Count of Interesting Subarrays

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Arrays

You are given a 0-indexed integer array nums, an integer modulo, and an integer k.

Your task is to find the count of subarrays that are interesting.

A subarray nums[l..r] is interesting if the following condition holds:

Let cnt be the number of indices i in the range [l, r] such that nums[i] % modulo == k. Then, cnt % modulo == k.

Return an integer denoting the count of interesting subarrays.

Note: A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [3,2,4], modulo = 2, k = 1
Output: 3
Explanation: In this example the interesting subarrays are: 
The subarray nums[0..0] which is [3]. 
- There is only one index, i = 0, in the range [0, 0] that satisfies nums[i] % modulo == k. 
- Hence, cnt = 1 and cnt % modulo == k.  
The subarray nums[0..1] which is [3,2].
- There is only one index, i = 0, in the range [0, 1] that satisfies nums[i] % modulo == k.  
- Hence, cnt = 1 and cnt % modulo == k.
The subarray nums[0..2] which is [3,2,4]. 
- There is only one index, i = 0, in the range [0, 2] that satisfies nums[i] % modulo == k. 
- Hence, cnt = 1 and cnt % modulo == k. 
It can be shown that there are no other interesting subarrays. So, the answer is 3.

Example 2:

Input: nums = [3,1,9,6], modulo = 3, k = 0
Output: 2
Explanation: In this example the interesting subarrays are: 
The subarray nums[0..3] which is [3,1,9,6]. 
- There are three indices, i = 0, 2, 3, in the range [0, 3] that satisfy nums[i] % modulo == k. 
- Hence, cnt = 3 and cnt % modulo == k. 
The subarray nums[1..1] which is [1]. 
- There is no index, i, in the range [1, 1] that satisfies nums[i] % modulo == k. 
- Hence, cnt = 0 and cnt % modulo == k. 
It can be shown that there are no other interesting subarrays. So, the answer is 2.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
1 <= modulo <= 10⁹
0 <= k < modulo

Try coding it in LeetCode

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:

What are the constraints on the size of the `nums` array and the values within it, as well as the values of `modulo` and `k`?
Can `modulo` be zero? If so, what should I do?
If no interesting subarrays exist, what value should I return?
Are overlapping subarrays counted as distinct subarrays if they meet the 'interesting' criteria?
Are the elements in the input array guaranteed to be integers?

Brute Force Solution

Approach

We want to find special groups within a list of numbers. The most straightforward approach is to simply check every possible group, one by one. We look at all possible starting points and all possible ending points for each group.

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

Start with the first number in the list and consider it as a group by itself.
Check if that single number group is 'interesting' based on our criteria.
Now consider the first two numbers as a group, and check if that group is 'interesting'.
Continue adding one more number to the group at a time, checking if each new group is 'interesting', until we've included all the numbers in the list.
Next, start with the second number in the list, and repeat the same process: check the single number group, then the group of two numbers, then the group of three, and so on.
Keep doing this, shifting the starting point to the next number in the list each time, until we've considered every possible group.
Finally, count how many of these groups were 'interesting' according to the rules.

Code Implementation

def count_interesting_subarrays_brute_force(numbers, modulo_value, k_value):
    number_of_interesting_subarrays = 0
    
    for start_index in range(len(numbers)):
        for end_index in range(start_index, len(numbers)):
            # Create subarray from start to end indices
            subarray = numbers[start_index:end_index+1]

            odd_count = 0
            for number in subarray:
                if number % 2 != 0:
                    odd_count += 1

            # Check if subarray is interesting
            if (odd_count * k_value) % modulo_value == 0:
                # Increase number of interesting subarrays
                number_of_interesting_subarrays += 1

    return number_of_interesting_subarrays

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible subarrays. The outer loop iterates 'n' times (where 'n' is the size of the input array), representing the start index of a subarray. The inner loop, for each start index, iterates up to 'n' times, representing the end index of the subarray. Thus, the number of operations is proportional to the sum of integers from 1 to n, which can be expressed as n * (n+1) / 2. This simplifies to O(n²).

Space Complexity

O(1) – The provided solution iterates through all possible subarrays using nested loops. It doesn't create any auxiliary data structures like lists, maps, or sets to store intermediate results or track visited elements. The operations only involve iterating and checking conditions on subarrays directly derived from the input list. Therefore, the space used is constant regardless of the input list's size N.

