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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 $O(1)$.

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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

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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;
    }
}

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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;
    }
};

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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
}

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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];
    }