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Find elements larger than half of the elements in an array

Given an array of n elements, the task is to find the elements that are greater than half of elements in an array. In case of of odd elements, we need to print elements larger than floor(n/2) elements where n is total number of elements in array.

Examples :

Input :  arr[] = {1, 6, 3, 4}
Output : 4 6

Input  :  arr[] = {10, 4, 2, 8, 9}
Output :  10 9 8


A naive approach is to take an element and compare with all other elements and if it is greater then increment the count and then check if count is greater than n/2 elements then print.

An efficient method is to sort the array in ascending order and then print last ceil(n/2) elements from sorted array.

Below is C++ implementation of this sorting based approach.

C/C++

// C++ program to find elements that are larger than
// half of the elements in array
#include <bits/stdc++.h>
using namespace std;
  
// Prints elements larger than n/2 element
void findLarger(int arr[], int n)
{
    // Sort the array in ascending order
    sort(arr, arr + n);
  
    // Print last ceil(n/2) elements
    for (int i = n-1; i >= n/2; i--)
        cout << arr[i] << " ";    
}
  
// Driver program 
int main() 
{
    int arr[] = {1, 3, 6, 1, 0, 9};
    int n = sizeof(arr)/sizeof(arr[0]);
    findLarger(arr, n);
    return 0;
}

Java

// Java program to find elements that are 
// larger than half of the elements in array
import java.util.*;
  
class Gfg
{
    // Prints elements larger than n/2 element
    static void findLarger(int arr[], int n)
    {
        // Sort the array in ascending order
        Arrays.sort(arr);
       
        // Print last ceil(n/2) elements
        for (int i = n-1; i >= n/2; i--)
            System.out.print(arr[i] + " ");  
    }
       
    // Driver program 
    public static void main(String[] args) 
    {
        int arr[] = {1, 3, 6, 1, 0, 9};
        int n = arr.length;
        findLarger(arr, n);
    }    
}
  
// This code is contributed by Raghav Sharma

/div>

Python

# Python program to find elements that are larger than
# half of the elements in array
# Prints elements larger than n/2 element
def findLarger(arr,n):
  
    # Sort the array in ascending order
    x = sorted(arr)
  
    # Print last ceil(n/2) elements
    for i in range(n/2,n):
        print(x[i]),
  
# Driver program
arr = [1, 3, 6, 1, 0, 9]
n = len(arr);
findLarger(arr,n)
  
# This code is contributed by Afzal Ansari

C#

// C# program to find elements 
// that are larger than half 
// of the elements in array
using System;
  
class GFG
{
    // Prints elements larger
    // than n/2 element
    static void findLarger(int []arr, 
                           int n)
    {
        // Sort the array 
        // in ascending order
        Array.Sort(arr);
      
        // Print last ceil(n/2) elements
        for (int i = n - 1; i >= n / 2; i--)
            Console.Write(arr[i] + " "); 
    }
      
    // Driver Code 
    public static void Main() 
    {
        int []arr = {1, 3, 6, 1, 0, 9};
        int n = arr.Length;
        findLarger(arr, n);
    
}
  
// This code is contributed
// by nitin mittal.

PHP

<?php
// PHP program to find elements
// that are larger than half of
// the elements in array
  
// Prints elements larger
// than n/2 element
function findLarger($arr, $n)
{
    // Sort the array in 
    // ascending order
    sort($arr);
  
    // Print last ceil(n/2) elements
    for ($i = $n - 1; $i >= $n / 2; $i--)
        echo $arr[$i] , " "
}
  
// Driver Code 
$arr = array(1, 3, 6, 1, 0, 9);
$n = count($arr);
findLarger($arr, $n);
  
// This code is contributed by anuj_67.
?>


Output :

9 6 3

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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