You are given an array
nums of non-negative integers.
nums is considered special if there exists a number
x such that there are exactly
x numbers in
nums that are greater than or equal to
x does not have to be an element in
x if the array is special, otherwise, return
-1. It can be proven that if
nums is special, the value for
x is unique.
Input: nums = [3,5] Output: 2 Explanation: There are 2 values (3 and 5) that are greater than or equal to 2.
Input: nums = [0,0] Output: -1 Explanation: No numbers fit the criteria for x. If x = 0, there should be 0 numbers >= x, but there are 2. If x = 1, there should be 1 number >= x, but there are 0. If x = 2, there should be 2 numbers >= x, but there are 0. x cannot be greater since there are only 2 numbers in nums.
Input: nums = [0,4,3,0,4] Output: 3 Explanation: There are 3 values that are greater than or equal to 3.
Input: nums = [3,6,7,7,0] Output: -1
1 <= nums.length <= 100
0 <= nums[i] <= 1000
Start with x = 0 and increase x to a most len(nums), and check if the total number of elements in list is equal to x.
class Solution: def specialArray(self, nums: List[int]) -> int: x = 0 while x <= len(nums): count = 0 for num in nums: if num >= x: count += 1 if count == x: return x x += 1 return -1
- Time Complexity: O(N^N).
- Space Complexity: O(1).