Special Array With X Elements Greater Than or Equal X

Easy

Special Array With X Elements Greater Than or Equal X

Easy

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 is the range of values for the integers in the input array?
Is the input array guaranteed to be non-empty?
If there are multiple possible values of 'X' that satisfy the condition, which one should I return?
Are all the elements in the input array integers or could they be floating-point numbers?
If no such 'X' exists, what value should I return?

Brute Force Solution

Approach

The brute force method for this problem is like guessing a number. We'll try every possible 'special' number, one at a time, and check if it works for the given set of numbers.

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

First, we'll start with the number 0 and check if it's the 'special' number.
To check, we count how many numbers in the given set are greater than or equal to 0. If that count is equal to 0, then 0 is our 'special' number and we're done.
If not, we move on to the next possibility, which is the number 1.
We repeat the counting process: how many numbers in the set are greater than or equal to 1? If the count equals 1, then 1 is our 'special' number.
We keep doing this, trying 2, then 3, then 4, and so on, each time counting how many numbers are greater than or equal to our guess.
We continue until we find a number where the count exactly matches the number we're guessing. That's our 'special' number.
If we try every possible number and never find a match, it means there is no 'special' number.

Code Implementation

def special_array_brute_force(numbers):
    # Iterate through all possible values of X
    for possible_special_number in range(len(numbers) + 1):

        count = 0
        # Count elements >= possible_special_number
        for number in numbers:
            if number >= possible_special_number:
                count += 1

        # If count matches possible_special_number, we found X
        if count == possible_special_number:
            return possible_special_number

    # No special number found if loop completes
    return -1

Big(O) Analysis

Time Complexity

O(n²) – The described algorithm iterates from 0 up to the largest possible value in the input array 'nums' of size 'n', which in the worst case could be proportional to 'n'. For each potential special number 'x', the algorithm iterates through the 'nums' array to count how many elements are greater than or equal to 'x'. This inner iteration takes O(n) time. Since the outer loop also iterates up to a value proportional to n, the total time complexity becomes O(n * n), or O(n²).

Space Complexity

O(1) – The algorithm iterates through potential 'special' numbers and counts elements greater than or equal to them. It doesn't create any auxiliary data structures like arrays, lists, or hash maps to store intermediate results. The only extra memory used is for variables to hold the current 'special' number being tested and the count of numbers greater than or equal to it. Since these variables require constant space regardless of the input array's size (N), the space complexity is O(1).

Optimal Solution

Approach

The challenge is to find a special number related to how many numbers in a list are greater than or equal to that number. We use a process that eliminates possibilities until we find the special number or determine it doesn't exist by focusing on arranging the list in a helpful way.

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

Arrange the list from largest to smallest. This lets us easily count how many numbers meet a certain condition.
Start by checking if 1 is the special number. See if there's at least one number in the list that's 1 or greater. If there isn't, 1 can't be it.
Then, check if 2 is the special number. See if there are at least two numbers in the list that are 2 or greater. Keep going like this, checking 3, 4, and so on.
As you go, if you find a number that works (like, there are exactly 5 numbers that are 5 or greater), that's your special number and you can stop.
If you get to a number that's bigger than the list size (for example, checking if 10 is the special number when your list only has 8 numbers), you can stop because that number can't be special.
If you've checked all possible numbers and haven't found one that works, the special number doesn't exist, and the answer is -1.

Code Implementation

def specialArray(numbers):
    numbers.sort(reverse=True)

    array_length = len(numbers)

    for potential_special_value in range(1, array_length + 1):
        # Check if 'potential_special_value' could be the special integer.
        count = 0
        for number in numbers:
            if number >= potential_special_value:
                count += 1

        #  If we find a match, return that number
        if count == potential_special_value:
            return potential_special_value

    # If no special integer is found, return -1
    return -1

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this algorithm is sorting the input array nums of size n. Sorting typically takes O(n log n) time using efficient algorithms like merge sort or quicksort. After sorting, we iterate implicitly from 1 up to n within the sorted array. While there is no explicit nested loop, the initial sort is the most expensive operation. The remaining steps to find the special number are O(n) at most, which is dominated by O(n log n).

Space Complexity

O(1) – The algorithm sorts the input array in place and then iterates through potential special numbers. No auxiliary data structures that scale with the input size N (the length of the input array) are used. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as an empty array cannot satisfy the condition.
Array with only one element	Check if the single element is greater than or equal to 1, return 1 if true, otherwise 0.
Array with all elements equal to 0	Return 0 since no element can satisfy the condition x >= x.
Array with all elements equal to the array length	Return array length, since all elements satisfy the condition.
Array with all elements greater than the array length	Return the array length because that many elements are greater or equal to the length.
Array with elements in descending order	The binary search should efficiently find the correct split point even in descending order.
Array with large numbers (potential integer overflow during calculations)	Ensure all calculations use appropriate data types (e.g., long) to prevent overflow.
No such 'X' exists where X elements are greater or equal to X	The binary search should converge to a point where low and high cross, indicating no such 'X' exists and return 0.