[Array] Product of Array Except Self - Medium

The “Product of Array Except Self” is a popular coding challenge. We need to find the product of all elements in an array except for the one at the current index. This means that for each index, we should get the product of all other elements. We can solve this problem well without using division. This way, we keep the solution fast and use less space.

In this article, we will look into the “Product of Array Except Self” problem closely. We will explain the problem and what we need to do. We will also find a good solution that does not use division. Then, we will show how to implement it in Java, Python, and C++. We will check the space and time complexity of our solutions. We will also look at common edge cases and answer some questions that people often ask about this topic.

  • Understanding Array Product of Array Except Self - Medium
  • Problem Statement and Requirements
  • Optimal Solution Without Division
  • Java Implementation of Product of Array Except Self
  • Python Implementation of Product of Array Except Self
  • C++ Implementation of Product of Array Except Self
  • Space Complexity Analysis of the Solution
  • Time Complexity Analysis of the Solution
  • Common Edge Cases and Their Handling
  • Frequently Asked Questions

Problem Statement and Requirements

The “Product of Array Except Self” problem asks us to create an output array. Each element at index i in this output array should be equal to the product of all the numbers in the input array except the number at i.

Requirements:

  1. We must solve the problem without using division.
  2. Our solution should be good in both time and space.
  3. The input is an array of integers. We need to return an array that has the same length as the input.

Example:

If we have an input array nums = [1, 2, 3, 4], the output should be:

Output: [24, 12, 8, 6]
  • Explanation:
    • For output[0]: the product is nums[1] * nums[2] * nums[3] which is 2 * 3 * 4 = 24
    • For output[1]: the product is nums[0] * nums[2] * nums[3] which is 1 * 3 * 4 = 12
    • For output[2]: the product is nums[0] * nums[1] * nums[3] which is 1 * 2 * 4 = 8
    • For output[3]: the product is nums[0] * nums[1] * nums[2] which is 1 * 2 * 3 = 6

Constraints:

  • The input array length must be at least 1 and no more than 10^5.
  • Each number in the input array must be between -30,000 and 30,000.

We can solve this problem well by going through the array two times. We will keep two separate arrays for the left and right products. It is very important to follow the constraints. We should also make our solution fast to work with bigger input sizes.

Optimal Solution Without Division

To solve the “Product of Array Except Self” problem without using division, we can use two passes through the array. This helps us keep the product of all elements to the left and right of each index. This way, we can get the result in O(n) time and use O(1) space, not counting the output array.

Steps:

  1. Initialize an output array for the product results.
  2. First pass (left products):
    • We go through the array from left to right.
    • We keep a variable to store the product of elements to the left of the current index.
    • We update the output array with the left product for each index.
  3. Second pass (right products):
    • Now, we go through the array from right to left.
    • We keep another variable for the product of elements to the right of the current index.
    • We multiply the current value in the output array by the right product.

Code Implementation

Here is a Python code for the optimal solution:

def product_except_self(nums):
    length = len(nums)
    output = [1] * length

    # First pass: calculate left products
    left_product = 1
    for i in range(length):
        output[i] = left_product
        left_product *= nums[i]

    # Second pass: calculate right products
    right_product = 1
    for i in range(length - 1, -1, -1):
        output[i] *= right_product
        right_product *= nums[i]

    return output

Example Usage

nums = [1, 2, 3, 4]
result = product_except_self(nums)
print(result)  # Output: [24, 12, 8, 6]

This solution helps us find the product of all elements except the one at the current index. It does this without using division. This is good for cases where we can’t or don’t want to use division.

We can use this method in many situations. It is like other problems with arrays. For example, problems like Array Rotate or Maximum Subarray need good array processing.

Java Implementation of Product of Array Except Self

We will show how to implement the “Product of Array Except Self” in Java. We will create a function that finds the product of all elements in an array except the one at each index. This solution does not use division. It works in linear time and has O(1) space complexity, not counting the output array.

Here is our step-by-step approach:

  1. First, we create an output array to store the result.
  2. Then, we go through the input array from left to right. We fill the output array with the products of all elements to the left of each index.
  3. Next, we go through the input array from right to left. We multiply the products in the output array by the products of all elements to the right of each index.

