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0238--product-of-array-except-self.js
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0238--product-of-array-except-self.js
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// 238. 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.
--- Examples
Input: nums = [1,2,3,4]
Output: [24,12,8,6]
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.)
*/
/**
* @param {number[]} nums
* @return {number[]}
*/
const productExceptSelf = function (nums) {
// initialize an array to keep track of products
const products = Array(nums.length).fill(1);
// for each number, find the product of all the numbers on its left
for (let i = 1; i < nums.length; i++) {
// products[i - 1] is the product of nums[0] to nums[i - 2]
// if we multiply it by nums[i - 1], we get the product of all the numbers on the left of nums[i]
products[i] = products[i - 1] * nums[i - 1];
}
let rightNumsProduct = 1;
// for each number, find the product of all the numbers on its right
// multiply the product of left nums and right nums to find the product except the number itself
for (let i = nums.length - 1; i >= 0; i--) {
products[i] = products[i] * rightNumsProduct;
rightNumsProduct *= nums[i];
}
return products;
};
// Time complexity = O(n)
// Space complexity = O(1)