Algorithm
LeetCode 75
Product of Array except Self

238. Product of Array Except Self

Tags

  • Array
  • Prefix Sum

Link

https://leetcode.com/problems/product-of-array-except-self/description/?envType=study-plan-v2&envId=leetcode-75 (opens in a new tab)

Question

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.)

Answer

JavaScript

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var productExceptSelf = function (nums) {
  const ans = new Array(nums.length).fill(1);
 
  let prefixProduct = 1;
  for (let i = 0; i < nums.length; i++) {
    ans[i] = prefixProduct;
    prefixProduct *= nums[i];
  }
 
  let suffixProduct = 1;
  for (let i = nums.length - 1; i >= 0; i--) {
    ans[i] *= suffixProduct;
    suffixProduct *= nums[i];
  }
 
  return ans;
};
/**
 * @param {number[]} nums
 * @return {number[]}
 */
var productExceptSelf = function (nums) {
  const n = nums.length;
  const left = new Array(n).fill(1);
  const right = new Array(n).fill(1);
 
  for (let i = 1; i < n; i++) {
    left[i] = left[i - 1] * nums[i - 1];
  }
 
  for (let i = n - 2; i >= 0; i--) {
    right[i] = right[i + 1] * nums[i + 1];
  }
 
  const result = [];
  for (let i = 0; i < n; i++) {
    result[i] = left[i] * right[i];
  }
 
  return result;
};
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