Added LCIS dynamic programming code. Feels good to code after all this time

Author: shiv

LCIS.py

@@ -0,0 +1,51 @@
+# SHIV's code
+# Longest Common Increasing Subsequence
+
+# Testing arrays
+a1 = [3,4,9,1]
+a2 = [5,3,8,9,10,2,1]
+a3 = [50, 3, 10, 7, 40, 80]
+a4 = [3, 10, 2, 1, 20]
+a5 = "AGGTAB"
+a6 = "GXTXAYB"
+
+# find the LIS for any list
+def LIS(a):
+	if len(a) < 1: return 1
+	least_index = 0
+	for ind, i in reversed(list(enumerate(a))):
+		#print(i, a[-1])
+		if i < a[-1]:
+			least_index=ind
+			break
+	return max(LIS(a[:-1]),1+LIS(a[:least_index]))
+
+# This func will now find LCS
+# Also a very nice way to see how to retrieve the elements of the efficient elements
+def LCS(a, b):
+	try:
+		if len(a) <1: return 0, []
+		if a[-1] == b[-1]: 
+			#cost, arr = LCS(a[:-1], b[:-1])
+			#print(arr, a[-1], cost)
+			cost, arr = LCS(a[:-1], b[:-1])
+			arr.append(a[-1])
+			return 1+cost, arr
+		else:
+			cost1, arr1 = LCS(a[:-1], b)
+			cost2, arr2 = LCS(a, b[:-1])
+			costs = [cost1, cost2]
+			arrs = [arr1, arr2]
+			ind = costs.index(max(costs))
+			return costs[ind], arrs[ind]
+	except:
+		return 0, []
+
+# This bro finally finds the LCIS
+def LCIS(a, b):
+	l, x = LCS(a, b)
+	cost = LIS(x)
+	return cost
+ 
+# Yayyyy!
+print(LCIS(a1, a2))