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Find a number K such that exactly K array elements are greater than or equal to K

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  • Last Updated : 06 Jan, 2022

Given an array a[] of size N, which contains only non-negative elements, the task is to find any integer K for which there are exactly K array elements that are greater than or equal to K. If no such K exists, then print -1.

Examples:

Input: a[] = {7, 8, 9, 0, 0, 1}
Output: 3
Explanation:
Since 3 is less than or equal to 7, 8, and 9, therefore, 3 is the answer.

Input: a[] = {0, 0}
Output: -1

 

Approach: The task is to find K such that the array elements are greater than or equal to K. Therefore, K cannot exceed the maximum element present in the array a[n]. Follow the steps below solve the problem:

  1. Traverse the array to find the largest array element, store it in a variable, say m.
  2. Initialize a counter variable, cnt to count the number of array elements greater than or equal to K.
  3. Iterate for possible values of K starting from 0 to m. Iterate over the array for each value and count the number of array elements greater than or equal to that value.
  4. If for any value, exactly K array elements are found to be greater than or equal to that value, print that value.
  5. If no such value is obtained after complete traversal of the array, print -1.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find K for which
// there are exactly K array
// elements greater than or equal to K
int zvalue(vector<int>& nums)
{
 
    // Finding the largest array element
    int m = *max_element(nums.begin(),
                         nums.end());
    int cnt = 0;
 
    // Possible values of K
    for (int i = 0; i <= m; i++) {
        cnt = 0;
 
        // Traverse the array
        for (int j = 0; j < nums.size(); j++) {
 
            // If current array element is
            // greater than or equal to i
            if (nums[j] >= i)
                cnt++;
        }
 
        // If i array elements are
        // greater than or equal to i
        if (cnt == i)
            return i;
    }
 
    // Otherwise
    return -1;
}
 
// Driver Code
int main()
{
    vector<int> nums = { 7, 8, 9, 0, 0, 1 };
    cout << zvalue(nums) << endl;
}

Java




// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to find K for which
// there are exactly K array
// elements greater than or equal to K
public static int zvalue(int[] nums)
{
     
    // Finding the largest array element
    int m = max_element(nums);
    int cnt = 0;
 
    // Possible values of K
    for(int i = 0; i <= m; i++)
    {
        cnt = 0;
 
        // Traverse the array
        for(int j = 0; j < nums.length; j++)
        {
             
            // If current array element is
            // greater than or equal to i
            if (nums[j] >= i)
                cnt++;
        }
 
        // If i array elements are
        // greater than or equal to i
        if (cnt == i)
            return i;
    }
 
    // Otherwise
    return -1;
}
 
// To find maximum Element
public static int max_element(int[] nums)
{
    int max = nums[0];
    for(int i = 1; i < nums.length; i++)
        max = Math.max(max, nums[i]);
         
    return max;
}
 
// Driver Code
public static void main(String args[])
{
    int[] nums = { 7, 8, 9, 0, 0, 1 };
     
    System.out.println(zvalue(nums));
}
}
 
// This code is contributed by hemanth gadarla

Python3




# Python3 program for the above approach
 
# Function to find K for which
# there are exactly K array
# elements greater than or equal to K
def zvalue(nums):
     
    # Finding the largest array element
    m = max(nums)
    cnt = 0
 
    # Possible values of K
    for i in range(0, m + 1, 1):
        cnt = 0
 
        # Traverse the array
        for j in range(0, len(nums), 1):
             
            # If current array element is
            # greater than or equal to i
            if (nums[j] >= i):
                cnt += 1
 
        # If i array elements are
        # greater than or equal to i
        if (cnt == i):
            return i
 
    # Otherwise
    return -1
 
# Driver Code
if __name__ == '__main__':
     
    nums =  [ 7, 8, 9, 0, 0, 1 ]
     
    print(zvalue(nums))
     
# This code is contributed by SURENDRA_GANGWAR

C#




// C# program for the
// above approach
using System;
class GFG{
 
// Function to find K for which
// there are exactly K array
// elements greater than or equal to K
public static int zvalue(int[] nums)
{
  // Finding the largest array element
  int m = max_element(nums);
  int cnt = 0;
 
  // Possible values of K
  for(int i = 0; i <= m; i++)
  {
    cnt = 0;
 
    // Traverse the array
    for(int j = 0;
            j < nums.Length; j++)
    {
      // If current array element is
      // greater than or equal to i
      if (nums[j] >= i)
        cnt++;
    }
 
    // If i array elements are
    // greater than or equal to i
    if (cnt == i)
      return i;
  }
 
  // Otherwise
  return -1;
}
 
// To find maximum Element
public static int max_element(int[] nums)
{
  int max = nums[0];
   
  for(int i = 1; i < nums.Length; i++)
    max = Math.Max(max, nums[i]);
   
  return max;
}
 
// Driver Code
public static void Main(String []args)
{
  int[] nums = {7, 8, 9, 0, 0, 1};
  Console.WriteLine(zvalue(nums));
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// javascript program for the above approach
 
// Function to find K for which
// there are exactly K array
// elements greater than or equal to K
function zvalue(nums)
{
     
    // Finding the largest array element
    var m = max_element(nums);
    var cnt = 0;
 
    // Possible values of K
    for(i = 0; i <= m; i++)
    {
        cnt = 0;
 
        // Traverse the array
        for(j = 0; j < nums.length; j++)
        {
             
            // If current array element is
            // greater than or equal to i
            if (nums[j] >= i)
                cnt++;
        }
 
