1What is the main purpose of keeping a binary search tree balanced?
Balanced Search Trees
Easy
A.To store duplicate keys
B.To keep its height small
C.To remove all leaf nodes
D.To sort keys in descending order
Correct Answer: To keep its height small
Explanation:
A balanced search tree keeps its height close to , allowing efficient search, insertion, and deletion.
Incorrect! Try again.
2What is the typical worst-case search time in a balanced binary search tree containing keys?
Balanced Search Trees
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Because the tree height is , a search follows at most levels.
Incorrect! Try again.
3Which operation is commonly used to restore balance in an AVL tree?
Balanced Search Trees
Easy
A.Digit counting
B.Tree rotation
C.Bucket merging
D.Linear probing
Correct Answer: Tree rotation
Explanation:
AVL trees use left and right rotations to restore balance after insertions or deletions.
Incorrect! Try again.
4Counting Sort is most suitable when the input consists of which kind of values?
Counting Sort
Easy
A.Floating-point values without bounds
B.Strings of arbitrary length
C.Graphs with weighted edges
D.Integers from a limited range
Correct Answer: Integers from a limited range
Explanation:
Counting Sort works by counting occurrences of keys, so it is effective when the integer key range is reasonably small.
Incorrect! Try again.
5What does the count array store during the basic counting phase of Counting Sort?
Counting Sort
Easy
A.The position of each swap
B.The depth of each element
C.The digits of each key
D.The frequency of each key
Correct Answer: The frequency of each key
Explanation:
Each count-array entry records how many times its corresponding key occurs in the input.
Incorrect! Try again.
6What is the time complexity of Counting Sort for elements with a key range of size ?
Counting Sort
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Counting Sort processes the input elements and the possible key values, giving time.
Incorrect! Try again.
7How does Radix Sort primarily process numbers?
Radix Sort
Easy
A.One tree level at a time
B.One adjacent pair at a time
C.One random pivot at a time
D.One digit position at a time
Correct Answer: One digit position at a time
Explanation:
Radix Sort repeatedly groups or sorts numbers according to individual digit positions.
Incorrect! Try again.
8In least-significant-digit Radix Sort, which digit is processed first?
Radix Sort
Easy
A.The middle digit
B.The leftmost digit
C.The rightmost digit
D.The largest digit
Correct Answer: The rightmost digit
Explanation:
Least-significant-digit Radix Sort starts with the rightmost digit and moves toward the left.
Incorrect! Try again.
9What property should the digit-sorting method used by Radix Sort have?
Radix Sort
Easy
A.It should use comparisons
B.It should be stable
C.It should be in-place
D.It should be recursive
Correct Answer: It should be stable
Explanation:
A stable digit sort preserves the ordering established for digits processed in earlier passes.
Incorrect! Try again.
10What is the first main step of Bucket Sort?
Bucket Sort
Easy
A.Compare every pair of elements
B.Distribute elements into buckets
C.Build a balanced search tree
D.Reverse the entire input array
Correct Answer: Distribute elements into buckets
Explanation:
Bucket Sort first places elements into buckets based on their values or value ranges.
Incorrect! Try again.
11After the individual buckets have been sorted, how is the final result formed?
Bucket Sort
Easy
A.Reverse the buckets independently
B.Concatenate the buckets in order
C.Select the largest bucket only
D.Merge the buckets randomly
Correct Answer: Concatenate the buckets in order
Explanation:
The sorted buckets are concatenated according to their value ranges to produce the sorted output.
Incorrect! Try again.
12Bucket Sort generally performs well when input values are distributed in what way?
Bucket Sort
Easy
A.Only as identical repeated values
B.Only in strictly decreasing order
C.Randomly across linked lists
D.Roughly uniformly across a range
Correct Answer: Roughly uniformly across a range
Explanation:
A roughly uniform distribution places similar numbers of elements into the buckets, supporting efficient sorting.
Incorrect! Try again.
13What special property does a monotonic stack maintain?
Monotonic Stack
Easy
A.Elements alternate between signs
B.Elements remain at fixed indices
C.Elements always have equal values
D.Elements remain in sorted order
Correct Answer: Elements remain in sorted order
Explanation:
A monotonic stack maintains its elements in either non-increasing or non-decreasing order.
