Problem Description

This problem asks you to design a data structure that maintains a stream of integers and can efficiently calculate the product of the last k integers in the stream.

You need to implement a ProductOfNumbers class with the following methods:

1. ProductOfNumbers(): Constructor that initializes an empty stream.

2. add(int num): Adds an integer num to the end of the stream.

3. getProduct(int k): Returns the product of the last k integers in the stream. You're guaranteed that the stream always has at least k numbers when this method is called.

The key insight for the solution is using a prefix product array. The solution maintains an array s where each element stores the cumulative product up to that position.

How the solution works:

• The array s is initialized with [1] as a base case.
• When adding a number:
  • If the number is 0, reset s to [1] (since any product involving 0 will be 0)
  • Otherwise, append s[-1] * num to s (multiply the last cumulative product by the new number)
• When getting the product of last k numbers:
  • If the length of s is less than or equal to k, return 0 (this means there was a 0 in the last k numbers)
  • Otherwise, return s[-1] // s[-k - 1] (divide the current cumulative product by the cumulative product from k positions back)

The division s[-1] // s[-k - 1] effectively gives us the product of elements from position [-k-1] to [-1], which are the last k elements.

Example walkthrough:

• After adding [3, 0, 2, 5, 4], the prefix product array would be reset at 0 and then built as [1, 2, 10, 40]
• getProduct(2) returns 40 // 2 = 20 (product of last 2: 5 * 4)
• getProduct(3) returns 40 // 1 = 40 (product of last 3: 2 * 5 * 4)
• getProduct(4) returns 0 (since array length 4 ≤ k=4, meaning the 0 is included)

This approach achieves O(1) time complexity for both add and getProduct operations.

Intuition

The naive approach would be to store all numbers in a list and calculate the product of the last k numbers each time getProduct is called. This would take O(k) time for each query, which isn't efficient when we have many queries.

The key observation is that we can use prefix products to calculate any range product in O(1) time. Think about prefix sums - if we know the sum from index 0 to i and from 0 to j, we can get the sum from i+1 to j by subtraction. Similarly, with products, we can use division.

For example, if we have numbers [2, 3, 4, 5] and their prefix products [2, 6, 24, 120]:

• To get the product of last 2 numbers (4 * 5), we divide 120 / 6 = 20
• To get the product of last 3 numbers (3 * 4 * 5), we divide 120 / 2 = 60

The pattern is: to get the product of last k numbers, we divide the total product by the product up to position n-k-1.

But what about zeros? A zero anywhere in the sequence makes all products involving it equal to 0. The clever trick here is to reset the prefix product array whenever we encounter a zero. This way:

• If we query for k numbers and our array has less than or equal to k elements, we know there was a zero within the last k numbers, so we return 0
• Otherwise, we can safely use our division formula

By starting our prefix array with [1] as a sentinel value, we handle edge cases elegantly - when we want the product of all stored numbers, we divide by s[0] which is 1.

This approach transforms an O(k) operation into O(1) by trading a small amount of space to store cumulative products, making both add and getProduct operations constant time.

Learn more about Math, Data Stream and Prefix Sum patterns.

Solution Approach

The solution uses a prefix product array to achieve O(1) time complexity for both operations.

Data Structure:

• We maintain a list s where s[i] represents the product of all numbers from the start up to index i
• Initialize s with [1] as a base value to handle edge cases

Implementation of add(num):

1. Check if num is 0:
   • If yes, reset s to [1]
   • This effectively "forgets" all previous numbers since any product involving 0 is 0
2. If num is not 0:
   • Calculate the new cumulative product: s[-1] * num
   • Append this value to s

Implementation of getProduct(k):

1. Check if len(s) <= k:
   • If true, return 0 (this means a zero was added within the last k numbers)
2. Otherwise:
   • Return s[-1] // s[-k - 1]
   • This divides the current total product by the product up to position n-k-1
   • The result is the product of elements from position n-k to n-1 (the last k elements)

Why this works:

• When there's no zero in the last k numbers, we have a continuous prefix product array of at least k+1 elements
• The product of numbers from index i to j equals prefix_product[j] / prefix_product[i-1]
• For the last k numbers, this becomes s[-1] / s[-k-1]

Handling zeros:

• Resetting to [1] when encountering 0 is crucial
• It creates a "checkpoint" - any query spanning across this point will correctly return 0
• The length check len(s) <= k elegantly handles this case

Time Complexity:

• add(): O(1) - just one multiplication and append operation
• getProduct(): O(1) - just one division operation
• Space: O(n) where n is the number of non-zero elements added since the last zero

Example Walkthrough

Let's walk through a concrete example to see how the prefix product approach works.

