Majority Element - GeeksforGeeks

Write a function which takes an array and prints the majority element (if it exists), otherwise prints “No Majority Element”. A majority element in an array A[] of size n is an element that appears more than n/2 times (and hence there is at most one such element).
Examples :

Input : {3, 3, 4, 2, 4, 4, 2, 4, 4}
Output : 4
Explanation: The frequency of 4 is 5 which is greater
than the half of the size of the array size. 

Input : {3, 3, 4, 2, 4, 4, 2, 4}
Output : No Majority Element
Explanation: There is no element whose frequency is
greater than the half of the size of the array size.

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

METHOD 1

Approach: The basic solution is to have two loops and keep track of the maximum count for all different elements. If maximum count becomes greater than n/2 then break the loops and return the element having maximum count. If the maximum count doesn’t become more than n/2 then majority element doesn’t exist.
Algorithm:
1. Create a variable to store the max count, count = 0
2. Traverse through the array from start to end.
3. For every element in the array run another loop to find the count of similar elements in the given array.
4. If the count is greater than the max count update the max count and store the index in another varaible.
5. If the maximum count is greater than the half the size of the array, print the element. Else print there is no majority element.

Implementation:

C++

// C++ program to find Majority  
// element in an array  
#include <bits/stdc++.h>  
using namespace std;  
  
// Function to find Majority element  
// in an array  
void findMajority(int arr[], int n)  
{  
    int maxCount = 0;  
    int index = -1; // sentinels  
    for(int i = 0; i < n; i++)  
    {  
        int count = 0;  
        for(int j = 0; j < n; j++)  
        {  
            if(arr[i] == arr[j])  
            count++;  
        }  
          
        // update maxCount if count of  
        // current element is greater  
        if(count > maxCount)  
        {  
            maxCount = count;  
            index = i;  
        }  
    }  
      
    // if maxCount is greater than n/2  
    // return the corresponding element  
    if (maxCount > n/2)  
    cout << arr[index] << endl;  
      
    else
        cout << "No Majority Element" << endl;  
}  
  
// Driver code  
int main()  
{  
    int arr[] = {1, 1, 2, 1, 3, 5, 1};  
    int n = sizeof(arr) / sizeof(arr[0]);  
      
    // Function calling  
    findMajority(arr, n);  
  
    return 0;  
}  

Java

// Java  program to find Majority  
// element in an array  
  
import java.io.*; 
  
class GFG { 
      
// Function to find Majority element  
// in an array  
static void findMajority(int arr[], int n)  
{  
    int maxCount = 0;  
    int index = -1; // sentinels  
    for(int i = 0; i < n; i++)  
    {  
        int count = 0;  
        for(int j = 0; j < n; j++)  
        {  
            if(arr[i] == arr[j])  
            count++;  
        }  
          
        // update maxCount if count of  
        // current element is greater  
        if(count > maxCount)  
        {  
            maxCount = count;  
            index = i;  
        }  
    }  
      
    // if maxCount is greater than n/2  
    // return the corresponding element  
    if (maxCount > n/2)  
    System.out.println (arr[index]);  
      
    else
    System.out.println ("No Majority Element");  
}  
  
// Driver code  
    public static void main (String[] args) { 
  
        int arr[] = {1, 1, 2, 1, 3, 5, 1};  
        int n = arr.length;  
      
    // Function calling  
    findMajority(arr, n);  
    } 
//This code is contributed by ajit.     
} 

Python 3

# Python 3 program to find Majority  
# element in an array 
  
# Function to find Majority  
# element in an array 
def findMajority(arr, n): 
  
    maxCount = 0; 
    index = -1 # sentinels 
    for i in range(n): 
      
        count = 0
        for j in range(n): 
          
            if(arr[i] == arr[j]): 
                count += 1
          
        # update maxCount if count of  
        # current element is greater 
        if(count > maxCount): 
          
            maxCount = count 
            index = i 
      
    # if maxCount is greater than n/2  
    # return the corresponding element  
    if (maxCount > n//2): 
        print(arr[index]) 
      
    else: 
        print("No Majority Element") 
  
