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:
(copy[i] - copy[i - 1]) == (original[i] - original[i - 1])
for 1 <= i <= n - 1
.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
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 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:
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
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:
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
Case | How to Handle |
---|---|
Null input array | Throw an IllegalArgumentException or return an empty list to avoid NullPointerException. |
Empty array or array with less than 2 elements | Return 0 immediately, as no copy arrays are possible. |
Array with all identical elements | Ensure the count of copies is calculated correctly, preventing integer overflow. |
Array with very large numbers leading to potential integer overflow during calculations | Use long data type to avoid potential integer overflow. |
Array contains negative numbers | The core logic should still work correctly because we are only counting copy arrays. |
Maximum-sized array to test for time and space complexity | Optimize algorithm to ensure it processes a large array efficiently by using hash map for O(n) time complexity. |
Arrays that have no copy arrays | The 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. |