238

238. Product of Array Except Self (Medium)

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.)

Companies:
Facebook, Amazon, Apple, Microsoft, Asana, Adobe, Nutanix, Lyft, Nvidia, Intel, VMware

Related Topics:
Array

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Solution 1.

// OJ: https://leetcode.com/problems/product-of-array-except-self/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<int> productExceptSelf(vector<int>& A) {
        vector<int> ans = A;
        int N = A.size(), prod = 1;
        for (int i = N - 2; i >= 0; --i) ans[i] = ans[i + 1] * A[i];
        for (int i = 0; i < N; ++i) {
            ans[i] = prod * (i + 1 < N ? ans[i + 1] : 1);
            prod *= A[i];
        }
        return ans;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/product-of-array-except-self/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<int> productExceptSelf(vector<int>& A) {
        int N = A.size(), left = 1, right = 1;
        vector<int> ans(N, 1);
        for (int i = 0, j = N - 1; i < N; ++i, --j) {
            ans[i] *= left;
            left *= A[i];
            ans[j] *= right;
            right *= A[j];
        }
        return ans;
    }
};