# Driver code 
if __name__ == "__main__": 
    arr = [1, 1, 2, 1, 3, 5, 1] 
    n = len(arr) 
      
    # Function calling  
    findMajority(arr, n) 
  
# This code is contributed  
# by ChitraNayal 

C#

// C#  program to find Majority  
// element in an array  
using System; 
  
public class GFG{ 
          
// Function to find Majority element  
// in an array  
static void findMajority(int []arr, int n)  
{  
    int maxCount = 0;  
    int index = -1; // sentinels  
    for(int i = 0; i < n; i++)  
    {  
        int count = 0;  
        for(int j = 0; j < n; j++)  
        {  
            if(arr[i] == arr[j])  
            count++;  
        }  
          
        // update maxCount if count of  
        // current element is greater  
        if(count > maxCount)  
        {  
            maxCount = count;  
            index = i;  
        }  
    }  
      
    // if maxCount is greater than n/2  
    // return the corresponding element  
    if (maxCount > n/2)  
    Console.WriteLine (arr[index]);  
      
    else
    Console.WriteLine("No Majority Element");  
}  
  
// Driver code  
    static public void Main (){ 
          
        int []arr = {1, 1, 2, 1, 3, 5, 1};  
        int n = arr.Length;  
      
        // Function calling  
        findMajority(arr, n);  
    } 
//This code is contributed by Tushil..  
} 

PHP

<?php 
// PHP program to find Majority  
// element in an array 
  
// Function to find Majority element 
// in an array 
function findMajority($arr, $n) 
{ 
    $maxCount = 0;  
    $index = -1; // sentinels 
    for($i = 0; $i < $n; $i++) 
    { 
        $count = 0; 
        for($j = 0; $j < $n; $j++) 
        { 
            if($arr[$i] == $arr[$j]) 
            $count++; 
        } 
          
        // update maxCount if count of  
        // current element is greater 
        if($count > $maxCount) 
        { 
            $maxCount = $count; 
            $index = $i; 
        } 
    } 
      
    // if maxCount is greater than n/2  
    // return the corresponding element  
    if ($maxCount > $n/2) 
        echo $arr[$index] . "\n"; 
    else
        echo "No Majority Element" . "\n"; 
} 
  
// Driver code 
$arr = array(1, 1, 2, 1, 3, 5, 1); 
$n = sizeof($arr); 
      
// Function calling  
findMajority($arr, $n); 
  
// This code is contributed  
// by Akanksha Rai 

Output:
```
1
```
Compelxity Analysis:
- Time Complexity: O(n*n).
  A nested loop is needed where both the loops traverse the array from start to end, so the time complexity is O(n^2).
- Auxiliary Space : O(1).
  As no extra space is required for any operation so the space complexity is constant.

METHOD 2 (Using Binary Search Tree)

Approach: Insert elements in BST one by one and if an element is already present then increment the count of the node. At any stage, if the count of a node becomes more than n/2 then return.
Algorithm:
1. Create a binary search tree, if same element is entered in the binary search tree the frequency of the node is increased.
2. traverse the array and insert the element in the binary search tree.
3. If the maximum frequency of any node is greater than the half the size of the array, then perform a inorder traversal and find the node with frequency greater than half
4. Else print No majority Element.

Implementation:

// C++ program to demonstrate insert operation in binary search tree.  
#include<bits/stdc++.h>   
using namespace std; 
  
struct node  
{  
    int key; 
    int c = 0; 
    struct node *left, *right;  
};  
  
// A utility function to create a new BST node  
struct node *newNode(int item)  
{  
    struct node *temp = (struct node *)malloc(sizeof(struct node));  
    temp->key = item; 
    temp->c = 1; 
    temp->left = temp->right = NULL;  
    return temp;  
}  
  
  
/* A utility function to insert a new node with given key in BST */
struct node* insert(struct node* node, int key,int &ma)  
{  
    /* If the tree is empty, return a new node */
    if (node == NULL)  
    { 
        if(ma==0) 
            ma=1; 
          
        return newNode(key);  
    } 
      
    /* Otherwise, recur down the tree */
    if (key < node->key)  
        node->left = insert(node->left, key, ma);  
    else if (key > node->key)  
        node->right = insert(node->right, key, ma);  
    else
        node->c++; 
          
