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# Product of Array Except Self LeetCode Solution

## Problem – Product of Array Except Self

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]`.

The product of any prefix or suffix of `nums` is guaranteed to fit in a 32-bit integer.

You must write an algorithm that runs in `O(n)` time and without using the division operation.

Example 1:

``````Input: nums = [1,2,3,4]
Output: [24,12,8,6]
``````

Example 2:

``````Input: nums = [-1,1,0,-3,3]
Output: [0,0,9,0,0]``````

Constraints:

• `2 <= nums.length <= 105`
• `-30 <= nums[i] <= 30`
• The product of any prefix or suffix of `nums` is guaranteed to fit in a 32-bit integer.

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.)

### Product of Array Except Self LeetCode Solution in Java

``````public class Solution {
public int[] productExceptSelf(int[] nums) {
int n = nums.length;
int[] res = new int[n];
res[0] = 1;
for (int i = 1; i < n; i++) {
res[i] = res[i - 1] * nums[i - 1];
}
int right = 1;
for (int i = n - 1; i >= 0; i--) {
res[i] *= right;
right *= nums[i];
}
return res;
}
``````

### Product of Array Except Self LeetCode Solution in C++

``````class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int prod = 1, zeroCnt = count(begin(nums), end(nums), 0);
if(zeroCnt > 1) return vector<int>(size(nums));               // Case-1
for(auto c : nums)
if(c) prod *= c;                                          // calculate product of all elements except 0
for(auto& c : nums)
if(zeroCnt) c = c ? 0 : prod;                             // Case-2
else c = prod / c;                                        // Case-3
return nums;
}
};
``````

### Product of Array Except Self LeetCode Solution in Python

``````class Solution:
def productExceptSelf(self, nums):
prod, zero_cnt = reduce(lambda a, b: a*(b if b else 1), nums, 1), nums.count(0)
if zero_cnt > 1: return [0]*len(nums)
for i, c in enumerate(nums):
if zero_cnt: nums[i] = 0 if c else prod
else: nums[i] = prod // c
return nums``````
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