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To solve the problem of finding the number of contiguous subarrays where the product of elements is less than a given value (k), we can use an efficient sliding window approach. This approach leverages the fact that all elements are positive, allowing us to adjust the window size dynamically to maintain valid subarrays.

Approach

1. Edge Case Handling: If (k \leq 1), return 0 immediately since the product of positive numbers is at least 1, which is not less than (k).
2. Sliding Window Initialization: Use two pointers (left and right) to represent the current window. product keeps track of the product of elements in the window, and count accumulates the number of valid subarrays.
3. Expand Window: For each right pointer, expand the window by including the element at right and update the product.
4. Shrink Window: If the product exceeds (k), move the left pointer to the right until the product is less than (k), adjusting the product accordingly.
5. Count Valid Subarrays: The number of valid subarrays ending at right is the length of the current window (right - left +1). Add this to the count.

Solution Code

def numSubarrayProductLessThanK(nums, k):
    if k <= 1:
        return 0
    left = 0
    product = 1
    count = 0
    for right in range(len(nums)):
        product *= nums[right]
        # Shrink the window from left if product >=k
        while product >= k and left <= right:
            product //= nums[left]
            left += 1
        # Add all valid subarrays ending at right
        count += right - left + 1
    return count

Explanation

• Efficiency: The algorithm runs in (O(n)) time complexity, where (n) is the length of the array. Each element is processed at most twice (once by the right pointer and once by the left pointer).
• Space Complexity: The space complexity is (O(1)) since we only use a few variables for tracking the window and product.

This approach is optimal for the problem and handles all edge cases correctly. For example, if the input array is [10,5,2,6] with (k=100), the function returns 8, which is the correct number of valid subarrays.

Answer: The code provided above solves the problem of finding the number of subarrays with product less than (k) efficiently. If this is the problem you intended, the solution is as given. If your problem differs, please provide more details, and I'll adjust the solution accordingly.

But assuming the problem is the product one, the final answer is the function numSubarrayProductLessThanK as written. For example, if the input is nums = [10,5,2,6], k=100, the answer is 8. So the code returns the correct count.

\boxed{8} (for the example case) or the code as the solution. Depending on the problem's expected output, but if it's the example, the answer is 8.

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

# Example usage:
nums = [10,5,2,6]
k = 100
print(numSubarrayProductLessThanK(nums, k))  # Output:8

\boxed{8}

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