# Product of Array Except Self LeetCode Solution

Here, We see Product of Array Except Self LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

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]

## Product of Array Except Self LeetCode SolutionC++

``````class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int n = nums.size();
vector<int> output;
for(int i=0; i<n; i++){
int product = 1;
for(int j=0; j<n; j++){
if(i == j) continue;
product *= nums[j];
}
output.push_back(product);
}
return output;
}
};```Code language: PHP (php)```

## Product of Array Except Self LeetCode SolutionJava

``````class Solution {
public int[] productExceptSelf(int[] nums) {
int n = nums.length;
int ans[] = new int[n];
for(int i = 0; i < n; i++) {
int pro = 1;
for(int j = 0; j < n; j++) {
if(i == j) continue;
pro *= nums[j];
}
ans[i] = pro;
}
return ans;
}
}```Code language: PHP (php)```

## Product of Array Except Self SolutionJavaScript

``````var productExceptSelf = function(nums) {
const n = nums.length;
const prefix = new Array(n).fill(1);
const suffix = new Array(n).fill(1);
for (let i = 1; i < n; i++) {
prefix[i] = prefix[i - 1] * nums[i - 1];
}
for (let i = n - 2; i >= 0; i--) {
suffix[i] = suffix[i + 1] * nums[i + 1];
}
const answer = [];
for (let i = 0; i < n; i++) {
answer[i] = prefix[i] * suffix[i];
}
};```Code language: JavaScript (javascript)```

## Product of Array Except Self SolutionPython

``````class Solution(object):
def productExceptSelf(self, nums):
n = len(nums)
prefix = [1] * n
suffix = [1] * n
for i in range(1, n):
prefix[i] = prefix[i - 1] * nums[i - 1]
for i in range(n - 2, -1, -1):
suffix[i] = suffix[i + 1] * nums[i + 1]
answer = [prefix[i] * suffix[i] for i in range(n)]