The Longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order. For example, the length of LIS for {10, 22, 9, 33, 21, 50, 41, 60, 80} is 6 and LIS is {10, 22, 33, 50, 60, 80}.
More Examples:
Input : arr[] = {3, 10, 2, 1, 20}
Output : Length of LIS = 3
The longest increasing subsequence is 3, 10, 20
Input : arr[] = {3, 2}
Output : Length of LIS = 1
The longest increasing subsequences are {3} and {2}
Input : arr[] = {50, 3, 10, 7, 40, 80}
Output : Length of LIS = 4
The longest increasing subsequence is {3, 7, 40, 80}
Optimal Substructure:
Let arr[0..n-1] be the input array and L(i) be the length of the LIS ending at index i such that arr[i] is the last element of the LIS.
Then, L(i) can be recursively written as:
L(i) = 1 + max( L(j) ) where 0 < j < i and arr[j] < arr[i]; or
L(i) = 1, if no such j exists.
To find the LIS for a given array, we need to return max(L(i)) where 0 < i < n.
Thus, we see the LIS problem satisfies the optimal substructure property as the main problem can be solved using solutions to subproblems.
Following is a simple recursive implementation of the LIS problem. It follows the recursive structure discussed above.
# A naive Python implementation of LIS problem """ To make use of recursive calls, this function must return two things: 1) Length of LIS ending with element arr[n-1]. We use max_ending_here for this purpose 2) Overall maximum as the LIS may end with an element before arr[n-1] max_ref is used this purpose. The value of LIS of full array of size n is stored in * max_ref which is our final result """ # global variable to store the maximum global maximum def _lis(arr, n ): # to allow the access of global variable global maximum # Base Case if n == 1 : return 1 # maxEndingHere is the length of LIS ending with arr[n-1] maxEndingHere = 1 """Recursively get all LIS ending with arr[0], arr[1]..arr[n-2] IF arr[n-1] is maller than arr[n-1], and max ending with arr[n-1] needs to be updated, then update it""" for i in xrange(1, n): res = _lis(arr, i) if arr[i-1] < arr[n-1] and res + 1 > maxEndingHere: maxEndingHere = res + 1 # Compare maxEndingHere with overall maximum. And # update the overall maximum if needed maximum = max(maximum, maxEndingHere) return maxEndingHere def lis(arr): # to allow the access of global variable global maximum # length of arr n = len(arr) # maximum variable holds the result maximum = 1 # The function _lis() stores its result in maximum _lis(arr, n) return maximum # Driver program to test the above function arr = [10, 22, 9, 33, 21, 50, 41, 60] n = len(arr) print "Length of lis is ", lis(arr) # This code is contributed by NIKHIL KUMAR SINGH |
Length of lis is 5
Overlapping Subproblems:
Considering the above implementation, following is recursion tree for an array of size 4. lis(n) gives us the length of LIS for arr[].
lis(4)
/ |
lis(3) lis(2) lis(1)
/ /
lis(2) lis(1) lis(1)
/
lis(1)
We can see that there are many subproblems which are solved again and again. So this problem has Overlapping Substructure property and recomputation of same subproblems can be avoided by either using Memoization or Tabulation. Following is a tabluated implementation for the LIS problem.
# Dynamic programming Python implementation of LIS problem # lis returns length of the longest increasing subsequence # in arr of size n def lis(arr): n = len(arr) # Declare the list (array) for LIS and initialize LIS # values for all indexes lis = [1]*n # Compute optimized LIS values in bottom up manner for i in range (1, n): for j in range(0, i): if arr[i] > arr[j] and lis[i]< lis[j] + 1 : lis[i] = lis[j]+1 # Initialize maximum to 0 to get the maximum of all # LIS maximum = 0 # Pick maximum of all LIS values for i in range(n): maximum = max(maximum, lis[i]) return maximum # end of lis function # Driver program to test above function arr = [10, 22, 9, 33, 21, 50, 41, 60] print "Length of lis is", lis(arr) # This code is contributed by Nikhil Kumar Singh |
Length of lis is 5
Please refer complete article on Dynamic Programming | Set 3 (Longest Increasing Subsequence) for more details!
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