Find elements larger than half of the elements in an array | Set 2

Given an array arr[] consisting of N positive integers, the task is to find the elements which are greater than at least half of the array elements.

Examples:

Input: arr[] = {1, 6, 3, 4}
Output: 4 6
Explanation:
Size of the array is 4. The elements which are greater than atleast N/2(= 4/2 = 2) elements of the array are 4 and 6.

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

Naive Approach: Please refer to the previous post of this article for the Naive Approach.

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

Sorting-based Approach: Please refer to the previous post of this article for the Sorting Based Approach.

Time Complexity: O(N*log N)
Auxiliary Space: O(1)

Approach: The above approach can also be optimized by using Hashing by keeping track of the count of array elements. After finding the frequency of each element, print elements in non-increasing order till N/2 elements have been printed. Follow the steps below to solve the problem:

• Initialize a variable, say mid as (N + 1)/2 to store the middle index of the array.
• Initialize a variable, say max to store the maximum element of the array.
• Initialize an array count[] to store the frequency of each element.
• Traverse the array, arr[] and increment count[arr[i]] by 1.
• Traverse the array count[] using the variable i from the back and perform the following steps:
  • Iterate until the value of the count[i] is positive and perform the following steps:
    • Print the value of i and decrease the value of count[i] and mid by 1.
    • If the value of mid is 0, then break out of the loop.
  • If the value of mid is 0, then break out of the loop.

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 the element that
// are larger than half of elements
// of the array
void findLarger(int arr[], int n)
{
    // Find the value of mid
    int mid = (n + 1) / 2;

    // Stores the maximum element
    int mx = *max_element(arr, arr + n);

    // Stores the frequency of each
    // array element
    int count[mx + 1] = { 0 };
    for (int i = 0; i < n; i++) {
        count[arr[i]]++;
    }

    // Traverse the array in the
    // reverse order
    for (int i = mx; i >= 0; i--) {

        while (count[i] > 0) {

            // Decrement the value
            // of count[i] and mid
            count[i]--;
            mid--;

            // Print the current element
            cout << i << ' ';

            // Check if the value of
            // mid is equal to 0
            if (mid == 0)
                break;
        }
        if (mid == 0)
            break;
    }
}

// Driver Code
int main()
{
    int arr[] = { 10, 4, 2, 8, 9 };
    int N = sizeof(arr) / sizeof(arr[0]);
    findLarger(arr, N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the element that
// are larger than half of elements
// of the array
static void findLarger(int arr[], int n)
{
    // Find the value of mid
    int mid = (n + 1) / 2;

    // Stores the maximum element
    int mx = Arrays.stream(arr).max().getAsInt();

    // Stores the frequency of each
    // array element
    int count[] = new int[mx+1];
    for (int i = 0; i < mx+1; i++) {
        count[i] = 0;
    }

    for (int i = 0; i < n; i++) {
        count[arr[i]]++;
    }

    // Traverse the array in the
    // reverse order
    for (int i = mx; i >= 0; i--) {

        while (count[i] > 0) {

            // Decrement the value
            // of count[i] and mid
            count[i]--;
            mid--;

            // Print the current element
            System.out.print(i + " ");

            // Check if the value of
            // mid is equal to 0
            if (mid == 0)
                break;
        }
        if (mid == 0)
            break;
    }
}


// Driver Code
public static void main(String[] args)
{
    int arr[] = { 10, 4, 2, 8, 9 };
    int N = arr.length;
    findLarger(arr, N);
}
}

// This code is contributed by sanjoy_62.

Python3

# Python program for the above approach

# Function to find the element that
# are larger than half of elements
# of the array
def findLarger(arr, n):
    # Find the value of mid
    mid = (n + 1) // 2

    # Stores the maximum element
    mx = max(arr)

    # Stores the frequency of each
    # array element
    count = [0]*(mx + 1)

    for i in range(n):
        count[arr[i]] += 1

    # Traverse the array in the
    # reverse order
    for i in range(mx,-1, -1):
        while (count[i] > 0):

            # Decrement the value
            # of count[i] and mid
            count[i] -= 1
            mid -= 1

            # Print current element
            print(i, end = " ")

            # Check if the value of
            # mid is equal to 0
            if (mid == 0):
                break
        if (mid == 0):
            break

# Driver Code
if __name__ == '__main__':
    arr = [10, 4, 2, 8, 9 ]
    N = len(arr)
    findLarger(arr, N)

# This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to find the element that
// are larger than half of elements
// of the array
static void findLarger(int []arr, int n)
{
    // Find the value of mid
    int mid = (n + 1) / 2;

    // Stores the maximum element
    int mx = Int32.MinValue;
    for(int i=0;i<n;i++){
        if(arr[i]>mx)
            mx = arr[i];
    }

    // Stores the frequency of each
    // array element
    int []count = new int[mx + 1];
    Array.Clear(count,0,mx+1);
    for (int i = 0; i < n; i++) {
        count[arr[i]]++;
    }

    // Traverse the array in the
    // reverse order
    for (int i = mx; i >= 0; i--) {

        while (count[i] > 0) {

            // Decrement the value
            // of count[i] and mid
            count[i]--;
            mid--;

            // Print the current element
            Console.Write(i+" ");

            // Check if the value of
            // mid is equal to 0
            if (mid == 0)
                break;
        }
        if (mid == 0)
            break;
    }
}

// Driver Code
public static void Main()
{
    int []arr = { 10, 4, 2, 8, 9 };
    int N = arr.Length;
    findLarger(arr, N);
}
}

// This code is contributed by SURENDRA_GANGWAR.

Javascript

<script>

        // JavaScript program for the above approach

        // Function to find the element that
        // are larger than half of elements
        // of the array
        function findLarger(arr, n) {
            // Find the value of mid
            let mid = (n + 1) / 2;

            // Stores the maximum element
            let mx = Math.max.apply(null, arr)

            // Stores the frequency of each
            // array element
            var count = new Array(mx + 1).fill(0);
            for (let i = 0; i < n; i++) {
                count[arr[i]]++;
            }

            // Traverse the array in the
            // reverse order
            for (let i = mx; i >= 0; i--) {

                while (count[i] > 0) {

                    // Decrement the value
                    // of count[i] and mid
                    count[i]--;
                    mid--;

                    // Print the current element
                    document.write(i + ' ');

                    // Check if the value of
                    // mid is equal to 0
                    if (mid == 0)
                        break;
                }
                if (mid == 0)
                    break;
            }
        }

        // Driver Code

        var arr = [10, 4, 2, 8, 9];
        var N = 5;
        findLarger(arr, N);

    </script>

Output:

10 9 8

Time Complexity: O(max(M², N)), where M is the maximum element of the array arr[] and N is the length of the given array.
Auxiliary Space: O(M)