Optimal Solution

Approach

The problem asks us to efficiently count specific sections within a larger list that meet a certain condition. The smart way to solve this is to keep track of how many sections fulfill the condition *up to each point* in the list and then use this information to quickly calculate the count of interesting sections without rechecking any section multiple times.

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

First, transform the original list into a new list. Each element in the new list will represent whether the corresponding element in the original list meets our special criteria (like being divisible by a certain number). This simplifies the problem.
Next, create another list that stores a running total. Each element in this list will be the sum of all the 'special' elements seen *up to that point* in the transformed list. This running total helps us count sections quickly.
Now, go through each possible starting point in the original list. For each starting point, consider all possible ending points.
Instead of counting the 'special' elements within each section from scratch, use the running total list. The difference between the running total at the ending point and the running total at the starting point (minus one) gives us the count of 'special' elements in that section.
Finally, check if the count of 'special' elements in the section satisfies the condition defined in the problem (like being divisible by a certain number). If it does, increase our overall count of 'interesting' sections.
By using the running total list, we avoid repeatedly counting the same 'special' elements, making the solution much faster than checking every section individually.

Code Implementation

def count_interesting_subarrays(array, modulo, target):
    modified_array = [1 if element % modulo == target else 0 for element in array]

    prefix_sum_array = [0]
    current_prefix_sum = 0
    for element in modified_array:
        current_prefix_sum += element
        prefix_sum_array.append(current_prefix_sum)

    interesting_subarray_count = 0

    # Iterate through all possible start indices
    for start_index in range(len(array)):
        # Iterate through all possible end indices
        for end_index in range(start_index, len(array)):

            # Efficiently calculate subarray sum using prefix sums
            subarray_sum = prefix_sum_array[end_index + 1] - prefix_sum_array[start_index]

            # Check if the subarray sum is interesting
            if subarray_sum % modulo == 0:
                interesting_subarray_count += 1

    return interesting_subarray_count

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible subarrays of the input array of size n. Specifically, it uses a nested loop structure where the outer loop considers each element as a potential starting point, and the inner loop iterates through all subsequent elements to define the ending point of the subarray. Therefore, for each of the n possible starting positions, there are approximately n ending positions to consider, leading to a total of approximately n * n/2 operations. This results in a time complexity of O(n²).

Space Complexity

O(N) – The algorithm creates a new list of size N to store whether each element meets the criteria, and another list of size N to store the running totals. These two lists are the dominant consumers of auxiliary space. The space used is therefore directly proportional to the input size N. Thus, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Empty input array	Return 0 immediately as there are no subarrays to consider.
Modulo is 0	Division by zero is undefined; return 0 or throw an IllegalArgumentException based on requirements.
k is not within the range [0, modulo)	Return 0 or throw an IllegalArgumentException as k must be a valid remainder when divided by modulo.
Single element array	Check if the single element satisfies the divisibility and remainder condition directly.
Large input array (performance)	Use a time complexity O(n) approach like prefix sum and hashmap to avoid timeouts.
Array contains negative numbers	Ensure the modulo operation handles negative numbers correctly (e.g., using (num % modulo + modulo) % modulo to ensure a positive remainder).
All elements are divisible by modulo	The prefix sum can increase rapidly, and the count can become large, which should be handled correctly by the chosen data type (long if necessary).
No interesting subarrays exist	The solution should correctly return 0 when no subarray satisfies the condition.

Empty input array

How to Handle:

Return 0 immediately as there are no subarrays to consider.

Modulo is 0

How to Handle:

Division by zero is undefined; return 0 or throw an IllegalArgumentException based on requirements.

k is not within the range [0, modulo)

How to Handle:

Return 0 or throw an IllegalArgumentException as k must be a valid remainder when divided by modulo.

Single element array

How to Handle:

Check if the single element satisfies the divisibility and remainder condition directly.

Large input array (performance)

How to Handle:

Use a time complexity O(n) approach like prefix sum and hashmap to avoid timeouts.

Array contains negative numbers

How to Handle:

Ensure the modulo operation handles negative numbers correctly (e.g., using (num % modulo + modulo) % modulo to ensure a positive remainder).

All elements are divisible by modulo

How to Handle:

The prefix sum can increase rapidly, and the count can become large, which should be handled correctly by the chosen data type (long if necessary).

No interesting subarrays exist

How to Handle:

The solution should correctly return 0 when no subarray satisfies the condition.