Check for Majority Element in a sorted array
Last Updated :
23 Jul, 2025
Given an array arr of N elements, A majority element in an array arr of size N is an element that appears more than N/2 times in the array. The task is to write a function say isMajority() that takes an array (arr[] ), array’s size (n) and a number to be searched (x) as parameters and returns true if x is a majority element (present more than n/2 times).
Examples:
Input: arr[] = {1, 2, 3, 3, 3, 3, 10}, x = 3
Output: True (x appears more than n/2 times in the given array)Input: arr[] = {1, 1, 2, 4, 4, 4, 6, 6}, x = 4
Output: False (x doesn't appear more than n/2 times in the given array)Input: arr[] = {1, 1, 1, 2, 2}, x = 1
Output: True (x appears more than n/2 times in the given array)
METHOD 1 (Using Linear Search): Linearly search for the first occurrence of the element, once you find it (let at index i), check the element at index i + n/2. If the element is present at i+n/2 then return 1 else return 0.
C++
/* C++ Program to check for majority element in a sorted array */
#include<bits/stdc++.h>
using namespace std;
bool isMajority(int arr[], int n, int x)
{
int i;
/* get last index according to n (even or odd) */
int last_index = n % 2 ? (n / 2 + 1): (n / 2);
/* search for first occurrence of x in arr[]*/
for (i = 0; i < last_index; i++)
{
/* check if x is present and is present more than n/2
times */
if (arr[i] == x && arr[i + n / 2] == x)
return 1;
}
return 0;
}
/* Driver code */
int main()
{
int arr[] ={1, 2, 3, 4, 4, 4, 4};
int n = sizeof(arr)/sizeof(arr[0]);
int x = 4;
if (isMajority(arr, n, x))
cout << x <<" appears more than "<<
n/2 << " times in arr[]"<< endl;
else
cout <<x <<" does not appear more than" << n/2 <<" times in arr[]" << endl;
return 0;
}
// This code is contributed by shivanisinghss2110
/* C Program to check for majority element in a sorted array */
# include <stdio.h>
# include <stdbool.h>
bool isMajority(int arr[], int n, int x)
{
int i;
/* get last index according to n (even or odd) */
int last_index = n%2? (n/2+1): (n/2);
/* search for first occurrence of x in arr[]*/
for (i = 0; i < last_index; i++)
{
/* check if x is present and is present more than n/2
times */
if (arr[i] == x && arr[i+n/2] == x)
return 1;
}
return 0;
}
/* Driver program to check above function */
int main()
{
int arr[] ={1, 2, 3, 4, 4, 4, 4};
int n = sizeof(arr)/sizeof(arr[0]);
int x = 4;
if (isMajority(arr, n, x))
printf("%d appears more than %d times in arr[]",
x, n/2);
else
printf("%d does not appear more than %d times in arr[]",
x, n/2);
return 0;
}
/* Program to check for majority element in a sorted array */
import java.io.*;
class Majority {
static boolean isMajority(int arr[], int n, int x)
{
int i, last_index = 0;
/* get last index according to n (even or odd) */
last_index = (n%2==0)? n/2: n/2+1;
/* search for first occurrence of x in arr[]*/
for (i = 0; i < last_index; i++)
{
/* check if x is present and is present more
than n/2 times */
if (arr[i] == x && arr[i+n/2] == x)
return true;
}
return false;
}
/* Driver function to check for above functions*/
public static void main (String[] args) {
int arr[] = {1, 2, 3, 4, 4, 4, 4};
int n = arr.length;
int x = 4;
if (isMajority(arr, n, x)==true)
System.out.println(x+" appears more than "+
n/2+" times in arr[]");
else
System.out.println(x+" does not appear more than "+
n/2+" times in arr[]");
}
}
/*This article is contributed by Devesh Agrawal*/
'''Python3 Program to check for majority element in a sorted array'''
def isMajority(arr, n, x):
# get last index according to n (even or odd) */
last_index = (n//2 + 1) if n % 2 != 0 else (n//2)
# search for first occurrence of x in arr[]*/
for i in range(last_index):
# check if x is present and is present more than n / 2 times */
if arr[i] == x and arr[i + n//2] == x:
return 1
# Driver program to check above function */
arr = [1, 2, 3, 4, 4, 4, 4]
n = len(arr)
x = 4
if (isMajority(arr, n, x)):
print ("% d appears more than % d times in arr[]"
%(x, n//2))
else:
print ("% d does not appear more than % d times in arr[]"
%(x, n//2))