Java Code Example

public class ProductOfArrayExceptSelf {
    public static int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        int[] output = new int[n];
        
        // Step 1: Calculate products of elements to the left of each index
        output[0] = 1; // First element has no left products
        for (int i = 1; i < n; i++) {
            output[i] = output[i - 1] * nums[i - 1];
        }
        
        // Step 2: Calculate products of elements to the right of each index
        int rightProduct = 1; // Start with no right products
        for (int i = n - 1; i >= 0; i--) {
            output[i] *= rightProduct; // Multiply with the right product
            rightProduct *= nums[i]; // Update the right product
        }
        
        return output;
    }
    
    public static void main(String[] args) {
        int[] nums = {1, 2, 3, 4};
        int[] result = productExceptSelf(nums);
        System.out.print("Product of Array Except Self: ");
        for (int value : result) {
            System.out.print(value + " ");
        }
    }
}

Explanation of the Code

  • Initialization: We make an array called output to store the results.
  • Left Product Calculation: In the first loop, we fill the output array with products of all elements to the left of each index.
  • Right Product Calculation: The second loop updates the output array by multiplying it with products of all elements to the right of each index.

This method gives us the result without using division. It also keeps good time and space performance.

If we want to learn more about array techniques, we can check articles like Array Two Sum and Array Rotate Array.

Python Implementation of Product of Array Except Self

To do the “Product of Array Except Self” in Python, we can use a two-step method. This helps us get the right answer without division. We want to make an output array. Each element at index i should be the product of all numbers in the input array, except the number at i.

Implementation

Here is a simple Python code for this solution:

def product_except_self(nums):
    length = len(nums)
    output = [1] * length

    # Calculate products of elements to the left of each index
    left_product = 1
    for i in range(length):
        output[i] = left_product
        left_product *= nums[i]

    # Calculate products of elements to the right of each index
    right_product = 1
    for i in range(length - 1, -1, -1):
        output[i] *= right_product
        right_product *= nums[i]

    return output

# Example usage
nums = [1, 2, 3, 4]
result = product_except_self(nums)
print(result)  # Output: [24, 12, 8, 6]

Explanation

  • Left Product Calculation: We go through the array from left to right. For each index, we keep the product of all the numbers before it.

  • Right Product Calculation: We go through the array from right to left. For each index, we multiply the current value in the output array with the product of all elements after it.

This method works in O(n) time and uses O(1) space. We do not use any extra arrays to keep temporary results, except for the output array.

If you want to learn more about similar problems, you can look at the article on Array Two Sum. It gives more tips on how to work with arrays.

C++ Implementation of Product of Array Except Self

To solve the “Product of Array Except Self” problem in C++, we can make a good solution without using division. The main idea is to use two passes to get the result.

Here is the code:

#include <vector>
using namespace std;

vector<int> productExceptSelf(vector<int>& nums) {
    int n = nums.size();
    vector<int> result(n, 1);

    // Calculate prefix products
    for (int i = 1; i < n; i++) {
        result[i] = result[i - 1] * nums[i - 1];
    }

    // Calculate suffix products and multiply with prefix products
    int suffix = 1;
    for (int i = n - 1; i >= 0; i--) {
        result[i] *= suffix;
        suffix *= nums[i];
    }

    return result;
}

Explanation of the Code:

  1. Initialization: We create a result vector and set it to 1.
  2. First Pass (Prefix Products):
    • We go through the array and save the product of all previous elements in the result array.
  3. Second Pass (Suffix Products):
    • We start from the end of the array. We keep a running product of elements after the current index and multiply it with the value in the result array.

Example Usage:

#include <iostream>

int main() {
    vector<int> nums = {1, 2, 3, 4};
    vector<int> result = productExceptSelf(nums);
    
    for (int r : result) {
        cout << r << " ";
    }
    return 0;
}

Output:

For the input {1, 2, 3, 4}, the output will be:

24 12 8 6

This solution runs in O(n) time and uses O(1) extra space, not counting the output array. The code solves the “Product of Array Except Self” problem in C++. If we want to read more about similar array problems, we can check Array Rotate Array - Medium.

Space Complexity Analysis of the Solution

The space complexity of the “Product of Array Except Self” problem can change based on how we solve it.

  1. Optimal Solution Without Division:
    • The best solution needs O(1) extra space if we think about the output array as part of the input. But if we cannot change the input array, we must think about the space used for the output array. In this case, it would be O(n). Here, n is the length of the input array.
  2. Intermediate Arrays:
    • If we use extra arrays to keep intermediate results, like left products and right products, the space complexity becomes O(n). This is because we have two extra arrays.

Space Complexity Breakdown:

  • Using output array only: O(n) to store results.
  • Using two extra arrays (left and right products): O(n) for both arrays, so total is O(n).
  • Using in-place modification with little extra space: O(1) if we do not count the output array.

Example:

For an input array nums = [1, 2, 3, 4]: - The output array would be [24, 12, 8, 6] if we store results. This needs O(n) space.

def productExceptSelf(nums):
    n = len(nums)
    output = [1] * n
    left_product = 1
    for i in range(n):
        output[i] = left_product
        left_product *= nums[i]
    
    right_product = 1
    for i in range(n - 1, -1, -1):
        output[i] *= right_product
        right_product *= nums[i]
    
    return output

In this example, the output array needs O(n) space. The extra variables left_product and right_product only need O(1).