Incorrect! Try again.
14Which problem is commonly solved using a monotonic stack?
Monotonic Stack
Easy
A.Finding a graph's minimum cut
B.Finding the next greater element
C.Building a minimum spanning tree
D.Multiplying two square matrices
Correct Answer: Finding the next greater element
Explanation:
A monotonic stack efficiently tracks candidates for the next greater or next smaller element.
Incorrect! Try again.
15Why can many monotonic-stack algorithms process an array in time?
Monotonic Stack
Easy
A.Each element is compared with every other element
B.The array is always sorted before processing
C.The stack stores only a single element
D.Each element is pushed and popped at most once
Correct Answer: Each element is pushed and popped at most once
Explanation:
Since each element enters and leaves the stack at most once, the total number of stack operations is linear.
Incorrect! Try again.
16Which problem is a common application of a monotonic queue?
Monotonic Queue
Easy
A.Computing matrix determinants
B.Finding all graph cycles recursively
C.Sorting strings alphabetically
D.Finding the maximum in each sliding window
Correct Answer: Finding the maximum in each sliding window
Explanation:
A monotonic queue keeps useful candidates in order, allowing each sliding-window maximum to be found efficiently.
Incorrect! Try again.
17Which data structure is commonly used to implement a monotonic queue?
Monotonic Queue
Easy
A.Binary search tree
B.Undirected graph
C.Double-ended queue
D.Singly linked stack
Correct Answer: Double-ended queue
Explanation:
A double-ended queue permits efficient removal from the front and back while maintaining monotonic order.
Incorrect! Try again.
18In a sorted array, where are two pointers commonly placed when searching for a pair with a target sum?
Two-Pointers Technique
Easy
A.Outside the array bounds
B.At the same middle position
C.At the two ends
D.At two random positions
Correct Answer: At the two ends
Explanation:
One pointer starts at the smallest value and the other at the largest value, and they move inward based on the sum.
Incorrect! Try again.
19What is the usual time complexity of the two-pointers method for finding a target-sum pair in a sorted array?
Two-Pointers Technique
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Each pointer moves across the array at most once, so the search takes time.
Incorrect! Try again.
20If the sum of the two pointed values in an ascending sorted array is less than the target, which pointer is usually moved?
Two-Pointers Technique
Easy
A.Neither pointer is moved
B.The left pointer moves right
C.The right pointer moves left
D.Both pointers move left
Correct Answer: The left pointer moves right
Explanation:
Moving the left pointer right selects a larger value, which can increase the current sum toward the target.
Incorrect! Try again.
21An empty AVL tree receives the keys , , and in that order. Which operation restores balance after the third insertion?
Balanced Search Trees
Medium
A.A single left rotation at node
B.A single right rotation at node
C.A left-right rotation at node
D.A right-left rotation at node
Correct Answer: A single right rotation at node
Explanation:
The insertion creates a left-left imbalance at node . A single right rotation makes the root, with children and .
Incorrect! Try again.
22A balanced binary search tree contains keys. What is the worst-case time complexity of searching for a key?
Balanced Search Trees
Medium
A.
B. because each visited node may require rebalancing during the search
C.
D.
Correct Answer:
Explanation:
A balanced search tree maintains height , and a search follows only one root-to-leaf path.
Incorrect! Try again.
23In a B-tree with minimum degree , what is the maximum number of keys that a node can contain?
Balanced Search Trees
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
A B-tree node can have at most children, so it can contain at most keys.
Incorrect! Try again.
24For the input array , what is the cumulative count for key in counting sort?
Counting Sort
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The values less than or equal to are , so the cumulative count is .
Incorrect! Try again.
25When cumulative counts are used, which implementation choice makes counting sort stable?
Counting Sort
Medium
A.Scan the input from left to right and reset each cumulative count after placement
B.Place each key directly at the index equal to its numeric value
C.Scan the count array from right to left before accumulating counts
D.Scan the input from right to left while placing elements
Correct Answer: Scan the input from right to left while placing elements
Explanation:
Right-to-left placement preserves the original relative order of equal keys while cumulative counts are decremented.