# This code is contributed by shreyanshi_arun.
// C# Program to check for majority
// element in a sorted array
using System;
class GFG {
static bool isMajority(int[] arr,
int n, int x)
{
int i, last_index = 0;
// Get last index according to
// n (even or odd)
last_index = (n % 2 == 0) ? n / 2 :
n / 2 + 1;
// Search for first occurrence
// of x in arr[]
for (i = 0; i < last_index; i++) {
// Check if x is present and
// is present more than n/2 times
if (arr[i] == x && arr[i + n / 2] == x)
return true;
}
return false;
}
// Driver code
public static void Main()
{
int[] arr = { 1, 2, 3, 4, 4, 4, 4 };
int n = arr.Length;
int x = 4;
if (isMajority(arr, n, x) == true)
Console.Write(x + " appears more than " +
n / 2 + " times in arr[]");
else
Console.Write(x + " does not appear more than " +
n / 2 + " times in arr[]");
}
}
// This code is contributed by Sam007
<script>
// Javascript Program to check for majority
// element in a sorted array
function isMajority(arr, n, x)
{
let i, last_index = 0;
// Get last index according to
// n (even or odd)
last_index = (n % 2 == 0) ?
parseInt(n / 2, 10) : parseInt(n / 2, 10) + 1;
// Search for first occurrence
// of x in arr[]
for (i = 0; i < last_index; i++) {
// Check if x is present and
// is present more than n/2 times
if (arr[i] == x && arr[i +
parseInt(n / 2, 10)] == x)
return true;
}
return false;
}
let arr = [ 1, 2, 3, 4, 4, 4, 4 ];
let n = arr.length;
let x = 4;
if (isMajority(arr, n, x) == true)
document.write(x + " appears more than " +
parseInt(n / 2, 10) + " times in arr[]");
else
document.write(x + " does not appear more than " +
parseInt(n / 2, 10) + " times in arr[]");
</script>
<?php
// PHP Program to check for
// majority element in a
// sorted array
// function returns majority
// element in a sorted array
function isMajority($arr, $n, $x)
{
$i;
// get last index according
// to n (even or odd)
$last_index = $n % 2? ($n / 2 + 1): ($n / 2);
// search for first occurrence
// of x in arr[]
for ($i = 0; $i < $last_index; $i++)
{
// check if x is present and
// is present more than n/2
// times
if ($arr[$i] == $x && $arr[$i + $n / 2] == $x)
return 1;
}
return 0;
}
// Driver Code
$arr = array(1, 2, 3, 4, 4, 4, 4);
$n = sizeof($arr);
$x = 4;
if (isMajority($arr, $n, $x))
echo $x, " appears more than "
, floor($n / 2), " times in arr[]";
else
echo $x, "does not appear more than "
, floor($n / 2), "times in arr[]";
// This code is contributed by Ajit
?>
Output4 appears more than 3 times in arr[]
Time Complexity: O(n)
Auxiliary Space: O(1)
METHOD 2 (Using Binary Search): Use binary search methodology to find the first occurrence of the given number. The criteria for binary search is important here.
C++
// C++ program to check for majority
// element in a sorted array
#include<bits/stdc++.h>
using namespace std;
// If x is present in arr[low...high]
// then returns the index of first
// occurrence of x, otherwise returns -1
int _binarySearch(int arr[], int low,
int high, int x);
// This function returns true if the x
// is present more than n/2 times in
// arr[] of size n
bool isMajority(int arr[], int n, int x)
{
// Find the index of first occurrence
// of x in arr[]
int i = _binarySearch(arr, 0, n - 1, x);
// If element is not present at all,
// return false
if (i == -1)
return false;
// Check if the element is present
// more than n/2 times
if (((i + n / 2) <= (n - 1)) &&
arr[i + n / 2] == x)
return true;
else
return false;
}
// If x is present in arr[low...high] then
// returns the index of first occurrence
// of x, otherwise returns -1
int _binarySearch(int arr[], int low,
int high, int x)
{
if (high >= low)
{
int mid = (low + high)/2; /*low + (high - low)/2;*/
/* Check if arr[mid] is the first occurrence of x.
