Merge pull request matthewsamuel95#549 from 96mohitm/dp-longest-increasing-subsequence-python

longest common subsequence added in python.

Author: matthewsamuel95

DP/LongestIncreasingSubsequence/LIC.py

@@ -0,0 +1,60 @@
+# 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 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 
+
+    # lenght 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))