1+ '''
2+ 238. Product of Array Except Self
3+
4+ Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i].
5+
6+ The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.
7+
8+ You must write an algorithm that runs in O(n) time and without using the division operation.
9+
10+
11+
12+ Example 1:
13+
14+ Input: nums = [1,2,3,4]
15+ Output: [24,12,8,6]
16+ Example 2:
17+
18+ Input: nums = [-1,1,0,-3,3]
19+ Output: [0,0,9,0,0]
20+
21+
22+ Constraints:
23+
24+ 2 <= nums.length <= 105
25+ -30 <= nums[i] <= 30
26+ The input is generated such that answer[i] is guaranteed to fit in a 32-bit integer.
27+
28+
29+ Follow up: Can you solve the problem in O(1) extra space complexity? (The output array does not count as extra space for space complexity analysis.)
30+ '''
31+ 32+ # Brute Force (Inbuilt math function approach)
33+ # Time Complexity: O(n^2)
34+ # Space Complexity: O(n)
35+ 36+ import math
37+ 38+ class Solution :
39+ def productExceptSelf (self , nums ):
40+ result = []
41+ 42+ for i in range (len (nums )):
43+ new_arr = nums [:i ] + nums [i + 1 :]
44+ res = math .prod (new_arr )
45+ result .append (res )
46+ return result
47+ 48+ 49+ # Brute Force (Without using inbuilt math function)
50+ # Time Complexity: O(n^2)
51+ # Space Complexity: O(n)
52+ 53+ '''
54+ In this approach, we will iterate through the array and for each element, we will calculate the product of all elements except the current element.
55+ We will store the result in the result array and return it.
56+ '''
57+ 58+ class Solution :
59+ def productExceptSelf (self , nums ):
60+ result = []
61+ 62+ for i in range (len (nums )):
63+ prod = 1
64+ for j in range (len (nums )):
65+ if i != j :
66+ prod *= nums [j ]
67+ result .append (prod )
68+ 69+ return result
70+ 71+ # Optimized Approach
72+ # Time Complexity: O(n)
73+ # Space Complexity: O(n)
74+ 75+ '''
76+ In this approach, we will first calculate the prefix product of the array by iterating from right to left and store it in an array pref[].
77+ Then we will calculate the suffix product of the array by iterating through left to right and store it in an array suff[].
78+
79+ Finally, we will calculate the product of pref[i]*suff[i] for each element of the array and store it in the result array.
80+ '''
81+ class Solution :
82+ def productExceptSelf (self , nums ):
83+ n = len (nums )
84+ prod = [1 ]* n
85+ pref = [1 ]* n
86+ suff = [1 ]* n
87+ 88+ prefix = 1
89+ 90+ for i in range (n ):
91+ pref [i ]= prefix
92+ prefix *= nums [i ]
93+ 94+ suffix = 1
95+ 96+ for i in range (n - 1 ,- 1 ,- 1 ):
97+ suff [i ]= suffix
98+ suffix *= nums [i ]
99+ 100+ for i in range (n ):
101+ prod [i ]= pref [i ]* suff [i ]
102+ 103+ return prod
104+ 105+ # Optimized Approach
106+ # Time Complexity: O(n)
107+ # Space Complexity: O(1)
108+ 109+ '''
110+ This approach calculates the product of all elements except nums[i] using two passes:
111+
112+ Prefix Pass (left to right) stores the product of all elements before nums[i].
113+
114+ Suffix Pass (right to left) multiplies it with the product of all elements after nums[i].
115+ This achieves O(n) time and O(1) extra space by updating the result in-place.
116+ '''
117+ 118+ class Solution :
119+ def productExceptSelf (self , nums ):
120+ n = len (nums )
121+ prod = [1 ]* n
122+ 123+ prefix = 1
124+ 125+ for i in range (n ):
126+ prod [i ]= prefix
127+ prefix *= nums [i ]
128+ 129+ suffix = 1
130+ 131+ for i in range (n - 1 ,- 1 ,- 1 ):
132+ prod [i ]*= suffix
133+ suffix *= nums [i ]
134+ 135+ return prod
136+ 137+ 138+ 139+ 140+ obj = Solution ()
141+ nums = [1 ,2 ,4 ,6 ]
142+ print (obj .productExceptSelf (nums ))
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