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.

OR

Given a array A[1,2,……,n] , calculate B[1,2….m] with B[i]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 i-1 {LIS(j)} if A[i]>A[j] for 1

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 (j-1)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

Code:

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[] =

Longest Increasing Subsequence - track the result

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If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.
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