@@ -9,7 +9,7 @@
 
 =========================================
 Need only one iteration, in each step add the current element to the current sum.
-When the sum is equal or less than 0, reset the sum to 0 and continue with adding. (we care only about the positive sums)
+When the sum is less than 0, reset the sum to 0 and continue with adding. (we care only about non-negative sums)
 After each addition, check if the current sum is greater than the max sum. (Called Kadane's algorithm)
     Time Complexity:    O(N)
     Space Complexity:   O(1)
@@ -21,20 +21,17 @@
 ############
 
 def max_subarray_sum(a):
-    i = 0
-    n = len(a)
     curr_sum = 0
     max_sum = 0
 
-    while i < n:
-        # restart the current sum if 0 or smaller, we care only about the positive sums
-        curr_sum = max(0, curr_sum + a[i])
+    for val in a:
+        # extend the current sum with the curren value; 
+        # reset it to 0 if it is smaller than 0, we care only about non-negative sums
+        curr_sum = max(0, curr_sum + val)
 
         # check if this is the max sum
         max_sum = max(max_sum, curr_sum)
 
-        i += 1
-
     return max_sum
 
 
@@ -52,4 +49,8 @@ def max_subarray_sum(a):
 
 # Test 3
 # Correct result => 7
-print(max_subarray_sum([-2, -5, 6, -2, -3, 1, 5, -6]))
+print(max_subarray_sum([-2, -5, 6, -2, -3, 1, 5, -6]))
+
+# Test 4
+# Correct result => 0
+print(max_subarray_sum([-6, -1]))