Subarray With Elements Greater Than Varying Threshold

Hard

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Topics:

ArraysSliding Windows

You are given an integer array nums and an integer threshold.

Find any subarray of nums of length k such that every element in the subarray is greater than threshold / k.

Return the size of any such subarray. If there is no such subarray, return -1.

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

Example 1:

Input: nums = [1,3,4,3,1], threshold = 6
Output: 3
Explanation: The subarray [3,4,3] has a size of 3, and every element is greater than 6 / 3 = 2.
Note that this is the only valid subarray.

Example 2:

Input: nums = [6,5,6,5,8], threshold = 7
Output: 1
Explanation: The subarray [8] has a size of 1, and 8 > 7 / 1 = 7. So 1 is returned.
Note that the subarray [6,5] has a size of 2, and every element is greater than 7 / 2 = 3.5. 
Similarly, the subarrays [6,5,6], [6,5,6,5], [6,5,6,5,8] also satisfy the given conditions.
Therefore, 2, 3, 4, or 5 may also be returned.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i], threshold <= 10⁹

Subarray With Elements Greater Than Varying Threshold

Hard

Google

3 views

Topics:

ArraysSliding Windows

You are given an integer array nums and an integer threshold.

Find any subarray of nums of length k such that every element in the subarray is greater than threshold / k.

Return the size of any such subarray. If there is no such subarray, return -1.

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

Example 1:

Input: nums = [1,3,4,3,1], threshold = 6
Output: 3
Explanation: The subarray [3,4,3] has a size of 3, and every element is greater than 6 / 3 = 2.
Note that this is the only valid subarray.

Example 2:

Input: nums = [6,5,6,5,8], threshold = 7
Output: 1
Explanation: The subarray [8] has a size of 1, and 8 > 7 / 1 = 7. So 1 is returned.
Note that the subarray [6,5] has a size of 2, and every element is greater than 7 / 2 = 3.5. 
Similarly, the subarrays [6,5,6], [6,5,6,5], [6,5,6,5,8] also satisfy the given conditions.
Therefore, 2, 3, 4, or 5 may also be returned.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i], threshold <= 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:

What are the possible ranges for values within `nums` and `threshold` arrays? Can they be negative, zero, or very large?
What happens if the `nums` array is shorter than the `threshold` array? Should I return 0, or is this an invalid input?
If no valid subarray exists (i.e., no subarray satisfies the threshold conditions), what value should I return?
Can you provide some example inputs and expected outputs to clarify the problem statement further, especially with edge cases and different lengths of valid subarrays?
Is the length of the `threshold` array guaranteed to be less than or equal to the length of the `nums` array?

Brute Force Solution

Approach

The brute force method involves checking every possible group of consecutive numbers within the given list. For each of these groups, we verify if it meets the criteria related to the threshold. We continue this process until we've assessed all possible groups.

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

Consider the first number in the list as a group by itself.
Check if every number in this group is bigger than the specified threshold for this group (threshold equals group size).
Now, include the next number to form a group of two consecutive numbers.
Again, check if every number in this new group is bigger than the group's threshold (threshold equals group size).
Continue adding consecutive numbers and checking each time until you reach the end of the list.
After considering all groups starting with the first number, move on to the second number in the list.
Repeat the process of forming groups, checking their thresholds, and moving through the entire list until all possible consecutive groups have been checked.
Record the length of the longest group that satisfied the condition.

Code Implementation

def find_longest_subarray(numbers):
    max_length = 0
    list_length = len(numbers)

    for start_index in range(list_length):
        for end_index in range(start_index, list_length):
            current_subarray = numbers[start_index : end_index + 1]
            subarray_length = len(current_subarray)

            # Check if subarray is empty
            if subarray_length == 0:
                continue

            is_valid = True

            # Validate the subarray elements.
            for element in current_subarray:

                if element <= subarray_length:
                    is_valid = False
                    break

            if is_valid:

                # Update the result.
                max_length = max(max_length, subarray_length)

    return max_length

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible subarrays of the input array of size n. For each starting element, the algorithm potentially considers subarrays of lengths from 1 up to n. This means that for each of the n starting positions, we are potentially doing n operations in the worst case by incrementing the subarray one element at a time. Thus, the total number of operations grows proportional to n * n, where n is the input array size. Therefore, the time complexity is O(n²).

Space Complexity

O(1) – The brute force approach, as described, iterates through subarrays without creating any auxiliary data structures like lists or maps to store intermediate results. It only uses a few variables to keep track of the start and end indices of the current subarray and the length of the longest valid subarray. Therefore, the space complexity is constant and independent of the input size N.

Optimal Solution

Approach

The most efficient way to solve this problem involves sliding a window to track eligible subarrays. Instead of recalculating for every possible subarray, we update the window dynamically, maintaining only the subarrays that meet the threshold requirement. This approach reduces redundant checks, saving a lot of time.

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

Start by examining the numbers one by one from left to right.
Imagine a 'window' that shows a continuous group of numbers.
For each number, see if it is bigger than the given threshold when you divide it by the length of the current window.
If the number *is* bigger than the threshold, expand the window to include it.
If the number is *not* bigger than the threshold, start a new window from the next number.
Keep track of the length of the longest window you find that satisfies the condition.
The length of the longest window represents the longest subarray that meets the requirements.

Code Implementation

def longest_subarray_length(numbers, threshold):
    longest_length = 0
    current_length = 0

    for number in numbers:
        # Check if the current number meets the threshold condition
        if number > threshold * (current_length + 1):
            current_length += 1
            longest_length = max(longest_length, current_length)

        # Reset current length when condition is not met.
        else:
            current_length = 0

    return longest_length

Big(O) Analysis

Time Complexity

O(n) – The solution iterates through the array once using a sliding window approach. For each element, it performs a constant number of operations to check the threshold condition and potentially expand or reset the window. Since each element is visited at most twice (once when the window expands and possibly once when it resets), the time complexity is directly proportional to the number of elements in the input array, n.

Space Complexity

O(1) – The algorithm described iterates through the input array, maintaining only a window of indices. No auxiliary data structures like arrays, lists, or hash maps are used to store intermediate results or visited elements. The algorithm solely uses a few variables to keep track of the current window's start, end and the maximum window length seen so far. Therefore, the space used is constant and independent of the input array size N.

Edge Cases

Case	How to Handle
nums is null or empty	Return 0 immediately, as there can be no subarray.
threshold is null or empty	Return 0 immediately, as there are no thresholds to compare against.
nums and threshold have different lengths	Consider only the minimum length between nums and threshold for iteration or return an error.
nums contains only numbers smaller or equal to the thresholds.	The result should be 0, indicating no valid subarray exists.
nums contains only positive numbers and threshold contains negative numbers.	The result should be the minimum length between the two arrays.
Very large input arrays (nums.length > 10^5)	Ensure the algorithm has O(n) or O(n log n) time complexity to avoid exceeding time limit.
threshold values are very large or very small integers	Ensure that the comparisons do not result in integer overflow/underflow issues.
A single element in nums is greater than all applicable thresholds.	The result should be 1, the length of that single element subarray.