Longest_Increasing_Subsequence.py

# 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[i-1] is smaller than arr[n-1], and max ending with
	arr[n-1] needs to be updated, then update it"""
	for i in range(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