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Find the Number of Copy Arrays

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Arrays

You are given an array original of length n and a 2D array bounds of length n x 2, where bounds[i] = [ui, vi].

You need to find the number of possible arrays copy of length n such that:

  1. (copy[i] - copy[i - 1]) == (original[i] - original[i - 1]) for 1 <= i <= n - 1.
  2. ui <= copy[i] <= vi for 0 <= i <= n - 1.

Return the number of such arrays.

Example 1:

Input: original = [1,2,3,4], bounds = [[1,2],[2,3],[3,4],[4,5]]

Output: 2

Explanation:

The possible arrays are:

  • [1, 2, 3, 4]
  • [2, 3, 4, 5]

Example 2:

Input: original = [1,2,3,4], bounds = [[1,10],[2,9],[3,8],[4,7]]

Output: 4

Explanation:

The possible arrays are:

  • [1, 2, 3, 4]
  • [2, 3, 4, 5]
  • [3, 4, 5, 6]
  • [4, 5, 6, 7]

Example 3:

Input: original = [1,2,1,2], bounds = [[1,1],[2,3],[3,3],[2,3]]

Output: 0

Explanation:

No array is possible.

Constraints:

  • 2 <= n == original.length <= 105
  • 1 <= original[i] <= 109
  • bounds.length == n
  • bounds[i].length == 2
  • 1 <= bounds[i][0] <= bounds[i][1] <= 109

Solution


Clarifying Questions

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:

  1. What data types will be present in the array, and can I expect any null or empty values within the array?
  2. What is the expected size range of the input array? Are there any memory limitations I should be aware of?
  3. If no such copy array exists, what should the function return?
  4. Are we looking for exact copies (same values in the same order), or should I consider permutations or subsets as "copy arrays"?
  5. Are there any constraints on the elements within the copied array, like a minimum or maximum value or uniqueness requirements?

Brute Force Solution

Approach

The brute force method is all about checking every single possibility to see if it fits what we're looking for. To find copied arrays, we will compare each array with all other arrays.

Here's how the algorithm would work step-by-step:

  1. Take the first array in the list.
  2. Compare this array to every other array in the list, one at a time.
  3. If the first array is identical to another array, mark that we found one identical array.
  4. Repeat this process by taking the second array and comparing it to every other array (including the first one).
  5. Continue this process for each array in the list, comparing it against all other arrays.
  6. Count how many times you marked that an array was identical to another array. This is the number of copy arrays.

Code Implementation

def find_number_of_copy_arrays_brute_force(list_of_arrays):
    number_of_copy_arrays = 0
    number_of_arrays = len(list_of_arrays)

    for first_array_index in range(number_of_arrays):
        for second_array_index in range(number_of_arrays):
            # Avoid comparing the same array to itself
            if first_array_index == second_array_index:
                continue

            # Compare each array to every other array
            if list_of_arrays[first_array_index] == list_of_arrays[second_array_index]:
                # If they are the same, increment the number of copies
                number_of_copy_arrays += 1

    # Divide by two to avoid double counting
    number_of_copy_arrays = number_of_copy_arrays // 2

    return number_of_copy_arrays

Big(O) Analysis

Time Complexity
O(n²)The provided algorithm iterates through each of the n arrays in the input list. For each array, it compares it to every other array in the list, resulting in n comparisons in the worst case. Therefore, we have a nested loop structure where the outer loop iterates n times and the inner loop iterates n times. This means the algorithm performs approximately n * n operations, which simplifies to O(n²).
Space Complexity
O(1)The described brute force method primarily involves iterating through the input arrays and comparing them directly. It does not require creating any auxiliary data structures like new arrays, hash maps, or sets to store intermediate results or track visited arrays. Only a few constant space variables might be used for loop indices and comparison flags. Therefore, the space complexity remains constant regardless of the number of arrays or their sizes.

Optimal Solution

Approach

The key is to identify repeated segments within the larger array. We can efficiently count these repeats by using a way to quickly tell if two array segments are identical without checking each element individually.

Here's how the algorithm would work step-by-step:

  1. First, create a way to represent each segment of the main array with a short, unique code.
  2. Now, go through each possible length of a 'copy' array, starting from length 1.
  3. For each length, slide a 'window' of that size across the main array.
  4. At each position, check if the segment under the window matches any previous segment of the same length.
  5. If a match is found (using the short code from before), we know we have another occurrence of that 'copy' array.
  6. Keep a count of all the distinct copy arrays you've found.

Code Implementation

def find_number_of_copy_arrays(main_array):
    array_length = len(main_array)
    distinct_copy_arrays_count = 0
    seen_hashes = set()

    for copy_array_length in range(1, array_length):
        for i in range(array_length - copy_array_length + 1):
            sub_array = main_array[i:i + copy_array_length]
            # Using a simple hash for segment identification.
            segment_hash = hash(tuple(sub_array))

            if segment_hash not in seen_hashes:
                # Keep track of distinct copy arrays.
                seen_hashes.add(segment_hash)
                distinct_copy_arrays_count += 1

    return distinct_copy_arrays_count

Big(O) Analysis

Time Complexity
O(n³)The algorithm iterates through all possible lengths of the copy array, which contributes a factor of n to the time complexity. For each length, it slides a window of that size across the main array, resulting in another factor of n. Within each window, the comparison of the segment to previously seen segments of the same length requires computing and comparing the unique codes, which takes O(n) in the worst case (if the hash function has collisions or if the comparison of the codes themselves takes linear time relative to the segment size). Therefore, the overall time complexity is O(n * n * n) = O(n³).
Space Complexity
O(N)The algorithm uses a technique to represent each segment of the array with a short unique code. This likely involves storing these codes for previously seen segments of each length to check for matches. The number of possible array segments is related to the size of the main array. In the worst case, storing all these codes would take space proportional to N, where N is the length of the main array. Therefore, the auxiliary space complexity is O(N).

Edge Cases

CaseHow to Handle
Null input arrayThrow an IllegalArgumentException or return an empty list to avoid NullPointerException.
Empty array or array with less than 2 elementsReturn 0 immediately, as no copy arrays are possible.
Array with all identical elementsEnsure the count of copies is calculated correctly, preventing integer overflow.
Array with very large numbers leading to potential integer overflow during calculationsUse long data type to avoid potential integer overflow.
Array contains negative numbersThe core logic should still work correctly because we are only counting copy arrays.
Maximum-sized array to test for time and space complexityOptimize algorithm to ensure it processes a large array efficiently by using hash map for O(n) time complexity.
Arrays that have no copy arraysThe algorithm should return 0 copies correctly when no copy arrays exist.
Array with extreme boundary values (Integer.MAX_VALUE, Integer.MIN_VALUE)Handle cases to prevent potential issues during arithmetic operations, possibly by converting to long type.