arr[mid] is first occurrence if x is one of
the following is true:
(i) mid == 0 and arr[mid] == x
(ii) arr[mid-1] < x and arr[mid] == x
*/
if ((mid == 0 || x > arr[mid - 1]) &&
(arr[mid] == x) )
return mid;
else if (x > arr[mid])
return _binarySearch(arr, (mid + 1),
high, x);
else
return _binarySearch(arr, low,
(mid - 1), x);
}
return -1;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 3, 3, 3, 10 };
int n = sizeof(arr) / sizeof(arr[0]);
int x = 3;
if (isMajority(arr, n, x))
cout << x << " appears more than "
<< n / 2 << " times in arr[]"
<< endl;
else
cout << x << " does not appear more than"
<< n / 2 << " times in arr[]" << endl;
return 0;
}
// This code is contributed by shivanisinghss2110
/* C Program to check for majority element in a sorted array */
# include <stdio.h>
# include <stdbool.h>
/* If x is present in arr[low...high] then returns the index of
first occurrence of x, otherwise returns -1 */
int _binarySearch(int arr[], int low, int high, int x);
/* This function returns true if the x is present more than n/2
times in arr[] of size n */
bool isMajority(int arr[], int n, int x)
{
/* Find the index of first occurrence of x in arr[] */
int i = _binarySearch(arr, 0, n-1, x);
/* If element is not present at all, return false*/
if (i == -1)
return false;
/* check if the element is present more than n/2 times */
if (((i + n/2) <= (n -1)) && arr[i + n/2] == x)
return true;
else
return false;
}
/* If x is present in arr[low...high] then returns the index of
first occurrence of x, otherwise returns -1 */
int _binarySearch(int arr[], int low, int high, int x)
{
if (high >= low)
{
int mid = (low + high)/2; /*low + (high - low)/2;*/
/* Check if arr[mid] is the first occurrence of x.
arr[mid] is first occurrence if x is one of the following
is true:
(i) mid == 0 and arr[mid] == x
(ii) arr[mid-1] < x and arr[mid] == x
*/
if ( (mid == 0 || x > arr[mid-1]) && (arr[mid] == x) )
return mid;
else if (x > arr[mid])
return _binarySearch(arr, (mid + 1), high, x);
else
return _binarySearch(arr, low, (mid -1), x);
}
return -1;
}
/* Driver program to check above functions */
int main()
{
int arr[] = {1, 2, 3, 3, 3, 3, 10};
int n = sizeof(arr)/sizeof(arr[0]);
int x = 3;
if (isMajority(arr, n, x))
printf("%d appears more than %d times in arr[]",
x, n/2);
else
printf("%d does not appear more than %d times in arr[]",
x, n/2);
return 0;
}
/* Java Program to check for majority element in a sorted array */
import java.io.*;
class Majority {
/* If x is present in arr[low...high] then returns the index of
first occurrence of x, otherwise returns -1 */
static int _binarySearch(int arr[], int low, int high, int x)
{
if (high >= low)
{
int mid = (low + high)/2; /*low + (high - low)/2;*/
/* Check if arr[mid] is the first occurrence of x.
arr[mid] is first occurrence if x is one of the following
is true:
(i) mid == 0 and arr[mid] == x
(ii) arr[mid-1] < x and arr[mid] == x
*/
if ( (mid == 0 || x > arr[mid-1]) && (arr[mid] == x) )
return mid;
else if (x > arr[mid])
return _binarySearch(arr, (mid + 1), high, x);
else
return _binarySearch(arr, low, (mid -1), x);
}
return -1;
}
/* This function returns true if the x is present more than n/2
times in arr[] of size n */
static boolean isMajority(int arr[], int n, int x)
{
/* Find the index of first occurrence of x in arr[] */
int i = _binarySearch(arr, 0, n-1, x);
/* If element is not present at all, return false*/
if (i == -1)
return false;
/* check if the element is present more than n/2 times */
if (((i + n/2) <= (n -1)) && arr[i + n/2] == x)
return true;
else
return false;
}
/*Driver function to check for above functions*/
public static void main (String[] args) {
int arr[] = {1, 2, 3, 3, 3, 3, 10};
int n = arr.length;
int x = 3;
if (isMajority(arr, n, x)==true)
System.out.println(x + " appears more than "+
n/2 + " times in arr[]");
else
System.out.println(x + " does not appear more than " +
n/2 + " times in arr[]");
}
}
/*This code is contributed by Devesh Agrawal*/
'''Python3 Program to check for majority element in a sorted array'''
# This function returns true if the x is present more than n / 2
# times in arr[] of size n */
def isMajority(arr, n, x):
# Find the index of first occurrence of x in arr[] */
i = _binarySearch(arr, 0, n-1, x)
# If element is not present at all, return false*/
if i == -1:
return False
# check if the element is present more than n / 2 times */
if ((i + n//2) <= (n -1)) and arr[i + n//2] == x:
return True
else:
return False
# If x is present in arr[low...high] then returns the index of
# first occurrence of x, otherwise returns -1 */
def _binarySearch(arr, low, high, x):
if high >= low:
mid = (low + high)//2 # low + (high - low)//2;
''' Check if arr[mid] is the first occurrence of x.
