238. Product of Array Except Self

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238. Product of Array Except Self

Description

Difficulty: Medium

Related Topics: Array, Prefix Sum

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:

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Input: nums = [1,2,3,4]
Output: [24,12,8,6]

Example 2:

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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.)

Hints/Notes

Solution

Language: C++

O(1) space

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class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int size = nums.size();
        vector<int> sufProduct(size, 1);
        for (int i = size - 2; i >= 0; i--) {
            sufProduct[i] = sufProduct[i + 1] * nums[i + 1];
        }
        int preProduct = 1;
        for (int i = 0; i < size; i++) {
            sufProduct[i] = preProduct * sufProduct[i];
            preProduct *= nums[i];
        }
        return sufProduct;
    }
};
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class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int size = nums.size();
        vector<int> preProduct(size + 1, 1);
        vector<int> sufProduct(size + 1, 1);
        for (int i = 0; i < size; i++) {
            preProduct[i + 1] = preProduct[i] * nums[i];
        }
        for (int i = size - 1; i >= 0; i--) {
            sufProduct[i] = sufProduct[i + 1] * nums[i];
        }
        vector<int> res(size, 0);
        for (int i = 0; i < size; i++) {
            res[i] = preProduct[i] * sufProduct[i + 1];
        }
        return res;
    }
};