C++ Solution to Coding Challenge 238. Product of Array Except Self

Problem statement

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 <= 10^5.

• -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.)

Solution 1: Compute the prefix and suffix products

To avoid division operation, you can compute the prefix product and the suffix one of nums[i].

Code

#include <vector>
#include <iostream>
using namespace std;
vector<int> productExceptSelf(vector<int>& nums) {
    const int n = nums.size();
    vector<int> prefix(n);
    prefix[0] = 1;
    for (int i = 1; i < n; i++) {
        prefix[i] = prefix[i - 1] * nums[i - 1];
    }
    vector<int> suffix(n);        
    suffix[n - 1] = 1;
    for (int i = n - 2; i >= 0; i--) {
        suffix[i] = suffix[i + 1] * nums[i + 1];
    }
    vector<int> answer(n);
    for (int i = 0; i < n; i++) {
        answer[i] = prefix[i] * suffix[i];
    }
    return answer;
}
void print(vector<int>& nums) {
    for (auto& v : nums) {
        cout << v << " ";
    }
    cout << endl;
}
int main() {
    vector<int> nums = {1, 2, 3, 4};
    auto answer = productExceptSelf(nums);
    print(answer);
    nums = {-1, 1, 0, -3, 3};
    answer = productExceptSelf(nums);
    print(answer);
}

Output:
24 12 8 6 
0 0 9 0 0

Complexity

• Runtime: O(n), where n = nums.length.

• Extra space: O(2n).

Solution 2: Use directly vector answer to store the prefix product

In the solution above you can use directly vector answer for prefix and merge the last two loops into one.

Code

#include <vector>
#include <iostream>
using namespace std;
vector<int> productExceptSelf(vector<int>& nums) {
    const int n = nums.size();
    vector<int> answer(n);
    answer[0] = 1;
    for (int i = 1; i < n; i++) {
        answer[i] = answer[i - 1] * nums[i - 1];
    }
    int suffix = 1;
    for (int i = n - 2; i >= 0; i--) {
        suffix *= nums[i + 1];
        answer[i] *= suffix;
    }
    return answer;
}
void print(vector<int>& nums) {
    for (auto& v : nums) {
        cout << v << " ";
    }
    cout << endl;
}
int main() {
    vector<int> nums = {1, 2, 3, 4};
    auto answer = productExceptSelf(nums);
    print(answer);
    nums = {-1, 1, 0, -3, 3};
    answer = productExceptSelf(nums);
    print(answer);
}

Output:
24 12 8 6 
0 0 9 0 0

Complexity

• Runtime: O(n), where n = nums.length.

• Extra space: O(1).

References

• https://leetcode.com/problems/product-of-array-except-self/

• https://www.leetsolve.com/238-product-of-array-except-self

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