Incorrect! Try again.
26A counting sort implementation must process integer keys ranging from through , inclusive. What is the minimum required size of the count array?
Counting Sort
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The number of distinct integer values is . An offset can map these values to indices through .
Incorrect! Try again.
27Using base- LSD radix sort, how many digit-processing passes are required to sort ?
Radix Sort
Medium
A. passes
B. passes
C. passes
D. passes
Correct Answer: passes
Explanation:
The largest number, , has three decimal digits, so the algorithm processes the ones, tens, and hundreds positions.
Incorrect! Try again.
28After the first, ones-digit pass of stable base- LSD radix sort, what is the order of ?
Radix Sort
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The numbers are grouped by ones digits , while stability preserves the original order within each group.
Incorrect! Try again.
29Why must the per-digit sorting method used by LSD radix sort be stable?
Radix Sort
Medium
A.It reduces the number of required passes from to
B.It preserves the ordering established by earlier digit passes
C.It allows the algorithm to compare complete keys whenever two current digits are equal
D.It ensures that every digit value occurs exactly once per pass
Correct Answer: It preserves the ordering established by earlier digit passes
Explanation:
Stability keeps elements with equal current digits in the order determined by less significant digits processed earlier.
Incorrect! Try again.
30Bucket sort divides the interval into equal buckets using index . Into which zero-based bucket is placed?
Bucket Sort
Medium
A.Bucket
B.Bucket
C.Bucket
D.Bucket
Correct Answer: Bucket
Explanation:
The bucket index is .
Incorrect! Try again.
31If all input values are placed into one bucket and that bucket is sorted using insertion sort, what is the worst-case running time?
Bucket Sort
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
When all values enter one bucket, insertion sort may take quadratic time on that bucket.
Incorrect! Try again.
32For temperatures , how many days must the temperature at index wait for a warmer day?
Monotonic Stack
Medium
A. days
B. days
C. days
D. day
Correct Answer: days
Explanation:
The temperature at index is , and the next warmer temperature is at index , a distance of .
Incorrect! Try again.
33In the array , what is the next strictly smaller element to the right of the value ?
Monotonic Stack
Medium
A.No smaller element
B.
C.
D.
Correct Answer:
Explanation:
Scanning right from , the first value encountered that is smaller than is .
Incorrect! Try again.
34What is the largest rectangle area in the histogram with bar heights ?
Monotonic Stack
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The bars of heights and form a rectangle of width and limiting height , giving area .
Incorrect! Try again.
35For the array with window size , what are the first three sliding-window maximums?
Monotonic Queue
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The first three windows are , , and , with maximums .
Incorrect! Try again.
36A deque is used to compute sliding-window maximums. Which invariant should be maintained?
Monotonic Queue
Medium
A.Indices are stored so their values increase from front to back
B.Indices are stored so their values decrease from front to back
C.All window indices are sorted by value using a full sort after each insertion
D.Only the newest index is retained because older indices cannot become maximums later
Correct Answer: Indices are stored so their values decrease from front to back
Explanation:
A decreasing deque keeps the current maximum at the front. Smaller trailing values are removed when a larger value arrives.
Incorrect! Try again.
37What are the sliding-window minimums for using window size ?
Monotonic Queue
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The four windows have minimums , , , and , respectively.
Incorrect! Try again.
38Using two pointers on the sorted array , which pair sums to the target ?
Two-Pointers Technique
Medium
A. and
B. and
C. and
D. and
Correct Answer: and
Explanation:
The pair and has sum . Two pointers find it by adjusting the endpoints according to the current sum.
Incorrect! Try again.
39A two-pointer algorithm removes duplicates in place from the sorted array . What unique-element count should it return?
Two-Pointers Technique
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The distinct values are , , and , so the compacted prefix has length .
Incorrect! Try again.
40In the container-with-most-water problem, the initial pointers reference heights and at the two ends of . Which pointer should move next?
Two-Pointers Technique
Medium
A.Move both pointers inward
B.Keep both pointers fixed and examine every interior height before deciding which endpoint to replace
C.Move the left pointer inward
D.Move the right pointer inward
Correct Answer: Move the left pointer inward
Explanation:
The area is limited by the shorter height, which is on the left. Moving the taller endpoint cannot improve the limiting height.
