Maximum value K such that array has at-least K elements that are >= K
Last Updated :
08 Mar, 2023
Given an array of positive integers, find maximum possible value K such that the array has at-least K elements that are greater than or equal to K. The array is unsorted and may contain duplicate values.
Examples :
Input: [2, 3, 4, 5, 6, 7]
Output: 4
Explanation : 4 elements [4, 5, 6, 7]
are greater than equal to 4
Input: [1, 2, 3, 4]
Output: 2
Explanation : 3 elements [2, 3, 4] are
greater than equal to 2
Input: [4, 7, 2, 3, 8]
Output: 3
Explanation : 4 elements [4, 7, 3, 8]
are greater than equal to 3
Input: [6, 7, 9, 8, 10]
Output: 5
Explanation : All 5 elements are greater
than equal to 5
Expected time complexity : O(n)
Method 1 [Simple : O(n2) time] :
Let size of input array be n. Let us consider following important observations.
- The maximum possible value of result can be n. We get the maximum possible value when all elements are greater than or equal to n. For example, if input array is {10, 20, 30}, n is 3. The value of result can’t be greater than 3.
- The minimum possible value would be 1. An example case when get this output is, when all elements are 1.
So we can run a loop from n to 1 and count greater elements for every value.
C++
#include <iostream>
using namespace std;
int findMaximumNum(unsigned int arr[], int n)
{
for (int i = n; i >= 1; i--)
{
int count = 0;
for (int j=0; j<n; j++)
if (i <= arr[j])
count++;
if (count >= i)
return i;
}
return 1;
}
int main()
{
unsigned int arr[] = {1, 2, 3, 8, 10 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << findMaximumNum(arr, n);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static int findMaximumNum(int arr[],
int n)
{
for (int i = n; i >= 1; i--)
{
int count = 0;
for (int j = 0; j < n; j++)
if (i <= arr[j])
count++;
if (count >= i)
return i;
}
return 1;
}
public static void main (String[] args)
{
int arr[] = {1, 2, 3, 8, 10 };
int n = arr.length;
System.out.println(findMaximumNum(arr, n));
}
}
|
Python3
def findMaximumNum(arr, n):
i = n
while(i >= 1):
count = 0
for j in range(0,n,1):
if (i <= arr[j]):
count += 1
if (count >= i):
return i
i -= 1
return 1
if __name__ == '__main__':
arr = [1, 2, 3, 8, 10]
n = len(arr)
print(findMaximumNum(arr, n))
|
C#
using System;
class GFG
{
static int findMaximumNum(int []arr,
int n)
{
for (int i = n; i >= 1; i--)
{
int count = 0;
for (int j = 0; j < n; j++)
if (i <= arr[j])
count++;
if (count >= i)
return i;
}
return 1;
}
static public void Main ()
{
int []arr = {1, 2, 3, 8, 10 };
int n = arr.Length;
Console.WriteLine(findMaximumNum(arr, n));
}
}
|
PHP
<?php
function findMaximumNum($arr, $n)
{
for ($i = $n; $i >= 1; $i--)
{
$count = 0;
for ($j = 0; $j < $n; $j++)
if ($i <= $arr[$j])
$count++;
if ($count >= $i)
return $i;
}
return 1;
}
$arr = array (1, 2, 3, 8, 10);
$n = sizeof($arr);
echo findMaximumNum($arr, $n);
?>
|
Javascript
<script>
function findMaximumNum(arr, n)
{
for (let i = n; i >= 1; i--)
{
let count = 0;
for (let j = 0; j < n; j++)
if (i <= arr[j])
count++;
if (count >= i)
return i;
}
return 1;
}
let arr = [1, 2, 3, 8, 10 ];
let n = arr.length;
document.write(findMaximumNum(arr, n));
</script>
|
Time Complexity : O(N2) ,here N is size of array.
Space Complexity : O(1) ,since no extra space required.
Method 2 [Efficient : O(n) time and O(n) extra space]
- The idea is to construct auxiliary array of size n + 1, and use that array to find count of greater elements in input array. Let the auxiliary array be freq[]. We initialize all elements of this array as 0.
- We process all input elements.
- If an element arr[i] is less than n, then we increment its frequency, i.e., we do freq[arr[i]]++.
- Else we increment freq[n].
- After step 2 we have two things.
- Frequencies of elements for elements smaller than n stored in freq[0..n-1].
- Count of elements greater than n stored in freq[n].
Finally, we process the freq[] array backwards to find the output by keeping sum of the values processed so far.
Below is implementation of above idea.
C++
#include <bits/stdc++.h>
using namespace std;
int findMaximumNum(unsigned int arr[], int n)
{
vector<int> freq(n+1, 0);
for (int i = 0; i < n; i++)
{
if (arr[i] < n)
freq[arr[i]]++;
else
freq[n]++;
}
int sum = 0;
for (int i = n; i > 0; i--)
{
sum += freq[i];
if (sum >= i)
return i;
}
}
int main()
{
unsigned int arr[] = {1, 2, 3, 8, 10 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << findMaximumNum(arr, n);
return 0;
}
|
Java
import java.util.Vector;
class GFG {
static int findMaximumNum(int arr[], int n) {
int[] freq=new int[n+1];
for (int i = 0; i < n + 1; i++) {
freq[i] = 0;
}
for (int i = 0; i < n; i++) {
if (arr[i] < n)
{
freq[arr[i]]++;
}
else {
freq[n]++;
}
}
int sum = 0;
for (int i = n; i > 0; i--) {
sum += freq[i];
if (sum >= i) {
return i;
}
}
return 0;
}
public static void main(String[] args) {
int arr[] = {1, 2, 3, 8, 10};
int n = arr.length;
System.out.println(findMaximumNum(arr, n));
}
}
|
Python3
def findMaximumNum(arr, n):
freq = [0 for i in range(n+1)]
for i in range(n):
if (arr[i] < n):
freq[arr[i]] += 1
else:
freq[n] += 1
sum = 0
for i in range(n,0,-1):
sum += freq[i]
if (sum >= i):
return i
arr = [1, 2, 3, 8, 10]
n = len(arr)
print(findMaximumNum(arr, n))
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static int findMaximumNum(int []arr, int n)
{
List<int> freq = new List<int>();
for (int i = 0; i < n + 1; i++)
{
freq.Insert(i, 0);
}
for (int i = 0; i < n; i++)
{
if (arr[i] < n)
{
freq.Insert(arr[i], freq[arr[i]] + 1);
}
else
{
freq.Insert(n, freq[n] + 1);
}
}
int sum = 0;
for (int i = n; i > 0; i--)
{
sum += freq[i];
if (sum >= i)
{
return i;
}
}
return 0;
}
public static void Main()
{
int []arr = {1, 2, 3, 8, 10};
int n = arr.Length;
Console.WriteLine(findMaximumNum(arr, n));
}
}
|
Javascript
<script>
function findMaximumNum(arr, n)
{
let freq = new Array(n + 1).fill(0);
for (let i = 0; i < n; i++) {
if (arr[i] < n)
freq[arr[i]]++;
else
freq[n]++;
}
let sum = 0;
for (let i = n; i > 0; i--) {
sum += freq[i];
if (sum >= i)
return i;
}
}
let arr = [1, 2, 3, 8, 10];
let n = arr.length;
document.write(findMaximumNum(arr, n));
</script>
|
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