Initial State:

• Start with s = [1] (our base case)

Operation 1: add(2)

• 2 is not zero, so calculate: s[-1] * 2 = 1 * 2 = 2
• Append to get s = [1, 2]

Operation 2: add(3)

• 3 is not zero, so calculate: s[-1] * 3 = 2 * 3 = 6
• Append to get s = [1, 2, 6]

Operation 3: add(4)

• 4 is not zero, so calculate: s[-1] * 4 = 6 * 4 = 24
• Append to get s = [1, 2, 6, 24]

Operation 4: getProduct(2)

• Check: len(s) = 4 is NOT ≤ k = 2, so proceed
• Calculate: s[-1] // s[-2-1] = s[-1] // s[-3] = 24 // 2 = 12
• This gives us the product of the last 2 numbers: 3 * 4 = 12 ✓

Operation 5: add(0)

• Number is zero, so reset: s = [1]
• All previous products are now "forgotten"

Operation 6: add(5)

• 5 is not zero, so calculate: s[-1] * 5 = 1 * 5 = 5
• Append to get s = [1, 5]

Operation 7: getProduct(3)

• Check: len(s) = 2 IS ≤ k = 3
• Return 0 (because there was a zero within the last 3 numbers)

Why the division formula works: When s = [1, 2, 6, 24], each element represents:

• s[0] = 1 (base)
• s[1] = 2 (product of first 1 number)
• s[2] = 6 (product of first 2 numbers: 2*3)
• s[3] = 24 (product of first 3 numbers: 234)

To get the product of numbers from position 2 to 3 (which are 3 and 4):

• We take s[3] / s[1] = 24 / 2 = 12
• This effectively "removes" the product of numbers before position 2

Solution Implementation

Time and Space Complexity

Time Complexity:

• __init__(): O(1) - Simply initializes a list with one element
• add(num): O(1) - Either resets the list to [1] when num == 0, or appends a single calculated value (self.s[-1] * num) to the list
• getProduct(k): O(1) - Performs a length check and at most one division operation between two list elements accessed by index

Overall, each operation has O(1) time complexity.

Space Complexity:

• The space complexity is O(n), where n is the total number of non-zero elements added since the last zero (or since initialization)
• The list self.s grows by one element for each non-zero number added via add()
• When a zero is added, the list resets to [1], effectively releasing previous memory
• In the worst case where no zeros are added, the list grows to size n + 1 (including the initial 1), resulting in O(n) space complexity

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow in Prefix Products

The prefix product array can grow extremely large when continuously multiplying positive integers. For example, adding just twenty 10s would result in 10^20, which exceeds typical integer limits in many languages.

Solution:

• In Python, integers have arbitrary precision, so overflow isn't an issue
• In languages like Java/C++, use long long or consider using modular arithmetic if the problem specifies a modulo value
• Alternatively, store logarithms of products and convert back when needed (though this introduces floating-point precision issues)

2. Incorrect Handling of Multiple Zeros

A common mistake is not properly resetting the prefix array when encountering zeros. Some might try to track zero positions separately or use complex logic to handle multiple zeros.

Why the current approach works:

# Incorrect approach - trying to track zeros separately
class ProductOfNumbers:
    def __init__(self):
        self.products = []
        self.zero_indices = []
  
    # This becomes unnecessarily complex!

Correct approach (as shown):

if num == 0:
    self.prefix_products = [1]  # Simple reset handles all cases

3. Off-by-One Error in Index Calculation

When calculating getProduct(k), it's easy to make indexing errors:

Common mistake:

# Wrong - this would get k+1 elements
return self.prefix_products[-1] // self.prefix_products[-k]

# Wrong - this would skip the last element
return self.prefix_products[-2] // self.prefix_products[-k-1]

Correct calculation:

return self.prefix_products[-1] // self.prefix_products[-k-1]

4. Not Initializing with [1]

Starting the prefix array with an empty list or [0] instead of [1] breaks the algorithm:

Wrong initialization:

def __init__(self):
    self.prefix_products = []  # Would require special handling for first element
    # OR
    self.prefix_products = [0]  # Would make all products 0

Correct initialization:

def __init__(self):
    self.prefix_products = [1]  # Serves as identity element for multiplication

5. Using Regular Division Instead of Integer Division

In Python 3, using / instead of // returns a float, which may cause issues:

Wrong:

return self.prefix_products[-1] / self.prefix_products[-k-1]  # Returns float

Correct:

return self.prefix_products[-1] // self.prefix_products[-k-1]  # Returns integer

6. Not Understanding Why len(s) <= k Returns 0

This condition is subtle but crucial. When len(prefix_products) <= k, it means:

• We've seen fewer than k non-zero numbers since the last zero
• Therefore, the last k numbers must include at least one zero
• So the product should be 0

Example to illustrate:

# After adding [2, 3, 0, 4, 5]
# prefix_products = [1, 4, 20]  (reset at 0, then built with 4, 5)
# getProduct(3) should return 0 because the 3rd-last number was 0
# len(prefix_products) = 3, k = 3, so 3 <= 3 is true → return 0 ✓