Incorrect! Try again.
41Which statement correctly compares AVL-tree rebalancing after a single insertion and after a single deletion? Count a double rotation as two primitive rotations.
Balanced Search Trees
Hard
A.Both operations may require rebuilding every ancestor subtree because restoring balance at one node can invalidate all descendants below it.
B.Insertion may require primitive rotations, while deletion requires at most two.
C.Both insertion and deletion require at most one primitive rotation regardless of tree height.
D.Insertion requires at most two primitive rotations, while deletion may require primitive rotations.
Correct Answer: Insertion requires at most two primitive rotations, while deletion may require primitive rotations.
Explanation:
After insertion, repairing the first unbalanced ancestor restores the subtree height, so at most one single or double rotation is needed. Deletion can reduce subtree height repeatedly, causing rebalancing at ancestors.
Incorrect! Try again.
42A red-black tree contains internal nodes, and its height is measured as the number of edges on the longest root-to-internal-node path. Which worst-case upper bound follows from the red-black invariants?
Balanced Search Trees
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
No path can contain consecutive red nodes, so its length is at most twice the black height. A subtree of black height has at least internal nodes, yielding .
Incorrect! Try again.
43An order-statistic tree stores . During a left rotation at node , let before the rotation. Which size updates are sufficient after the structural pointers are changed?
Balanced Search Trees
Hard
A.Swap and without inspecting any child sizes.
B.Set to the old , then increment by one.
C.Recompute the sizes of all nodes from to the root because a rotation changes the number of descendants at every ancestor.
D.Set to the old , then recompute from its new children.
Correct Answer: Set to the old , then recompute from its new children.
Explanation:
After rotation, roots exactly the subtree previously rooted at , so it inherits the old size of . The new subtree rooted at must then be recomputed from its new children.
Incorrect! Try again.
44Consider records , where subscripts distinguish records with equal keys. Counting sort obtains cumulative counts for keys . Scanning the input from right to left and decrementing after each placement produces which output?
Counting Sort
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The right-to-left scan places later equal-key records into later available positions. It therefore preserves the original relative order of equal keys and makes counting sort stable.
Incorrect! Try again.
45Records have integer fields and must be sorted lexicographically with as the primary key and as the secondary key. Both fields have small ranges. Which sequence of stable counting-sort passes is correct?
Counting Sort
Hard
A.Sort unstably by , then sort stably by .
B.Sort stably by , then sort stably by .
C.Sort stably by , then sort stably by .
D.Sort both fields independently and concatenate their resulting index orders.
Correct Answer: Sort stably by , then sort stably by .
Explanation:
The least significant field is sorted first. The stable pass on then groups records by primary key while preserving their established order by within each group.
Incorrect! Try again.
46An array contains integer keys with . A direct shifted counting sort allocates one counter for every integer in this range. What are its worst-case time and auxiliary-space complexities?
Counting Sort
Hard
A. time and space
B. time and space
C. time and space
D. time and space
Correct Answer: time and space
Explanation:
Direct counting sort costs time and auxiliary space, where is the key-range size. Here , so the range dominates both costs.
Incorrect! Try again.
47An LSD radix sort processes digits from least significant to most significant. What happens if the per-digit sorting procedure correctly orders each digit but is not stable?
Radix Sort
Hard
A.The algorithm remains correct whenever all keys have the same number of digits.
B.The algorithm becomes equivalent to MSD radix sort because an unstable pass automatically gives priority to the most recently processed digit.
C.The algorithm remains correct but its running time increases by one digit pass.
D.The algorithm may destroy ordering established by earlier passes and need not sort correctly.
Correct Answer: The algorithm may destroy ordering established by earlier passes and need not sort correctly.
Explanation:
Stability preserves the ordering by less significant digits among records sharing the current digit. Without it, a later pass can arbitrarily reorder those records and invalidate the final lexicographic digit order.
Incorrect! Try again.
48A radix-sort implementation orders fixed-width -bit words as unsigned integers, but the input must be sorted in ascending two's-complement signed order. Which preprocessing transformation permits the unsigned radix sort to produce the required order?
