Problem Description

You have a 0-indexed integer array nums that undergoes a cyclic transformation process starting at minute 0:

Phase 1 - Removal Phase: Every minute, the leftmost element is removed from nums until the array becomes empty. This takes n minutes (where n is the length of the original array).

Phase 2 - Restoration Phase: Every minute, one element is appended back to the end of nums in the same order they were removed, until the original array is restored. This also takes n minutes.

This entire cycle of 2n minutes repeats indefinitely.

For example, with nums = [0,1,2]:

• Minute 0: [0,1,2] (original)
• Minute 1: [1,2] (removed 0)
• Minute 2: [2] (removed 1)
• Minute 3: [] (removed 2)
• Minute 4: [0] (added back 0)
• Minute 5: [0,1] (added back 1)
• Minute 6: [0,1,2] (added back 2, cycle complete)
• Minute 7: [1,2] (cycle repeats...)

You're given a 2D array queries where each queries[j] = [time_j, index_j] asks: "What is the value at position index_j in the array at minute time_j?"

For each query, return:

• The value nums[index_j] if index_j is a valid index at minute time_j
• -1 if index_j is out of bounds at minute time_j

The solution leverages the cyclic nature of the transformation. Since the pattern repeats every 2n minutes, we can use modulo arithmetic (t % 2n) to determine the state of the array at any given time:

• When t < n: The array is in the removal phase with n - t elements remaining, starting from index t of the original array
• When t > n: The array is in the restoration phase with t - n elements restored, which are the first t - n elements of the original array
• When t = n: The array is empty
• When t = 0 or t = 2n: The array is in its original state

Intuition

The key insight is recognizing that the array transformation follows a predictable, cyclic pattern that repeats every 2n minutes. Instead of simulating the entire process minute by minute, we can directly calculate what the array looks like at any given time.

Let's think about what happens during one complete cycle:

• For the first n minutes (removal phase), we're removing elements from the left, which is equivalent to the array "sliding left" - the element at position 0 in the current array was originally at position t in the original array
• For the next n minutes (restoration phase), we're building the array back up from empty, adding elements in their original order

Since the pattern repeats every 2n minutes, we can use modulo arithmetic to map any time t to its equivalent position within a single cycle: t % (2n).

Now, for any query at time t asking for index i:

During removal phase (t < n):

• The array has n - t elements remaining
• These elements are the last n - t elements of the original array
• The element at index i in the current array corresponds to index i + t in the original array
• So if i < n - t, the answer is nums[i + t]

During restoration phase (t > n):

• The array has t - n elements (we've added back t - n elements)
• These are the first t - n elements of the original array in their original positions
• So if i < t - n, the answer is nums[i]

At the boundary (t = n):

• The array is empty, so any query returns -1

This mathematical relationship allows us to answer each query in O(1) time without actually simulating the transformation process.

Solution Approach

The implementation follows a direct calculation approach based on the cyclic pattern we identified:

1. Initialize the answer array: Create an array ans of length m (number of queries) and initialize all values to -1. This handles the default case where an index is out of bounds.

2. Process each query: For each query (t, i) in the queries array:

   a. Normalize the time: Calculate t %= 2 * n to map any time to its position within a single cycle of 2n minutes.

   b. Check removal phase: If t < n, the array is in the removal phase:

   • The array currently has n - t elements
   • These elements start from index t of the original array
   • If i < n - t (index is valid), then ans[j] = nums[i + t]
   • The mapping i + t gives us the corresponding index in the original array

   c. Check restoration phase: If t > n, the array is in the restoration phase:

   • The array currently has t - n elements
   • These are the first t - n elements of the original array
   • If i < t - n (index is valid), then ans[j] = nums[i]
   • No offset is needed since elements are in their original positions

   d. Handle empty array: When t = n, the array is empty, so we keep the default value of -1.

3. Return the result: After processing all queries, return the ans array.

The algorithm runs in O(m) time where m is the number of queries, with O(1) time per query. The space complexity is O(m) for storing the answer array.

The elegance of this solution lies in avoiding simulation - we directly compute the state of the array at any given time using mathematical relationships, making it highly efficient even for large time values.

Example Walkthrough

Let's walk through the solution with nums = [3, 5, 7] and queries = [[2, 0], [4, 1], [5, 2]].

First, we identify that n = 3, so one complete cycle takes 2n = 6 minutes.

