You are given a stream of n videos, each represented by a distinct number from 1 to n that you need to "upload" to a server. You need to implement a data structure that calculates the length of the longest uploaded prefix at various points in the upload process.
We consider i to be an uploaded prefix if all videos in the range 1 to i (inclusive) have been uploaded to the server. The longest uploaded prefix is the maximum value of i that satisfies this definition.
Implement the LUPrefix class:
LUPrefix(int n) Initializes the object for a stream of n videos.void upload(int video) Uploads video to the server.int longest() Returns the length of the longest uploaded prefix defined above.Example 1:
Input
["LUPrefix", "upload", "longest", "upload", "longest", "upload", "longest"]
[[4], [3], [], [1], [], [2], []]
Output
[null, null, 0, null, 1, null, 3]
Explanation
LUPrefix server = new LUPrefix(4); // Initialize a stream of 4 videos.
server.upload(3); // Upload video 3.
server.longest(); // Since video 1 has not been uploaded yet, there is no prefix.
// So, we return 0.
server.upload(1); // Upload video 1.
server.longest(); // The prefix [1] is the longest uploaded prefix, so we return 1.
server.upload(2); // Upload video 2.
server.longest(); // The prefix [1,2,3] is the longest uploaded prefix, so we return 3.
Constraints:
1 <= n <= 1051 <= video <= nvideo are distinct.2 * 105 calls in total will be made to upload and longest.longest.When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The brute force approach is like trying every possible scenario. For this problem, it involves checking prefixes of uploaded files one by one to see how long the valid sequence is.
Here's how the algorithm would work step-by-step:
def longest_uploaded_prefix_brute_force(uploaded_files, number_of_files):
longest_prefix = 0
for i in range(1, number_of_files + 1):
# Check each prefix of uploaded files.
is_valid_prefix = True
for j in range(1, i + 1):
if j not in uploaded_files:
is_valid_prefix = False
break
# If the prefix is valid, update the longest prefix.
if is_valid_prefix:
longest_prefix = i
return longest_prefixThe problem involves finding the longest sequence of uploaded file pieces that are contiguous. The optimal strategy involves keeping track of which pieces have been uploaded and efficiently determining the length of the longest starting sequence. It avoids repeatedly checking the same pieces.
Here's how the algorithm would work step-by-step:
def longest_uploaded_prefix(number_of_files, uploaded_files):
uploaded = [False] * (number_of_files + 1)
for file_index in uploaded_files:
uploaded[file_index] = True
longest_prefix = 0
# Start checking from the beginning of the file.
for i in range(1, number_of_files + 1):
# Stop if we find a piece that hasn't been uploaded.
if not uploaded[i]:
break
longest_prefix += 1
# The number of pieces checked is the length of longest prefix.
return longest_prefix| Case | How to Handle |
|---|---|
| Null or empty 'uploaded' array | Return 0 immediately, as no prefix can be formed. |
| 'uploaded' array with only one element | Return 0 if the element is not 1, otherwise return 1 if it is 1. |
| 'uploaded' array with a large number of elements (close to maximum integer size) | Ensure the solution uses a data structure and algorithm with optimal time complexity to avoid timeouts, like a set. |
| 'uploaded' array contains duplicate values | The set will only store unique uploaded IDs, so duplicates should not affect the correctness of the prefix calculation. |
| 'uploaded' array contains out-of-order IDs (e.g., [3, 1, 2]) | The algorithm should correctly identify that only ID 1 is uploaded first, then 2 when added, then 3 when added. |
| 'uploaded' array missing some IDs in the expected sequence (e.g., [1, 3, 4]) | The algorithm should correctly identify the longest contiguous prefix, stopping at 1 in this case since 2 is missing. |
| 'uploaded' array with extreme boundary values (close to MAX_INT) | The algorithm should still work correctly as long as it correctly verifies all integers up to the uploaded value. |
| Very large N with memory usage considerations | Using a boolean array or set data structure should be efficient in space, but consider the constraints of the problem. |