        // If i array elements are
        // greater than or equal to i
        if (cnt == i)
            return i;
    }
 
    // Otherwise
    return -1;
}
 
// To find maximum Element
function max_element(nums)
{
    var max = nums[0];
    for(i = 1; i < nums.length; i++)
        max = Math.max(max, nums[i]);
         
    return max;
}
 
// Driver Code
nums = [ 7, 8, 9, 0, 0, 1 ];
     
document.write(zvalue(nums));
 
// This code contributed by shikhasingrajput
 
</script>

Output

3

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach:

  1. Find the maximum element in the array and store it.
  2. Create a count array with size of (maximum element +1) and initialize to 0.
  3. Traverse over the given array and increment the element count of the corresponding index in the count array
    • i.e., if (nums[i] == j) then count[j]=count[j]+1
  4. Find the suffix sum of the count array
    1. Traverse through the count array from right to left and update the count value with the sum of all the remaining array elements count to the right of it
  5. Again traverse through the count array, for any index == count[index], return index
  6. Else return -1

C++




#include <bits/stdc++.h>
using namespace std;
int solve(vector<int>& nums)
{
    // Finding the maximum element
    int max = *max_element(nums.begin(), nums.end());
    // initialising the count to 0 for all indices
    int count[max + 1] = { 0 };
    // incrementing count of corresponding element in the
    // count array
    for (int i = 0; i < nums.size(); i++) {
        count[nums[i]]++;
    }
    // corner case
    if (count[max] == max)
        return max;
    // finding suffix sum
    for (int j = max - 1; j >= 0; j--) {
        count[j] += count[j + 1];
        // Checking the index after updating the count
        if (j == count[j]) {
            return j;
        }
    }
    return -1;
}
// Driver Code
int main()
{
    vector<int> v = { 2, 0, 0 };
    cout << solve(v);
}
 
// This code is contributed by Maneesh Gupta NVSS

Java




// Java code to implement the above approach
import java.io.*;
import java.util.Arrays;
 
class GFG {
 
  public static int solve(int nums[])
  {
 
    // Finding the maximum element
    int max = Arrays.stream(nums).max().getAsInt();
 
    // initialising the count to 0 for all indices
    int count[] = new int[max + 1];
    Arrays.fill(count, 0);
 
    // incrementing count of corresponding element in the
    // count array
    for (int i = 0; i < nums.length; i++) {
      count[nums[i]]++;
    }
 
    // corner case
    if (count[max] == max)
      return max;
 
    // finding suffix sum
    for (int j = max - 1; j >= 0; j--) {
      count[j] += count[j + 1];
 
      // Checking the index after updating the count
      if (j == count[j]) {
        return j;
      }
    }
    return -1;
  }
 
  // Driver Code
  public static void main (String[] args)
  {
    int v[] = new int[]{ 2, 0, 0 };
    System.out.println(solve(v));
  }
}
 
// This code is contributed by Shubham Singh

Python3




def solve(nums):
     
    # Finding the maximum element
    maxx = max(nums)
     
    # initialising the count to 0 for all indices
    count = [0]*(maxx + 1)
     
    # incrementing count of corresponding element in the
    # count array
    for i in range(len(nums)):
        count[nums[i]] += 1
         
    # corner case
    if (count[maxx] == maxx):
        return maxx
         
    # finding suffix sum
    for j in range(maxx - 1, -1, -1):
        count[j] += count[j + 1]
         
        # Checking the index after updating the count
        if (j == count[j]):
            return j
    return -1
 
# Driver Code
v = [ 2, 0, 0 ]
print(solve(v))
 
# This code is contributed by ShubhamSingh

C#




// C# code to implement the above approach
using System;
using System.Linq;
 
public class GFG{
 
  public static int solve(int[] nums)
  {
 
    // Finding the maximum element
    int max = nums.Max();
 
    // initialising the count to 0 for all indices
    int[] count = new int[max + 1];
 
    // incrementing count of corresponding element in the
    // count array
    for (int i = 0; i < nums.Length; i++) {
      count[nums[i]]++;
    }
 
    // corner case
    if (count[max] == max)
      return max;
 
    // finding suffix sum
    for (int j = max - 1; j >= 0; j--) {
      count[j] += count[j + 1];
 
      // Checking the index after updating the count
      if (j == count[j]) {
        return j;
      }
    }
    return -1;
  }
 
  // Driver Code
  public static void Main ()
  {
    int[] v = new int[]{ 2, 0, 0 };
    Console.Write(solve(v));
  }
}
 
// This code is contributed by Shubham Singh

Javascript




<script>
// Javascript code to implement the above approach
 
function solve(nums)
{
 
    // Finding the maximum element
    var max = Math.max.apply(Math, nums);
     
    // initialising the count to 0 for all indices
    var count = new Array(max+1).fill(0);
     
    // incrementing count of corresponding element in the
    // count array
    for (var i = 0; i < nums.length; i++) {
      count[nums[i]]++;
    }
     
    // corner case
    if (count[max] == max)
      return max;
     
    // finding suffix sum
    for (var j = max - 1; j >= 0; j--) {
      count[j] += count[j + 1];
     
      // Checking the index after updating the count
      if (j == count[j]) {
        return j;
      }
    }
    return -1;
}
 
// Driver Code
var v = [ 2, 0, 0 ];
document.write(solve(v));
 
// This code is contributed by Shubham Singh
</script>

Output

1

Time complexity : O (N) where N is maximum(nums.size(), max element)

Space complexity : O ( max element)


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