# Longest Increasing Subsequenceby Isai Damier, Android Engineer @ Google

```/***********************************************************************
* Author: Isai Damier
* Title: Longest Increasing Subsequence
* Project: geekviewpoint
* Package: algorithms
*
* Statement:
*   Given a sequence of numbers, find a longest increasing subsequence.
*
*
*  Time Complexity: O(n^2)
*
* Sample Input: {8,1,2,3,0,5}
* Sample Output: {1,2,3,5}
*
* DEFINITION OF SUBSEQUENCE:
*   A sequence is a particular order in which related objects follow
*   each other (e.g. DNA, Fibonacci). A sub-sequence is a sequence
*   obtained by omitting some of the elements of a larger sequence.
*
*   SEQUENCE       SUBSEQUENCE     OMISSION
*   {3,1,2,5,4}     {1,2}            3,5,4
*   {3,1,2,5,4}     {3,1,4}          2,5
*
*   SEQUENCE       NOT SUBSEQUENCE   REASON
*   {3,1,2,5,4}     {4,2,5}           4 should follow 5
*
* STRATEGY:
*   Illustrating by finding a longest increasing subsequence
*   of {8,1,2,3,0,5}:
*
*   - Start by finding all subsequences of size 1: {8},{1},{2},{3},{0},{5};
*     each element is its own increasing subsequence.
*
*   - Since we already have the solutions for the size 1 subsequences,
*     we can use them in solving for the size two subsequences. For
*     instance, we already know that 0 is the smallest element of an
*     increasing subsequence of size 1, i.e. the subsequence {0}.
*     Therefore, all we need to get a subsequence of size 2 is add an
*     element greater than 0 to {0}: {0,5}. The other size 2
*     subsequences are: {1,2}, {1,3}, {1,5}, {2,3}, {2,5}, {3,5}.
*
*   - Now we use the size 2 subsequences to get the size 3 subsequences:
*     {1,2,3}, {1,2,5}, {1,3,5}, {2,3,5}
*
*   - Then we use the size 3 subsequences to get the size 4 subsequences:
*     {1,2,3,5}. Since there are no size 5 solutions, we are done.
*
* SUMMARY:
*   Instead of directly solving the big problem, we solved a smaller
*   version and then 'copied and pasted' the solution of the subproblem
*   to find the solution to the big problem. To make the 'copy and paste'
*   part easy, we use a table (i.e. array) to track the subproblems
*   and their solutions. This strategy as a whole is called Dynamic
*   Programming. The tabling part is known as memoization, which means
*   writing memo.
*
*   To recognize whether you can use dynamic programming on a problem,
*   look for the following two traits: optimal substructures and
*   overlapping subproblems.
*
*   Optimal Substructures: the ability to 'copy and paste' the solution
*     of a subproblem plus an additional trivial amount of work so to
*     solve a larger problem. For example, we were able to use {1,2}
*     itself an optimal solution to the problem {8,1,2} to get {1,2,3}
*     as an optimal solution to the problem {8,1,2,3}.
*
*   Overlapping Subproblems: Okay. So in our approach the solution grew
*     from left to right: {1} to {1,2} to {1,2,3} etc. But in reality
*     we could have solved the problem using recursion trees so that
*     for example {1,2} could be reached either from {1} or from {2}.
*     That wouldn't really be a problem except we would be solving for
*     {1,2} more than once. Anytime a recursive solution would lead to
*     such overlaps, the bet is dynamic programming is the way to go.
*
*          {1}                 {2}
*         / | \               / | \
*        /  |  \             /  |  \
*       /   |   \           /   |   \
*   {1,2} {1,3} {1,5}   {1,2} {2,3} {2,5}
*
* NOTE:
* Dynamic Programming = Optimal Substructures + Overlapping Subproblems
* Divide and Conquer = Optimal Substructures - Overlapping Subproblems
*   see merge sort: http://www.geekviewpoint.com/java/sorting/mergesort
*
* Alternate coding: Not really much difference here, just another code
*   that some readers will find more intuitive:
*
*   int[] m = new int[A.length];
*   Arrays.fill(m, 1);
*
*   for(int x=1; x <A.length; x++)
*     for(int y = 0; y < x; y++)
*       if(m[y] >= m[x] && A[y] < A[x])
*         m[x]++;
*
*   int max = m;
*   for (int i = 0; i < m.length; i++)
*     if (max < m[i])
*       max = m[i];
*
*   List<Integer> result = new ArrayList<Integer>();
*   for (int i = m.length-1; i >=0 ; i--)
*     if (max == m[i]) {
*       max--;
*     }
*
*   Collections.reverse(result);
*   return result;
*
**********************************************************************/
package algorithm.programming.dynamic;

import java.util.ArrayList;
import java.util.List;

public class LongestIncreasingSubsequence {

public Integer[] LIS(Integer[] A) {
int[] m = new int[A.length];
//Arrays.fill(m, 1);//not important here
for (int x = A.length - 2; x >= 0; x--) {
for (int y = A.length - 1; y > x; y--) {
if (A[x] < A[y] && m[x] <= m[y]) {
m[x]++;//or use m[x] = m[y] + 1;
}
}
}
int max = m;
for (int i = 1; i < m.length; i++) {
if (max < m[i]) {
max = m[i];
}
}
List<Integer> result = new ArrayList<Integer>();
for (int i = 0; i < m.length; i++) {
if (m[i] == max) {
max--;
}
}
return result.toArray(new Integer);
}
}
```
```package algorithm.programming.dynamic;

import java.util.Arrays;
import static org.junit.Assert.*;

import org.junit.After;
import org.junit.Before;
import org.junit.Test;

public class LongestIncreasingSubsequenceTest {

LongestIncreasingSubsequence seq;

@Before
public void setUp() throws Exception {
seq = new LongestIncreasingSubsequence();
}

@After
public void tearDown() throws Exception {
seq = null;
}

@Test
public void testLIS() {
Integer[] A = {1, 2, 3, 4, 5, 6, 7, 8, 9};
Integer[] B = {9, 8, 7, 6, 5, 4, 3, 2, 1};
Integer[] C = {50, 2, 45, 3, 5, 44, 1, 8, 40, 12, 37, 15};
Integer[] D = {2, 3, 5, 8, 12, 37};

Integer[] actual = seq.LIS(A);
assertTrue(Arrays.equals(A, actual));

assertEquals(1, seq.LIS(B).length);

actual = seq.LIS(C);
assertTrue(Arrays.equals(D, actual));
}
}```