arr[mid] is first occurrence if x is one of the following
is true:
(i) mid == 0 and arr[mid] == x
(ii) arr[mid-1] < x and arr[mid] == x'''
if (mid == 0 or x > arr[mid-1]) and (arr[mid] == x):
return mid
elif x > arr[mid]:
return _binarySearch(arr, (mid + 1), high, x)
else:
return _binarySearch(arr, low, (mid -1), x)
return -1
# Driver program to check above functions */
arr = [1, 2, 3, 3, 3, 3, 10]
n = len(arr)
x = 3
if (isMajority(arr, n, x)):
print ("% d appears more than % d times in arr[]"
% (x, n//2))
else:
print ("% d does not appear more than % d times in arr[]"
% (x, n//2))
# This code is contributed by shreyanshi_arun.
// C# Program to check for majority
// element in a sorted array */
using System;
class GFG {
// If x is present in arr[low...high]
// then returns the index of first
// occurrence of x, otherwise returns -1
static int _binarySearch(int[] arr, int low,
int high, int x)
{
if (high >= low) {
int mid = (low + high) / 2;
//low + (high - low)/2;
// Check if arr[mid] is the first
// occurrence of x. arr[mid] is
// first occurrence if x is one of
// the following is true:
// (i) mid == 0 and arr[mid] == x
// (ii) arr[mid-1] < x and arr[mid] == x
if ((mid == 0 || x > arr[mid - 1]) &&
(arr[mid] == x))
return mid;
else if (x > arr[mid])
return _binarySearch(arr, (mid + 1),
high, x);
else
return _binarySearch(arr, low,
(mid - 1), x);
}
return -1;
}
// This function returns true if the x is
// present more than n/2 times in arr[]
// of size n
static bool isMajority(int[] arr, int n, int x)
{
// Find the index of first occurrence
// of x in arr[]
int i = _binarySearch(arr, 0, n - 1, x);
// If element is not present at all,
// return false
if (i == -1)
return false;
// check if the element is present
// more than n/2 times
if (((i + n / 2) <= (n - 1)) &&
arr[i + n / 2] == x)
return true;
else
return false;
}
//Driver code
public static void Main()
{
int[] arr = { 1, 2, 3, 3, 3, 3, 10 };
int n = arr.Length;
int x = 3;
if (isMajority(arr, n, x) == true)
Console.Write(x + " appears more than " +
n / 2 + " times in arr[]");
else
Console.Write(x + " does not appear more than " +
n / 2 + " times in arr[]");
}
}
// This code is contributed by Sam007
<script>
// Javascript Program to check for majority
// element in a sorted array */
// If x is present in arr[low...high]
// then returns the index of first
// occurrence of x, otherwise returns -1
function _binarySearch(arr, low, high, x)
{
if (high >= low) {
let mid = parseInt((low + high) / 2, 10);
//low + (high - low)/2;
// Check if arr[mid] is the first
// occurrence of x. arr[mid] is
// first occurrence if x is one of
// the following is true:
// (i) mid == 0 and arr[mid] == x
// (ii) arr[mid-1] < x and arr[mid] == x
if ((mid == 0 || x > arr[mid - 1]) && (arr[mid] == x))
return mid;
else if (x > arr[mid])
return _binarySearch(arr, (mid + 1), high, x);
else
return _binarySearch(arr, low, (mid - 1), x);
}
return -1;
}
// This function returns true if the x is
// present more than n/2 times in arr[]
// of size n
function isMajority(arr, n, x)
{
// Find the index of first occurrence
// of x in arr[]
let i = _binarySearch(arr, 0, n - 1, x);
// If element is not present at all,
// return false
if (i == -1)
return false;
// check if the element is present
// more than n/2 times
if (((i + parseInt(n / 2, 10)) <= (n - 1)) && arr[i + parseInt(n / 2, 10)] == x)
return true;
else
return false;
}
let arr = [ 1, 2, 3, 3, 3, 3, 10 ];
let n = arr.length;
let x = 3;
if (isMajority(arr, n, x) == true)
document.write(x + " appears more than " + parseInt(n / 2, 10) + " times in arr[]");
else
document.write(x + " does not appear more than " + parseInt(n / 2, 10) + " times in arr[]");
</script>
Output3 appears more than 3 times in arr[]
Time Complexity: O(log n)
Auxiliary Space: O(1)
Algorithmic Paradigm: Divide and Conquer
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