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You are given a non-negative integer array nums. The special array is defined as an array where there exists a number x such that there are exactly x elements in the array that are greater than or equal to x.

Return the smallest possible value of x. If no such x exists, return -1.

Clarifying Questions

  1. What is the range of values for the elements in nums?
    • The values in nums are non-negative integers and could be very large.
  2. What is the length of the array nums?
    • The length of nums could vary from 1 to 1000.
  3. Is it important to consider edge cases like an empty array?
    • No, since nums will have at least one element as per constraint 1 <= nums.length.

Strategy

  1. Sort the Array:
    • Start by sorting the array nums. This helps in determining the number of elements greater than or equal to any number x efficiently.
  2. Iterate Through Possible Values of x:
    • Iterate over the possible values of x from 0 to the length of the array. For each x, we need to check if there are exactly x elements greater than or equal to x.
  3. Use Binary Search Logic:
    • For a given x, find the first position in the sorted array where the element is greater than or equal to x using binary search.
    • The number of elements greater than or equal to x can then be determined by the length of the array minus this position.
  4. Check the Condition for Each x:
    • If we find any such x that meets the condition, return that x.
    • If no such x is found, return -1.

Code

def specialArray(nums):
    nums.sort()
    n = len(nums)
    
    for x in range(n + 1):
        # Find the number of elements greater than or equal to `x`
        idx = binary_search(nums, x)
        if n - idx == x:
            return x
    return -1

def binary_search(nums, x):
    left, right = 0, len(nums)
    while left < right:
        mid = (left + right) // 2
        if nums[mid] >= x:
            right = mid
        else:
            left = mid + 1
    return left

# Example Usage:
nums = [3, 5, 0, 4, 3]
print(specialArray(nums))  # Output might be 3

Time Complexity

  1. Sorting the Array:
    • The time complexity for sorting is ( O(n \log n) ).
  2. Iterating Through Possible Values of x:
    • We iterate from 0 to n, resulting in ( O(n) ) operations.
  3. Binary Search for Position:
    • Each binary search operation takes ( O(\log n) ). Since it is repeated n+1 times, it gives ( O(n \log n) ).

Overall, the dominant factor remains the sorting step, resulting in a final time complexity of ( O(n \log n) ).

This approach ensures that we efficiently identify the smallest possible x or determine that no such x exists.

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