset(tailTable, 0, sizeof(tailTable[0])*size);
tailTable[0] = A[0];
len = 1;
for( int i = 1; i < size; i++ ) {
if( A[i] < tailTable[0] )
tailTable[0] = A[i];
else if( A[i] > tailTable[len-1] )
tailTable[len++] = A[i];
else
tailTable[CeilIndex(tailTable, -1, len-1, A[i])] = A[i];
}
delete[] tailTable;
return len;
}
int main() {
int A[] = { 2, 5, 3, 7, 11, 8, 10, 13, 6 };
int n = ARRAY_SIZE(A);
printf("Length of Longest Increasing Subsequence is %d\n",
LongestIncreasingSubsequenceLength(A, n));
return 0;
}
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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).
This post has been taken from http://www.geeksforgeeks.org/longest-monotonically-increasing-subsequence-size-n-log-n/.. I happened to encounter this post after going through a LIS question.
Veni,vedi,vici 