Radix Sort
Hard
A.Reverse the bytes within every key.
B.Complement all bits of every key.
C.XOR every key with .
D.Take the absolute value of every key.
Correct Answer: XOR every key with .
Explanation:
Flipping the sign bit maps signed order monotonically into unsigned order: negative values move to the lower unsigned half, and nonnegative values move to the upper half while preserving order within each half.
Incorrect! Try again.
49An LSD radix sort handles -bit nonnegative integers using bits per digit and counting sort for each pass. Its cost is . Which choice yields time and auxiliary space?
Radix Sort
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
This choice gives four passes and counters, so the total cost is . Larger listed digits require or counters, while needs passes.
Incorrect! Try again.
50Independent keys follow a known continuous distribution with cumulative distribution function , and can be evaluated in time. Which bucket assignment makes buckets approximately equiprobable and supports expected sorting time with insertion-sorted buckets?
Bucket Sort
Hard
A.Assign to bucket regardless of the support of the distribution.
B.Assign to bucket , clamping the endpoint if needed.
C.Assign to a bucket selected by repeatedly comparing it with every previously inserted key and then balancing all bucket sizes.
D.Assign to bucket determined only by the binary length of .
Correct Answer: Assign to bucket , clamping the endpoint if needed.
Explanation:
For continuous , the probability integral transform makes uniform on . Equal-width buckets in this transformed space therefore have probability about , giving constant expected occupancy.
Incorrect! Try again.
51Suppose bucket receives each independently drawn key with probability , and each bucket is sorted by insertion sort. With buckets and keys, which condition on the probabilities characterizes an expected linear bound through the bucket-squared-cost analysis?
Bucket Sort
Hard
A. only
B.
C. only
D.
Correct Answer:
Explanation:
If is bucket occupancy, then . Thus the insertion-sort work is expected exactly when the collision probability sum is .
Incorrect! Try again.
52To compute the sum of all subarray minimums by assigning each subarray to exactly one occurrence of its minimum, let and be boundary indices for . Which asymmetric tie rule and contribution formula are valid?
Monotonic Stack
Hard
A. and are both previous and next smaller-or-equal; contribute .
B. is previous greater-or-equal and is next strictly greater; contribute .
C. is previous strictly smaller and is next smaller-or-equal; contribute .
D. and are both previous and next strictly smaller; contribute .
Correct Answer: is previous strictly smaller and is next smaller-or-equal; contribute .
Explanation:
Using strict comparison on one side and non-strict comparison on the other gives unique ownership when minima are duplicated. The two distance factors count valid left and right endpoint choices.
Incorrect! Try again.
53In the standard largest-rectangle-in-a-histogram algorithm, index is popped when processing index because . After the pop, the new stack top is index . What maximal width is associated with height ?
Monotonic Stack
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
After is popped, is the nearest smaller boundary on the left and is the first smaller boundary on the right. The usable interval is , of width .
Incorrect! Try again.
54For the circular next-strictly-greater-element problem, an algorithm scans indices modulo for iterations. It pushes each original index only during the first iterations and pops while the current value is strictly greater than the value at the stored index. Which claim is correct?
Monotonic Stack
Hard
A.Its time is because the second traversal repeats all stack comparisons.
B.Its time is whenever the array is nonincreasing.
C.Its time is because every original index is pushed once and popped at most once.
D.It incorrectly handles equal values because equal elements must always be treated as strictly greater in a circular traversal.
Correct Answer: Its time is because every original index is pushed once and popped at most once.
Explanation:
Although there are scan iterations, stack operations are amortized: each of the indices is pushed once and popped at most once. Equal values correctly remain unresolved when a strictly greater element is required.
Incorrect! Try again.
55For the shortest nonempty subarray with sum at least , let be prefix sums and maintain candidate indices in a deque. Which processing rule is correct for each ?
Monotonic Queue
Hard
A.Pop front while , then pop back while , and finally append .
B.Append first, sort the deque by index, and remove only candidates whose prefix sums equal .
C.Pop back while , then pop front while , and finally append .
