Problem Description

You need to find a special number x for a given array of non-negative integers. The array is considered special if there exists a number x such that exactly x numbers in the array are greater than or equal to x.

Key points to understand:

• The number x doesn't have to be an element in the array itself
• You need to count how many numbers in the array are >= x
• If this count equals x, then the array is special with that value of x
• Return x if such a value exists, otherwise return -1
• The value of x is guaranteed to be unique if it exists

For example, if nums = [3, 5], then x = 2 makes the array special because there are exactly 2 numbers (both 3 and 5) that are greater than or equal to 2.

The solution checks all possible values of x from 1 to the length of the array. For each candidate x, it counts how many elements in nums are greater than or equal to x. If the count equals x, that's the answer. The range [1..len(nums)] is sufficient because:

• If x = 0, then 0 elements must be >= 0, but all non-negative integers are >= 0, so this is impossible unless the array is empty
• If x > len(nums), then more than len(nums) elements must be in the array, which is impossible

Intuition

The key insight is recognizing that x has an upper bound. Since we're looking for exactly x numbers that are >= x, the maximum possible value of x is the length of the array. Why? Because we can't have more numbers in the array than the array's size itself.

Similarly, x has a lower bound of 1 (since x = 0 would require 0 elements to be >= 0, but all non-negative integers satisfy this condition, creating a contradiction).

This gives us a finite search space: x must be in the range [1, len(nums)].

Since the search space is small (at most n values to check), we can afford to try every possible value of x. For each candidate value, we simply count how many elements in the array are greater than or equal to it.

The brute force approach works well here because:

1. The search space is bounded and small (linear in the size of the array)
2. For each candidate x, counting takes O(n) time
3. The total time complexity O(n²) is acceptable for typical array sizes

The uniqueness guarantee mentioned in the problem means we don't need to worry about multiple valid answers - as soon as we find an x where the count equals x, we can return it immediately.

Learn more about Binary Search and Sorting patterns.

Solution Approach

The solution uses a brute force enumeration approach, systematically checking each possible value of x from 1 to n (where n is the length of the array).

Here's the step-by-step implementation:

1. Enumerate possible values of x: We loop through x in the range [1, len(nums) + 1). This range covers all possible valid values since:

   • x cannot be 0 (as explained earlier)
   • x cannot exceed the array length (we can't have more than n elements in an array of size n)

2. Count qualifying elements: For each candidate x, we count how many elements in nums are greater than or equal to x:

   cnt = sum(v >= x for v in nums)

   This uses a generator expression with Python's sum() function. The expression v >= x evaluates to True (1) or False (0), so summing these boolean values gives us the count.

3. Check the special condition: If cnt == x, we've found our answer and immediately return x. The uniqueness property guaranteed by the problem means we don't need to check further.

4. Return -1 if no solution exists: If we've checked all possible values of x from 1 to n and none satisfy the condition, the array is not special, so we return -1.

The algorithm's time complexity is O(n²) where n is the length of the array:

• Outer loop runs n times
• Inner counting operation takes O(n) time for each iteration

The space complexity is O(1) as we only use a constant amount of extra space for variables x and cnt.

Example Walkthrough

Let's trace through the solution with nums = [3, 5]:

Step 1: Try x = 1

• Count how many numbers in [3, 5] are >= 1
• Both 3 >= 1 and 5 >= 1 are true
• Count = 2
• Is count (2) equal to x (1)? No, 2 ≠ 1
• Continue searching

Step 2: Try x = 2

• Count how many numbers in [3, 5] are >= 2
• Both 3 >= 2 and 5 >= 2 are true
• Count = 2
• Is count (2) equal to x (2)? Yes, 2 = 2
• Found our answer! Return 2

The array is special with x = 2 because exactly 2 numbers (3 and 5) are greater than or equal to 2.

Let's also see a case where no solution exists with nums = [1]:

Step 1: Try x = 1

• Count how many numbers in [1] are >= 1
• 1 >= 1 is true
• Count = 1
• Is count (1) equal to x (1)? Yes, 1 = 1
• Found our answer! Return 1

Now consider nums = [0, 0] where no solution exists:

Step 1: Try x = 1

• Count how many numbers in [0, 0] are >= 1
• Both 0 >= 1 are false
• Count = 0
• Is count (0) equal to x (1)? No, 0 ≠ 1

Step 2: Try x = 2

• Count how many numbers in [0, 0] are >= 2
• Both 0 >= 2 are false
• Count = 0
• Is count (0) equal to x (2)? No, 0 ≠ 2

All possible values of x have been checked. Return -1.

Solution Implementation

Time and Space Complexity

The time complexity is O(n²), where n is the length of the array nums. This is because:

• The outer loop iterates from 1 to len(nums) + 1, which gives us n iterations
• For each iteration of the outer loop, the sum(v >= x for v in nums) operation iterates through all n elements of the array
• Therefore, we have n × n = n² total operations

The space complexity is O(1). This is because:

• We only use a constant amount of extra space for variables x and cnt
• The generator expression (v >= x for v in nums) doesn't create a new list but evaluates lazily
• No additional data structures are created that scale with the input size

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Forgetting to check x = 0

While the current solution starts from x = 1, some might wonder if x = 0 is a valid case. Actually, x = 0 means exactly 0 elements should be >= 0. Since all elements are non-negative integers, this is only possible for an empty array. The current solution handles this correctly by starting from 1, but it's worth explicitly checking if needed:

def specialArray(self, nums: List[int]) -> int:
    # Edge case: empty array could have x = 0
    if not nums:
        return 0
  
    for x in range(1, len(nums) + 1):
        count = sum(value >= x for value in nums)
        if count == x:
            return x
  
    return -1

2. Inefficient counting leads to timeout for large inputs

The current O(n²) solution recounts elements for every possible x value. A more efficient approach sorts the array first and uses binary search or clever counting:

def specialArray(self, nums: List[int]) -> int:
    nums.sort(reverse=True)  # Sort in descending order
  
    for i in range(len(nums)):
        # i elements have been seen so far (all >= nums[i])
        # We need exactly x elements >= x
        x = i + 1  # Number of elements >= nums[i]
      
        # Check if x is valid
        if x <= nums[i] and (i == len(nums) - 1 or x > nums[i + 1]):
            return x
  
    return -1

3. Misunderstanding the problem: thinking x must be in the array

A critical misconception is assuming x must be one of the array elements. For example, with nums = [3, 5], the answer is x = 2, which isn't in the array. The corrected understanding:

# WRONG: Only checking array elements
def specialArray_wrong(self, nums: List[int]) -> int:
    for x in nums:  # This misses valid x values not in the array
        count = sum(value >= x for value in nums)
        if count == x:
            return x
    return -1

# CORRECT: Checking all possible values in range
def specialArray(self, nums: List[int]) -> int:
    for x in range(len(nums) + 1):  # Checks all possible x values
        count = sum(value >= x for value in nums)
        if count == x:
            return x
    return -1

4. Off-by-one error in the range

Some might mistakenly use range(1, len(nums)) instead of range(1, len(nums) + 1), missing the case where all elements could be greater than or equal to x = len(nums):

# WRONG: Missing x = len(nums)
for x in range(1, len(nums)):  # Stops at len(nums) - 1
    ...

# CORRECT: Including x = len(nums)
for x in range(1, len(nums) + 1):  # Includes len(nums)
    ...

Example where this matters: nums = [100, 100, 100] has x = 3 as the answer (all 3 elements are >= 3).