algoadvance

Given an array of integers nums and an integer k, return the number of contiguous subarrays where the product of all the elements in the subarray is strictly less than k.

Clarifying Questions

1. What should be returned if no subarrays meet the requirement?
   • Return 0.
2. Can the elements in nums be negative or zero?
   • The elements will be positive integers as products involving zero or negative numbers can significantly complicate the problem constraints.
3. What is the length range of the nums array?
   • The length of nums can vary, but typically it can be assumed to be reasonably large (e.g., up to (10^4)).
4. What should we return if k is less than or equal to 1?
   • If k is less than or equal to 1, it’s impossible to have any product less than k since all elements are positive integers. Thus, return 0.

Strategy

1. Sliding Window Technique:
   • We will use a two-pointer approach to maintain a sliding window and expand or shrink this window to satisfy the condition of the product being less than k.
   • Initialize two pointers, start and end, both at the beginning of the array.
   • Maintain a variable prod to keep track of the product of elements in the current window.
   • Iterate end from 0 to the end of the array:
     • Multiply prod by the value at nums[end].
     • If prod becomes greater than or equal to k, increment start until prod is less than k, adjusting prod accordingly.
   • Count the number of valid subarrays that end at position end by adding (end - start + 1) to the result.

Code

def numSubarrayProductLessThanK(nums, k):
    if k <= 1:
        return 0
    
    start = 0
    prod = 1
    count = 0
    
    for end in range(len(nums)):
        prod *= nums[end]
        
        while prod >= k and start <= end:
            prod //= nums[start]
            start += 1
        
        count += (end - start + 1)
    
    return count

Time Complexity

• Time Complexity: O(N), where N is the number of elements in nums.
  • In the worst case, each element is processed twice (once by end and potentially once by start).
• Space Complexity: O(1), as the algorithm uses a constant amount of extra space regardless of the input size.

Conclusion

This solution efficiently counts the number of valid subarrays with a product less than k using a sliding window approach. The algorithm runs in linear time, making it suitable for large input sizes.

Try our interview co-pilot at AlgoAdvance.com

• Got blindsided by a question you didn’t expect?

• Spend too much time studying?

• Or simply don’t have the time to go over all 3000 questions?