    //find the max count 
    ma = max(ma, node->c); 
      
    /* return the (unchanged) node pointer */
    return node;  
}  
// A utility function to do inorder traversal of BST  
void inorder(struct node *root,int s)  
{  
    if (root != NULL)  
    {  
        inorder(root->left,s); 
          
        if(root->c>(s/2)) 
            printf("%d \n", root->key);  
          
        inorder(root->right,s);  
    }  
}  
  
// Driver Program to test above functions  
int main()  
{  
    int a[] = {1, 3, 3, 3, 2}; 
    int size = (sizeof(a))/sizeof(a[0]); 
      
    struct node *root = NULL;  
    int ma=0; 
      
    for(int i=0;i<size;i++) 
    { 
        root = insert(root, a[i],ma);  
    } 
      
    if(ma>(size/2)) 
        inorder(root,size); 
    else 
        cout<<"No majority element\n"; 
    return 0;  
}  

Output:

Compelxity Analysis:
- Time Complexity : If a Binary Search Tree is used then time complexity will be O(n^2). If a self-balancing-binary-search tree is used then it will be O(nlogn)
- Auxiliary Space : O(n).
  As extra space is needed to store the array in tree.

METHOD 3 (Using Moore’s Voting Algorithm): This method only works when the majority element does exist in the array. In the problem definition, it is said that the majority element may or may not exist but for applying this approach let’s assume that the majority element does exist in the given input array.

Approach: This is a two-step process.
- The first step gives the element that maybe the majority element in the array. If there is a majority element in an array, then this step will definitely return majority element, otherwise, it will return candidate for majority element.
- Check if the element obtained from the above step is majority element. This step is necessary as there might be no majority element.
Step 1: Finding a Candidate
The algorithm for the first phase that works in O(n) is known as Moore’s Voting Algorithm. Basic idea of the algorithm is that if each occurrence of an element e can be cancelled with all the other elements that are different from e then e will exist till end if it is a majority element.

Step 2: Check if the element obtained in step 1 is majority element or not.
Traverse through the array and check if the count of the element found is greater than half the size of the array, then print the answer else print “No majority element”.
Algorithm:
1. Loop through each element and maintains a count of majority element, and a majority index, maj_index
2. If the next element is same then increment the count if the next element is not same then decrement the count.
3. if the count reaches 0 then changes the maj_index to the current element and set the count again to 1.
4. Now again traverse through the array and find the count of majority element found.
5. If the count is greater than half the size of the array, print the element
6. Else print that there is no majority element

Implementation:

C++

/* C++ Program for finding out  
   majority element in an array */
#include <bits/stdc++.h> 
using namespace std; 
  
/* Function to find the candidate for Majority */
int findCandidate(int a[], int size) 
{ 
    int maj_index = 0, count = 1; 
    for (int i = 1; i < size; i++) 
    { 
        if (a[maj_index] == a[i]) 
            count++; 
        else
            count--; 
        if (count == 0) 
        { 
            maj_index = i; 
            count = 1; 
        } 
    } 
    return a[maj_index]; 
} 
  
/* Function to check if the candidate 
   occurs more than n/2 times */
bool isMajority(int a[], int size, int cand) 
{ 
    int count = 0; 
    for (int i = 0; i < size; i++) 
      
    if (a[i] == cand) 
    count++; 
          
    if (count > size/2) 
    return 1; 
      
    else
    return 0; 
} 
  
/* Function to print Majority Element */
void printMajority(int a[], int size) 
{ 
   /* Find the candidate for Majority*/
   int cand = findCandidate(a, size); 
  
   /* Print the candidate if it is Majority*/
   if (isMajority(a, size, cand)) 
   cout << " " << cand << " "; 
     
   else
   cout << "No Majority Element"; 
} 
  
  
/* Driver function to test above functions */
int main() 
{ 
    int a[] = {1, 3, 3, 1, 2}; 
    int size = (sizeof(a))/sizeof(a[0]); 
      
