Find element in a sorted array whose frequency is greater than or equal to n/2.

Given a sorted array of length n, find the number in array that appears more than or equal to n/2 times. It is given that such element always exists.

Examples: 

Input :  2 3 3 4
Output : 3

Input :  3 4 5 5 5 
Output : 5

Input : 1 1 1 2 3
Output : 1

To find that number, we traverse the array and check the frequency of every element in array if it is greater than or equals to n/2 but it requires extra space and time complexity will be O(n). 

But we can see that the if there is number that comes more than or equal to n/2 times in a sorted array, then that number must be present at the position n/2 i.e. a[n/2].

Implementation:

C++

// C++ code to find majority element in a
// sorted array
#include <iostream>
using namespace std;

int findMajority(int arr[], int n)
{
    return arr[n / 2];
}

int main()
{
    int arr[] = { 1, 2, 2, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findMajority(arr, n);
    return 0;
}

Java

// Java code to find majority element in a
// sorted array
public class Test {

    public static int findMajority(int arr[], int n)
    {
        return arr[n / 2];
    }
    public static void main(String args[])
    {
        int arr[] = { 1, 2, 2, 3 };
        int n = arr.length;
        System.out.println(findMajority(arr, n));
    }
}

Python 3

# Python 3 code to find
# majority element in a
# sorted array

def findMajority(arr, n):

    return arr[int(n / 2)]

# Driver Code
arr = [1, 2, 2, 3]
n = len(arr)
print(findMajority(arr, n))

# This code is contributed by Smitha.

C#

// C# code to find majority element in a
// sorted array
using System;

public class GFG {

    public static int findMajority(int []arr, int n)
    {
        return arr[n / 2];
    }

    // Driver code
    public static void Main()
    {

        int []arr = { 1, 2, 2, 3 };
        int n = arr.Length;

        Console.WriteLine(findMajority(arr, n));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP code to find majority
// element in a sorted array

function findMajority($arr, $n)
{
    return $arr[intval($n / 2)];
}

    // Driver Code
    $arr = array(1, 2, 2, 3);
    $n = count($arr);
    echo findMajority($arr, $n);

// This code is contributed by Sam007
?>

Javascript

<script>

// Javascript code to find majority element in a
// sorted array

function findMajority(arr,  n)
{
    return arr[(Math.floor(n / 2))];
}

// driver code

    let arr = [ 1, 2, 2, 3 ];
    let n = arr.length;
    document.write(findMajority(arr, n));

</script>

Output

2

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

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