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.

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Product of Array Except Self 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 Solution C++

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 Solution Java

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 Solution JavaScript

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];
    }
    return answer;
};Code language: JavaScript (javascript)

Product of Array Except Self Solution Python

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)]
        return answer