README_EN.md

238. Product of Array Except Self

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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.

 

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:

• Could you solve it in O(n) time complexity and without using division?
• Could you solve it with O(1) constant space complexity? (The output array does not count as extra space for space complexity analysis.)

Solutions

Python3

class Solution:
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        n = len(nums)
        output = [1 for _ in nums]
        left = right = 1
        for i in range(n):
            output[i] = left
            left *= nums[i]
        for i in range(n - 1, -1, -1):
            output[i] *= right
            right *= nums[i]
        return output

Java

class Solution {
    public int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        int[] output = new int[n];
        for (int i = 0, left = 1; i < n; ++i) {
            output[i] = left;
            left *= nums[i];
        }
        for (int i = n - 1, right = 1; i >= 0; --i) {
            output[i] *= right;
            right *= nums[i];
        }
        return output;
    }
}

JavaScript

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

Go

func productExceptSelf(nums []int) []int {
	n := len(nums)

	l := make([]int, n)
	l[0] = 1
	for i := 1; i < n; i++ {
		l[i] = l[i-1] * nums[i-1]
	}

	r := make([]int, n)
	r[n-1] = 1
	for i := n - 2; i >= 0; i-- {
		r[i] = r[i+1] * nums[i+1]
	}

	ans := make([]int, n)
	for i := 0; i < n; i++ {
		ans[i] = l[i] * r[i]
	}

	return ans
}

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