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.

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.)

Solutions

Python3

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

Java

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

JavaScript

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

TypeScript

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

function productExceptSelf(nums: number[]): number[] {
    return nums.map((_, i) =>
        nums.reduce((pre, val, j) => pre * (i === j ? 1 : val), 1),
    );
}

Go

func productExceptSelf(nums []int) []int {
	n := len(nums)
	ans := make([]int, n)
	left, right := 1, 1
	for i := 0; i < n; i++ {
		ans[i] = left
		left *= nums[i]
	}
	for i := n - 1; i >= 0; i-- {
		ans[i] *= right
		right *= nums[i]
	}
	return ans
}

C++

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n = nums.size();
        vector<int> ans(n);
        for (int i = 0, left = 1; i < n; ++i) {
            ans[i] = left;
            left *= nums[i];
        }
        for (int i = n - 1, right = 1; i >= 0; --i) {
            ans[i] *= right;
            right *= nums[i];
        }
        return ans;
    }
};

Rust

impl Solution {
    pub fn product_except_self(nums: Vec<i32>) -> Vec<i32> {
        let mut dp_left = vec![1_i32; nums.len()];
        let mut dp_right = vec![1_i32; nums.len()];
        for i in 1..nums.len() {
            dp_left[i] = dp_left[i - 1] * nums[i - 1];
        }
        for i in (0..(nums.len() - 1)).rev() {
            dp_right[i] = dp_right[i + 1] * nums[i + 1];
        }
        dp_left
            .into_iter()
            .enumerate()
            .map(|(i, x)| x * dp_right[i])
            .collect()
    }
}

impl Solution {
    pub fn product_except_self(nums: Vec<i32>) -> Vec<i32> {
        let n = nums.len();
        let mut l = 1;
        let mut r = 1;
        let mut res = vec![0; n];
        for i in 0..n {
            res[i] = l;
            l *= nums[i];
        }
        for i in (0..n).rev() {
            res[i] *= r;
            r *= nums[i];
        }
        res
    }
}

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