Initial Setup:

• Create answer array: ans = [-1, -1, -1]

Query 1: [time=2, index=0]

• Normalize time: t = 2 % 6 = 2
• Since t = 2 < n = 3, we're in the removal phase
• Array has n - t = 3 - 2 = 1 element remaining
• Check if index = 0 < 1: Yes, it's valid
• The element at index 0 in current array maps to index 0 + 2 = 2 in original array
• ans[0] = nums[2] = 7
• Current array state: [7] (elements at indices 0 and 1 were removed)

Query 2: [time=4, index=1]

• Normalize time: t = 4 % 6 = 4
• Since t = 4 > n = 3, we're in the restoration phase
• Array has t - n = 4 - 3 = 1 element restored
• Check if index = 1 < 1: No, index 1 is out of bounds
• ans[1] remains -1
• Current array state: [3] (only first element has been restored)

Query 3: [time=5, index=2]

• Normalize time: t = 5 % 6 = 5
• Since t = 5 > n = 3, we're in the restoration phase
• Array has t - n = 5 - 3 = 2 elements restored
• Check if index = 2 < 2: No, index 2 is out of bounds (we only have indices 0 and 1)
• ans[2] remains -1
• Current array state: [3, 5] (first two elements have been restored)

Final Result: ans = [7, -1, -1]

This example demonstrates how we directly calculate the answer for each query without simulating the entire transformation process, using the mathematical relationships between the current array state and the original array.

Solution Implementation

Time and Space Complexity

The time complexity is O(m), where m is the length of the queries array. The algorithm iterates through each query exactly once, and for each query, it performs constant-time operations: computing t % (2 * n), comparing values, and potentially accessing an array element. Since the loop runs m times and each iteration takes O(1) time, the overall time complexity is O(m).

The space complexity is O(1) when excluding the space consumed by the answer array ans. The algorithm only uses a fixed amount of extra space for variables n, m, j, t, and i, regardless of the input size. If we include the answer array, the space complexity would be O(m) since we create an array of size m to store the results.

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

Common Pitfalls

1. Forgetting to Handle the Empty Array State

A critical pitfall is not properly handling when time_in_cycle == array_length. At this exact moment, the array is completely empty, and any index query should return -1.

Incorrect Implementation:

if time_in_cycle <= array_length:  # Wrong! Includes the empty state
    if target_index < array_length - time_in_cycle:
        result[query_idx] = nums[target_index + time_in_cycle]

Why it fails: When time_in_cycle == array_length, the expression array_length - time_in_cycle equals 0, but trying to access nums[target_index + array_length] will cause an index out of bounds error.

Correct Implementation:

if time_in_cycle < array_length:  # Strict inequality excludes empty state
    if target_index < array_length - time_in_cycle:
        result[query_idx] = nums[target_index + time_in_cycle]

2. Incorrect Time Normalization for Negative or Zero-Length Arrays

Another pitfall is not considering edge cases in time normalization or array length.

Problematic Code:

time_in_cycle = time % (2 * array_length)  # Fails when array_length is 0

Solution: Add a guard clause for empty input arrays:

if array_length == 0:
    return [-1] * num_queries
time_in_cycle = time % (2 * array_length) if array_length > 0 else 0

3. Off-by-One Errors in Index Mapping

Mixing up the index mapping during the removal phase is common. The key insight is that when t elements have been removed, the current array starts from index t of the original array.

Wrong Mapping:

# Incorrect: forgetting to add the offset
if time_in_cycle < array_length:
    if target_index < array_length - time_in_cycle:
        result[query_idx] = nums[target_index]  # Missing offset!

Correct Mapping:

if time_in_cycle < array_length:
    if target_index < array_length - time_in_cycle:
        result[query_idx] = nums[target_index + time_in_cycle]  # Correct offset

4. Misunderstanding the Restoration Phase Bounds

During restoration, the array has time_in_cycle - array_length elements, not time_in_cycle elements.

Incorrect Bound Check:

elif time_in_cycle > array_length:
    if target_index < time_in_cycle:  # Wrong bound!
        result[query_idx] = nums[target_index]

Correct Bound Check:

elif time_in_cycle > array_length:
    if target_index < time_in_cycle - array_length:  # Correct bound
        result[query_idx] = nums[target_index]

These pitfalls highlight the importance of carefully tracking array bounds and state transitions throughout the cyclic transformation process.