    // Function calling 
    printMajority(a, size); 
      
    return 0; 
} 

C

/* Program for finding out majority element in an array */
# include<stdio.h> 
# define bool int 
  
int findCandidate(int *, int); 
bool isMajority(int *, int, int); 
  
/* Function to print Majority Element */
void printMajority(int a[], int size) 
{ 
  /* Find the candidate for Majority*/
  int cand = findCandidate(a, size); 
  
  /* Print the candidate if it is Majority*/
  if (isMajority(a, size, cand)) 
    printf(" %d ", cand); 
  else
    printf("No Majority Element"); 
} 
  
/* Function to find the candidate for Majority */
int findCandidate(int a[], int size) 
{ 
    int maj_index = 0, count = 1; 
    int i; 
    for (i = 1; i < size; i++) 
    { 
        if (a[maj_index] == a[i]) 
            count++; 
        else
            count--; 
        if (count == 0) 
        { 
            maj_index = i; 
            count = 1; 
        } 
    } 
    return a[maj_index]; 
} 
  
/* Function to check if the candidate occurs more than n/2 times */
bool isMajority(int a[], int size, int cand) 
{ 
    int i, count = 0; 
    for (i = 0; i < size; i++) 
      if (a[i] == cand) 
         count++; 
    if (count > size/2) 
       return 1; 
    else
       return 0; 
} 
  
/* Driver function to test above functions */
int main() 
{ 
    int a[] = {1, 3, 3, 1, 2}; 
    int size = (sizeof(a))/sizeof(a[0]); 
    printMajority(a, size); 
    getchar(); 
    return 0; 
} 

Java

/* Program for finding out majority element in an array */
  
class MajorityElement  
{ 
    /* Function to print Majority Element */
    void printMajority(int a[], int size)  
    { 
        /* Find the candidate for Majority*/
        int cand = findCandidate(a, size); 
  
        /* Print the candidate if it is Majority*/
        if (isMajority(a, size, cand)) 
            System.out.println(" " + cand + " "); 
        else 
            System.out.println("No Majority Element"); 
    } 
  
    /* Function to find the candidate for Majority */
    int findCandidate(int a[], int size)  
    { 
        int maj_index = 0, count = 1; 
        int i; 
        for (i = 1; i < size; i++)  
        { 
            if (a[maj_index] == a[i]) 
                count++; 
            else
                count--; 
            if (count == 0) 
            { 
                maj_index = i; 
                count = 1; 
            } 
        } 
        return a[maj_index]; 
    } 
  
    /* Function to check if the candidate occurs more 
       than n/2 times */
    boolean isMajority(int a[], int size, int cand)  
    { 
        int i, count = 0; 
        for (i = 0; i < size; i++)  
        { 
            if (a[i] == cand) 
                count++; 
        } 
        if (count > size / 2)  
            return true; 
        else
            return false; 
    } 
  
    /* Driver program to test the above functions */
    public static void main(String[] args)  
    { 
        MajorityElement majorelement = new MajorityElement(); 
        int a[] = new int[]{1, 3, 3, 1, 2}; 
        int size = a.length; 
        majorelement.printMajority(a, size); 
    } 
} 
  
// This code has been contributed by Mayank Jaiswal 

Python

# Program for finding out majority element in an array 
  
# Function to find the candidate for Majority 
def findCandidate(A): 
    maj_index = 0
    count = 1
    for i in range(len(A)): 
        if A[maj_index] == A[i]: 
            count += 1
        else: 
            count -= 1
        if count == 0: 
            maj_index = i 
            count = 1
    return A[maj_index] 
  
# Function to check if the candidate occurs more than n/2 times 
def isMajority(A, cand): 
    count = 0
    for i in range(len(A)): 
        if A[i] == cand: 
            count += 1
    if count > len(A)/2: 
        return True
    else: 
        return False
  
# Function to print Majority Element 
def printMajority(A): 
    # Find the candidate for Majority 
    cand = findCandidate(A) 
      
    # Print the candidate if it is Majority 
    if isMajority(A, cand) == True: 
        print(cand) 
    else: 
        print("No Majority Element") 
  
