Added forward and backward loop approach for 152 Maximum Product Sub-array

Author: mahesh-3pillar

README.md

@@ -7,25 +7,39 @@ https://sebinsua.com/assets/posts/algorithmic-bathwater/algorithmic-bathwater-1.
 # 152 Maximum Product Sub-array
 
 Intuition: 
-1. If array can have at most one negative, we can take same approach as #053 Maximum subarray
+1. If array can have at most one negative, we can take same approach as Kadane's Algo in #053 Maximum subarray
 2. When there are more then one negative number in sub-array,
    1. Positive number multiplication with negative number will make it negative
    2. Negative number multiplication with negative number will make it positive 
-   3. In that case need to tack, calculate and get  min and max both for each element 
+3. There are two approach
+   1. First find the max product via kadane's algo form left to right and then max from right to left
+      1. return the max of both
+   2. track min and max 
 
+Approach #1 : Loop from both side in Kadane algo
+1. Loop from left to right 
+   1. find currProduct of  currMax * n
+   2. find max of currProduct and currMax
+   3. if currProduct ==0, rest to 1
+2. Loop from right to left and repeat above steps
+3. compare the max of both the loop and return the max
 
-Approach 
+
+Approach #2 : Track max and min in Kadane algo
 1. track max, min. Initialize it with 1
 2. track result and init with nums[0]
 3. for each number in array
-   1. store temp  = max * n // `because max will change after calculating the max`
+   1. store temp  = max // `because max will change after calculating the max`
    2. calculate max via max(max *n, min *n, n)
    3. calculate min via min(temp * n, min * n, n)
    4. calculate res  = max(res, max)
 4. return res
 
 
 
+
+
+
 # 238. Product of Array Except Self
 Intuition: at every element, product to array except self will be the product of all the element before that element and product of all the element after that element.
 

leetcode-py/152-maximum-product-sub-array-1.py

@@ -0,0 +1,23 @@
+class Solution:
+    def maxProduct(self, nums: List[int]) -> int:
+        
+        result  = float('-inf')
+        
+        prod = 1
+        for n in nums:
+            prod *=  n
+            result = max(result, prod)
+            if prod == 0:
+                prod = 1
+        
+        prod = 1
+        for i in range(len(nums)-1, -1, -1):
+            
+            prod *= nums[i]
+            result = max(result, prod)
+            if prod == 0:
+                prod = 1
+
+        return result
+
+