Script Output: The entire longest increasing subsequence instead of it's length

dynamic_programming/longest_increasing_subsequence.py

@@ -1,12 +1,41 @@
-"""
-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 
-"""
-def LIS(arr):
-    n= len(arr)
-    lis = [1]*n
-
-    for i in range(1, n):
-        for j in range(0, i):
-            if arr[i] > arr[j] and lis[i] <= lis[j]:
-                lis[i] = lis[j] + 1
-    return max(lis)
+'''
+Author  : Mehdi ALAOUI
+
+This is a pure Python implementation of Dynamic Programming solution to the longest increasing subsequence of a given sequence.
+
+The problem is  :
+Given an ARRAY, to find the longest and increasing sub ARRAY in that given ARRAY and return it.
+Example: [10, 22, 9, 33, 21, 50, 41, 60, 80] as input will return [10, 22, 33, 41, 60, 80] as output
+'''
+
+def longestSub(ARRAY): 			#This function is recursive
+
+	ARRAY_LENGTH = len(ARRAY)
+	if(ARRAY_LENGTH <= 1):  	#If the array contains only one element, we return it (it's the stop condition of recursion)
+		return ARRAY
+								#Else
+	PIVOT=ARRAY[0]
+	isFound=False
+	i=1
+	LONGEST_SUB=[]
+	while(not isFound and i<ARRAY_LENGTH):
+		if (ARRAY[i] < PIVOT):
+			isFound=True
+			TEMPORARY_ARRAY = [ element for element in ARRAY[i:] if element >= ARRAY[i] ]
+			TEMPORARY_ARRAY = longestSub(TEMPORARY_ARRAY)
+			if ( len(TEMPORARY_ARRAY) > len(LONGEST_SUB) ):
+				LONGEST_SUB = TEMPORARY_ARRAY
+		else:
+			i+=1
+
+	TEMPORARY_ARRAY = [ element for element in ARRAY[1:] if element >= PIVOT ]
+	TEMPORARY_ARRAY = [PIVOT] + longestSub(TEMPORARY_ARRAY)
+	if ( len(TEMPORARY_ARRAY) > len(LONGEST_SUB) ):
+		return TEMPORARY_ARRAY
+	else:
+		return LONGEST_SUB
+
+#Some examples
+
+print(longestSub([4,8,7,5,1,12,2,3,9]))
+print(longestSub([9,8,7,6,5,7]))