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++
#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
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
def findMajority(arr, n):
return arr[ int (n / 2 )]
arr = [ 1 , 2 , 2 , 3 ]
n = len (arr)
print (findMajority(arr, n))
|
C#
using System;
public class GFG {
public static int findMajority( int []arr, int n)
{
return arr[n / 2];
}
public static void Main()
{
int []arr = { 1, 2, 2, 3 };
int n = arr.Length;
Console.WriteLine(findMajority(arr, n));
}
}
|
PHP
<?php
function findMajority( $arr , $n )
{
return $arr [ intval ( $n / 2)];
}
$arr = array (1, 2, 2, 3);
$n = count ( $arr );
echo findMajority( $arr , $n );
?>
|
Javascript
<script>
function findMajority(arr, n)
{
return arr[(Math.floor(n / 2))];
}
let arr = [ 1, 2, 2, 3 ];
let n = arr.length;
document.write(findMajority(arr, n));
</script>
|
Time Complexity : O(1)
Auxiliary Space: O(1)
Related Articles :
Majority element in an unsorted array
Check for majority element in a sorted array
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Last Updated :
24 Mar, 2023
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