Merge pull request #202 from Uchiha-Itachi0/Adding-solution-to-450-DS…

…A-Array-in-python

Adding solution to 450 dsa array in python

1]. DSA/3]. 450 DSA by ( Love Babbar Bhaiya )/Python/01]. Array/_08_i_)_Kadane's_Algorithm.py

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+# Find the contiguous sub-array(containing at least one number) which has the maximum sum
+"""
+EXAMPLE:
+    INPUT : [1, 2, 3, -2, 5]
+    OUTPUT : 9
+    EXPLANATION :
+    Because sub-array (1,2,3,-2,5) has maximum sum among all sub-array.
+    For example : sub-array (1,2,3) has sum 6
+                  sub-array (1,2,3,-2) has sum 4
+                  sub-array (3,-2,5) has sum 6
+                  and so on..................
+                  Final max sum will be 9 and hence we return it
+"""
+
+"""
+------------------------------IMPORTANT NOTE----------------------------------------
+
+This algorithm gives the result without updating the actual array.
+So if constraind is given that you don't have to update the array then this is the algorithm you are looking for
+"""
+
+"""
+------------------------------EXPLANATION----------------------------------------
+
+The main idea is that we create 2 variable current_max and max_so_far
+we keep adding all the array element in current_max if at any point current_max
+became grater than max_so_far we change the value of max_so_far to current_max value
+Also if at any point current_so far became negative we reassigned it to 0
+
+For example 
+arr = [-2, 3, 2, -2]
+current_max = 0
+max_so_far = -inf   (-infinite)  
+Now we loop through arr and keep adding arr value to current_max
+
+-----------1st iteration----------
+current_max = -2, maximum_so_far = -inf   (current_max is negative and greater than max_so_far)
+Therefore current_max = 0, maximum_so_far = -2
+
+-----------2nd iteration----------
+current_max = 3 , maximum_so_far = 2   (current_max is positive and its value is > maximum_so_far)
+Therefore maximum_so_far = 3  
+
+-----------3rd iteration----------
+current_max = 5, maximum_so_far = 3
+Therefore maximum_so_far = 5
+
+-----------4th iteration----------
+current_max = 3, maximum_so_far = 5
+Since current_max < maximum_so_far therefore maximum_so_far = 5
+
+At last we return maximum_so_far    
+"""
+
+
+# Implementation of above approach
+
+def max_sub_array(arr):
+    """
+    Time Complexity : O(n)
+    Space Complexity : O(1)
+    """
+    current_max = 0
+    maximum_so_far = float('-inf')
+    for i in arr:
+        current_max += i
+        if current_max > maximum_so_far:
+            maximum_so_far = current_max
+        if current_max < 0:
+            current_max = 0
+
+    return maximum_so_far
+
+
+print(max_sub_array([-2, 3, 2, -2]))