Product of Array Except Self - codepath/compsci_guides GitHub Wiki

Problem Highlights

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

  • Established a set (2-3) of test cases to verify their own solution later.
  • Established a set (1-2) of edge cases to verify their solution handles complexities.
  • Have fully understood the problem and have no clarifying questions.
  • Have you verified any Time/Space Constraints for this problem?
  • Could the input parameter be Null?
    • Let’s assume no input will be Null. The minimum input has two numbers.
  • What is the time and space complexity requirements?
    • O(N) time and O(1) space complexity. Where N is the size of array. (The output array does not count as extra space for space complexity analysis.)
HAPPY CASE
Input: nums = [1,2,3,4]
Output: [24,12,8,6]

Input: nums = [-1,1,0,-3,3]
Output: [0,0,9,0,0]

EDGE CASE
Input: nums = [-3,3]
Output: [3,-3]

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

  • Sort
    • Does sorting help us achieve what we need in order to solve the problem?
  • Two pointer solutions (left and right pointer variables)
    • Two pointer may help us.
  • Storing the elements of the array in a HashMap or a Set
    • In reversing, hashing elements and storing them may not yield an optimal solution.
  • Traversing the array with a sliding window
    • Will viewing pieces of the input at a time help us?

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Use two pointer from both ends of the array to generate the product from both sides of each number to achieve the product except self.

1) Create product from the left side of each num and store in output array. 
    a) Store product into output array
    b) Multiply product with the current number to develop left product
2) Create product from the right side of each num and multiply with the left product stored in output array.
    a) Multiply right product with the left product stored in output array
    b) Multiply product with the current number to develop right product
3) Return the output array

⚠️ Common Mistakes

  • The problem is easy to complete with O(N) space, but without any new space, this problem rely on a creative use of the array data structure. This creative use of the array data structure can be tricky.

4: I-mplement

Implement the code to solve the algorithm.

class Solution:
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        # Create product from the left side of each num and store in output array.
        leftProduct = 1
        output = [1] * len(nums)
        for i in range(len(nums)):
            # Store product into output array
            output[i] = leftProduct
            # Multiply product with the current number to develop left product
            leftProduct *= nums[i]
        
        # Create product from the right side of each num and multiply with the left product stored in output array.
        rightProduct = 1
        for j in range(len(nums) - 1,-1,-1):
            # Multiply right product with the left product stored in output array.
            output[j] *= rightProduct
            # Mutiply product with the current number to develop right product
            rightProduct *= nums[j]
        
        # Return the output array
        return output

class Solution {
    public int[] productExceptSelf(int[] nums) {
        int[] result = new int[nums.length];

        // Create product from the left and right side of each num and store in output array.
        for (int i = 0; i < result.length; i++) result[i] = 1;
            int left = 1, right = 1;
            for (int i = 0, j = nums.length - 1; i < nums.length - 1; i++, j--) {
                // Store product into output array
                left *= nums[i];
                // Multiply right product with the left product stored in output array.
                right *= nums[j];
                // Multiply product with the current number to develop left product
                result[i + 1] *= left;
                // Mutiply product with the current number to develop right product
                result[j - 1] *= right;
            }
        // Return the output array
        return result;
    }
}

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

  • Trace through your code with an input to check for the expected output
  • Catch possible edge cases and off-by-one errors

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Assume N represents the number of items in the array.

  • Time Complexity: O(N), traversing done on every number of the array twice.
  • Space Complexity: O(1), when the output array does not count as extra space for space complexity analysis.