238. Product of Array Except Self

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

Solution 1: Two Passes

We define two variables $left$ and $right$, which represent the product of all elements to the left and right of the current element respectively. Initially, $left=1$, $right=1$. Define an answer array $ans$ of length $n$.

We first traverse the array from left to right, for the $i$th element we update $ans[i]$ with $left$, then $left$ multiplied by $nums[i]$.

Then, we traverse the array from right to left, for the $i$th element, we update $ans[i]$ to $ans[i] \times right$, then $right$ multiplied by $nums[i]$.

After the traversal, the array ans is the answer.

The time complexity is $O(n)$, where $n$ is the length of the array nums. Ignore the space consumption of the answer array, the space complexity is $O(1)$.

Python3JavaC++GoTypeScriptRustJavaScriptC#PHP

class Solution:
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        n = len(nums)
        ans = [0] * n
        left = right = 1
        for i, x in enumerate(nums):
            ans[i] = left
            left *= x
        for i in range(n - 1, -1, -1):
            ans[i] *= right
            right *= nums[i]
        return ans

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;
    }
}

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; --i) {
            ans[i] *= right;
            right *= nums[i];
        }
        return ans;
    }
};

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

function productExceptSelf(nums: number[]): number[] {
    const n = nums.length;
    const ans: number[] = 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;
}

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

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var productExceptSelf = function (nums) {
    const n = nums.length;
    const 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;
};

public 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;
    }
}

class Solution {
    /**
     * @param Integer[] $nums
     * @return Integer[]
     */
    function productExceptSelf($nums) {
        $n = count($nums);
        $ans = [];
        for ($i = 0, $left = 1; $i < $n; ++$i) {
            $ans[$i] = $left;
            $left *= $nums[$i];
        }
        for ($i = $n - 1, $right = 1; $i >= 0; --$i) {
            $ans[$i] *= $right;
            $right *= $nums[$i];
        }
        return $ans;
    }
}

Solution 2

TypeScript

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

Comments