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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