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160. Dynamic Programming  Longest Increasing Subsequence
Objective: The longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence in a given array such that all elements of the subsequence are sorted in increasing order.
Example:
int [] A = { 1, 12, 7, 0, 23, 11, 52, 31, 61, 69, 70, 2 }; length of LIS is 7 and LIS is {1, 12, 23, 52, 61, 69, 70}.
Approach:
Optimal Substructure:
LIS(i)  Length of longest increasing subsequence which includes element A[i] as its last element.
LIS(i) = 1 + Max _{j=1 to i1} {LIS(j)} if A[i]>A[j] for 1<j</j
Overlapping Subproblems:
If we solve using recursion for calculating the solution for jth index we will be solving the subproblems again which we had solved earlier while solving the solution for (j1)th index.
Example: A[] = {3, 4, 1, 5} i=1 , LIS(1) = 1 i=2 , LIS(2) = 1+ Max(LIS(1)) = 1 +1 =2 (4>3) i=3 , LIS(3) = 1 (1<3, 1<4) i=4 , LIS(4) = 1+ Max(LIS(1),LIS(2), LIS(3)) = 1 + Max(1,2,1) = 3
Output:
Output: Longest Increasing subsequence: 7 Actual Elements: 1 7 11 31 61 69 70
NOTE: To print the Actual elements 
 find the index which contains the longest sequence, print that index from main array.
 Start moving backwards and pick all the indexes which are in sequence (descending).
 If longest sequence for more than one indexes, pick any one.
From our code LIS[] =