In summary, the space complexity analysis of the “Product of Array Except Self” solution depends on if we see the output array as part of the input and if we use extra storage. The best way for space is to calculate the products in-place. This gives O(1) extra space complexity when we include the output array in the input.

Time Complexity Analysis of the Solution

We can look at the time complexity of the “Product of Array Except Self” problem by looking at the steps in the algorithm we use to find the output array.

  1. Initial Pass to Calculate Prefix Products:
    • First, we go through the array one time to find the prefix products. This takes O(n) time. Here, n is the length of the input array.
  2. Second Pass to Calculate Suffix Products:
    • Next, we go through the array again to find the suffix products. At the same time, we calculate the final product for each index. This also takes O(n) time.

So, since we do both passes one after the other, the total time complexity is:

  • O(n)

We get this good time complexity without using division. This makes our method work well even if the array has zeros. The algorithm quickly finds the product of all elements except itself for each index in linear time.

Common Edge Cases and Their Handling

When we implement the “Product of Array Except Self” algorithm, we need to think about some edge cases. These cases can change if our solution is correct or not. Here are some common edge cases and how we can handle them:

  1. Single Element Array:
    • Input: [1]
    • Output: [1] (The product of an empty array is usually 1).
    • Handling: We return an array of the same size filled with 1.
  2. Array with Zeros:
    • Input: [0, 1, 2, 3]
    • Output: [6, 0, 0, 0] (Only the product of non-zero numbers).
    • Handling: We count how many zeros are in the input array. If we have more than one zero, we give an array of zeros. If we have one zero, all places except the zero’s position should be zero. The zero’s position will have the product of all non-zero numbers.
  3. Array with All Zeros:
    • Input: [0, 0, 0]
    • Output: [0, 0, 0].
    • Handling: Like the last case, we return an array of zeros.
  4. Large Numbers:
    • Input: [100000, 200000, 300000]
    • Output: [60000000000, 30000000000, 20000000000] (We need to be careful of large integer overflow).
    • Handling: We use data types that can handle large numbers like long in Java or int in Python which handles large integers automatically.
  5. Negative Numbers:
    • Input: [-1, -2, -3, -4]
    • Output: [24, 12, 8, 6] (Negative numbers give a positive product).
    • Handling: Our algorithm should work the same no matter the sign of the numbers.
  6. Mixed Positive and Negative Numbers:
    • Input: [1, -2, 3, -4]
    • Output: [-24, 12, -8, 6].
    • Handling: We make sure the algorithm calculates products correctly with mixed signs.
  7. Performance Considerations:
    • Inputs with a very big size (for example, Array length = 10^5).
    • Handling: Our solution must be O(n) in time and O(1) in space (not counting the output array) to work well with large inputs.

By taking care of these edge cases, we make the “Product of Array Except Self” implementation much stronger. A good function should handle these cases without issues.

For more insights on problems with arrays, you can check the Array Rotate Array - Medium article.

Frequently Asked Questions

1. What is the “Product of Array Except Self” problem in arrays?

The “Product of Array Except Self” problem asks us to create an output array. Each element at index i in this array is the product of all numbers in the input array except nums[i]. We often see this problem in coding interviews. It shows the need for good solutions that do not use division.

2. How can I solve the “Product of Array Except Self” problem without using division?

To solve the “Product of Array Except Self” problem without division, we can do two passes over the input array. In the first pass, we keep a running product of all elements to the left of each index. In the second pass, we multiply this product with another running product of elements to the right. This way we get an output array with the products we want.

3. What is the time complexity of the optimal solution for the “Product of Array Except Self”?

The best solution for the “Product of Array Except Self” problem runs in O(n) time. Here, n is the length of the input array. We do this by making two passes through the array. One pass calculates left products and the other for right products. This makes our solution fast and able to handle large inputs.

4. Can you explain the space complexity of the “Product of Array Except Self” solution?

The space complexity for the best solution to the “Product of Array Except Self” problem is O(1), not counting the output array. We only need a small amount of extra space for variables that hold the running products. This makes our algorithm space-efficient.

5. What are some common edge cases to consider when implementing the “Product of Array Except Self”?

When we implement the “Product of Array Except Self,” we should think about edge cases. This includes arrays with zeros, negative numbers, or arrays with just one element. For example, if the input array has one zero, the output should show that all other products are zero. If there are many zeros, the output will be all zeros too.

For more challenges about arrays, check out the Array: Two Sum - Easy or look at the Array: Best Time to Buy and Sell Stock - Easy for more practice.