D.Remove every negative prefix sum before testing feasibility because negative prefixes cannot begin a minimum-length qualifying subarray.
Correct Answer: Pop front while , then pop back while , and finally append .
Explanation:
Feasible front indices are removed after updating the minimum length because later endpoints only make them longer. Back indices with greater or equal prefix sums are dominated by the newer index .
Incorrect! Try again.
56A deque is used to compute sliding-window maxima of width . To keep the newest index among equal maximum values, which update rule should be applied at index ?
Monotonic Queue
Hard
A.Expire indices above , pop the front while , then append .
B.Expire indices below , pop the back while , then append .
C.Retain every equal value and linearly search the entire deque after each window shift to identify the newest maximum.
D.Expire indices below , pop the back while , then append .
Correct Answer: Expire indices below , pop the back while , then append .
Explanation:
Indices smaller than are outside the current window. Popping values less than or equal to maintains decreasing values and discards older equal values in favor of the newer index.
Incorrect! Try again.
57Consider with given. Which monotonic-deque order computes all states in time?
Monotonic Queue
Hard
A.Read the back for the minimum, expire indices greater than , pop front while , then append .
B.Expire indices less than , read the front for the minimum, pop back while , then append .
C.Sort all indices in the current window by both index and cost after computing every new dynamic-programming state.
D.Append before computing , then use the deque front even if it refers to the same state.
Correct Answer: Expire indices less than , read the front for the minimum, pop back while , then append .
Explanation:
Expiration enforces the valid predecessor interval. Increasing deque values place the minimum at the front, and dominated states are removed from the back; every index is inserted and removed at most once.
Incorrect! Try again.
58For an array of strictly positive integers, a sliding window counts subarrays whose product is strictly less than . Which rule correctly handles all values of ?
Two-Pointers Technique
Hard
A.If , return ; otherwise shrink while the product is greater than and add .
B.For every , shrink while the product is at most and add the current window length.
C.For , retain zero-length windows as valid and count one additional subarray for every right endpoint.
D.If , return ; otherwise shrink while the product is at least and add .
Correct Answer: If , return ; otherwise shrink while the product is at least and add .
Explanation:
Every nonempty product is at least , so no valid subarray exists when . Otherwise, after shrinking, all suffixes ending at have product below .
Incorrect! Try again.
59In the container-with-most-water problem, pointers start at both ends. Why is moving the pointer at the shorter line a safe elimination step?
Two-Pointers Technique
Hard
A.The taller line can never participate in any globally optimal container.
B.Moving the shorter line guarantees that the next area is strictly larger.
C.The shorter line must be smaller than every line located between the two current pointers, so it cannot be part of an optimal pair.
D.Keeping the shorter line while reducing width cannot produce a larger area, regardless of the new opposite height.
Correct Answer: Keeping the shorter line while reducing width cannot produce a larger area, regardless of the new opposite height.
Explanation:
The current shorter line bounds the height. Pairing it with any inward line reduces the width without increasing that bound, so no better solution using that fixed shorter endpoint is possible.
Incorrect! Try again.
60A sorted array may contain duplicates. To count index pairs with and in time, what should be done when the two pointer values form a valid sum?
Two-Pointers Technique
Hard
A.If the values differ, add both run lengths; if equal, add and continue scanning every interior pair.
B.If the values differ, multiply their run lengths; if equal, add and terminate.
C.Multiply the distances of both pointers from the array endpoints.
D.Always add one pair and move both pointers by exactly one position.
Correct Answer: If the values differ, multiply their run lengths; if equal, add and terminate.
Explanation:
Different endpoint values contribute every cross-combination of their duplicate runs. If the endpoint values are equal and sum to , all remaining values are equal, so every pair among the remaining elements is valid.
Incorrect! Try again.
Did this save you a night before the exam?
LPU Notes is free, and it stays free. Ads cover part of the server bill.
The rest comes out of a student's own pocket: the domain, the storage,
and keeping the site up through the weeks everyone needs it at once.
The payment button didn't load. An ad blocker or a filtered network is the usual reason.
to try again.
Nothing here is ever locked, and nothing unlocks. Chip in only if it was worth it.
What it pays for →