238. Product of Array Except Self

Problem 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.)

Python Solution

class Solution:
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        n = len(nums)
        answer = [1] * n
        
        # Calculate prefix products
        prefix = 1
        for i in range(n):
            answer[i] = prefix
            prefix *= nums[i]
            
        # Calculate suffix products and combine with prefix products
        suffix = 1
        for i in range(n - 1, -1, -1):
            answer[i] *= suffix
            suffix *= nums[i]
            
        return answer

Java Solution

class Solution {
    public int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        int[] answer = new int[n];
        
        // Calculate prefix products
        answer[0] = 1;
        for (int i = 1; i < n; i++) {
            answer[i] = answer[i-1] * nums[i-1];
        }
        
        // Calculate suffix products and combine with prefix products
        int suffix = 1;
        for (int i = n - 1; i >= 0; i--) {
            answer[i] *= suffix;
            suffix *= nums[i];
        }
        
        return answer;
    }
}

C++ Solution

class Solution {
public:
    vector productExceptSelf(vector& nums) {
        int n = nums.size();
        vector answer(n);
        
        // Calculate prefix products
        answer[0] = 1;
        for (int i = 1; i < n; i++) {
            answer[i] = answer[i-1] * nums[i-1];
        }
        
        // Calculate suffix products and combine with prefix products
        int suffix = 1;
        for (int i = n - 1; i >= 0; i--) {
            answer[i] *= suffix;
            suffix *= nums[i];
        }
        
        return answer;
    }
};

JavaScript Solution

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var productExceptSelf = function(nums) {
    const n = nums.length;
    const answer = new Array(n);
    
    // Calculate prefix products
    answer[0] = 1;
    for (let i = 1; i < n; i++) {
        answer[i] = answer[i-1] * nums[i-1];
    }
    
    // Calculate suffix products and combine with prefix products
    let suffix = 1;
    for (let i = n - 1; i >= 0; i--) {
        answer[i] *= suffix;
        suffix *= nums[i];
    }
    
    return answer;
};

C# Solution

public class Solution {
    public int[] ProductExceptSelf(int[] nums) {
        int n = nums.Length;
        int[] answer = new int[n];
        
        // Calculate prefix products
        answer[0] = 1;
        for (int i = 1; i < n; i++) {
            answer[i] = answer[i-1] * nums[i-1];
        }
        
        // Calculate suffix products and combine with prefix products
        int suffix = 1;
        for (int i = n - 1; i >= 0; i--) {
            answer[i] *= suffix;
            suffix *= nums[i];
        }
        
        return answer;
    }
}