Product of Array Except Self

238. Product of Array Except Self 🔗

Date: Jun 03, 2024

Tags: prefix-sum

Description

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.)

Optimal

function productExceptSelf(nums: number[]): number[] {
  const len = nums.length;
  const result: number[] = Array(len).fill(1);

  let x = 1;

  for (let i = 0; i < len; i++) {
    result[i] = x;
    x *= nums[i];
  }

  x = 1;
  for (let i = len - 1; i >= 0; i--) {
    const product = result[i] * x;
    result[i] = Object.is(-0, product) ? 0 : product;
    x *= nums[i];
  }

  return result;
}

export { productExceptSelf };

Brute-Force

function productExceptSelf(nums: number[]): number[] {
  const result: number[] = Array.from({ length: nums.length }).map(() => 1);

  for (let i = 0; i < nums.length; i++) {
    for (let j = 0; j < nums.length; j++) {
      if (i !== j) {
        result[i] *= nums[j];
      }
    }
  }

  return result.map((x) => (Object.is(-0, x) ? 0 : x));
}

export { productExceptSelf };