**K Numbers Greater Than or Equal to K - Google Top Interview Questions**

### Problem Statement :

You are given a list of non-negative integers nums. If there are exactly k numbers in nums that are greater than or equal to k, return k. Otherwise, return -1. Constraints 1 ≤ n ≤ 100,000 where n is the length of nums Example 1 Input nums = [5, 3, 0, 9] Output 3 Explanation There are exactly 3 numbers that's greater than or equal to 3: [5, 3, 9].

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& nums) { // Time and Space: O(N)
int n = nums.size();
vector<int> bucket(n + 1, 0);
for (int value : nums) {
if (value < n)
bucket[value]++;
else
bucket[n]++;
}
int k = 0;
for (int i = n; i >= 0; i--) {
k += bucket[i];
if (k == i) return k;
}
return -1;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums) {
Arrays.sort(nums);
for (int i = nums.length - 1; i > 0; i--) {
int search = Arrays.binarySearch(nums, i);
search = search < 0 ? -search - 1 : search;
while (search - 1 >= 0 && search < nums.length && nums[search - 1] >= i) search--;
if (nums.length - search == i)
return i;
}
return -1;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums):
n = len(nums)
count = [0 for i in range(n + 1)]
for num in nums:
if num >= n:
count[n] += 1
else:
count[num] += 1
for i in range(len(count) - 2, -1, -1):
count[i] += count[i + 1]
for num in range(1, n + 1):
if count[num] == num:
return num
return -1
```

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