# Driver program to test above functions 
A = [1, 3, 3, 1, 2] 
printMajority(A)  

C#

// C# Program for finding out majority element in an array 
using System; 
  
class GFG 
{ 
    /* Function to print Majority Element */
    static void printMajority(int []a, int size)  
    { 
        /* Find the candidate for Majority*/
        int cand = findCandidate(a, size); 
  
        /* Print the candidate if it is Majority*/
        if (isMajority(a, size, cand)) 
            Console.Write(" " + cand + " "); 
        else
            Console.Write("No Majority Element"); 
    } 
  
    /* Function to find the candidate for Majority */
    static int findCandidate(int []a, int size)  
    { 
        int maj_index = 0, count = 1; 
        int i; 
        for (i = 1; i < size; i++)  
        { 
            if (a[maj_index] == a[i]) 
                count++; 
            else
                count--; 
                  
            if (count == 0) 
            { 
                maj_index = i; 
                count = 1; 
            } 
        } 
        return a[maj_index]; 
    } 
  
    // Function to check if the candidate  
    // occurs more than n/2 times 
    static bool isMajority(int []a, int size, int cand)  
    { 
        int i, count = 0; 
        for (i = 0; i < size; i++)  
        { 
            if (a[i] == cand) 
                count++; 
        } 
        if (count > size / 2)  
            return true; 
        else
            return false; 
    } 
  
    // Driver Code 
    public static void Main()  
    { 
          
        int []a = {1, 3, 3, 1, 2}; 
        int size = a.Length; 
        printMajority(a, size); 
    } 
} 
  
// This code is contributed by Sam007 

PHP

<?php 
// PHP Program for finding out majority  
// element in an array  
  
// Function to find the candidate  
// for Majority  
function findCandidate($a, $size) 
{ 
    $maj_index = 0; 
    $count = 1; 
    for ($i = 1; $i < $size; $i++) 
    { 
        if ($a[$maj_index] == $a[$i]) 
            $count++; 
        else
            $count--; 
        if ($count == 0) 
        { 
            $maj_index = $i; 
            $count = 1; 
        } 
    } 
    return $a[$maj_index]; 
} 
  
// Function to check if the candidate 
// occurs more than n/2 times  
function isMajority($a, $size, $cand) 
{ 
    $count = 0; 
    for ($i = 0; $i < $size; $i++) 
      
    if ($a[$i] == $cand) 
    $count++; 
          
    if ($count > $size / 2) 
    return 1; 
      
    else
    return 0; 
} 
  
// Function to print Majority Element  
function printMajority($a, $size) 
{ 
    /* Find the candidate for Majority*/
    $cand = findCandidate($a, $size); 
      
    /* Print the candidate if it is Majority*/
    if (isMajority($a, $size, $cand)) 
        echo " ", $cand, " "; 
    else
        echo "No Majority Element"; 
} 
  
// Driver Code 
$a = array(1, 3, 3, 1, 2); 
$size = sizeof($a); 
  
// Function calling 
printMajority($a, $size); 
  
// This code is contributed by jit_t 
?> 

Output:

No Majority Element

Complexity Analysis:
- Time Complexity: O(n).
  As two traversal of the array is needed, so the time complexity is linear.
- Auxiliary Space : O(1).
  As no extra space is required.

METHOD 4 (Using Hashmap):

Approach: This method is somewhat similar to Moore voting algorithm in terms of time complexity, but in this case, there is no need for the second step of Moore voting algorithm. But as usual, here space complexity becomes O(n).
In Hashmap(key-value pair), at value, maintain a count for each element(key) and whenever the count is greater than half of the array length, return that key(majority element).
Algorithm:
1. Create a hashmap to store a key-value pair, i.e. element-frequency pair.
2. Traverse the array from start to end.
3. For every element in the array, insert the element in the hashmap if the element does not exist as key, else fetch the value of the key ( array[i] ) and increase the value by 1
4. If the count is greater than half then print the majority element and break.
5. If no majority element is found print “No Majority element”

Implementation:

C++

/* C++ program for finding out majority  
element in an array */
#include <bits/stdc++.h> 
using namespace std; 
  
void findMajority(int arr[], int size) 
{ 
    unordered_map<int, int> m; 
    for(int i = 0; i < size; i++) 
        m[arr[i]]++; 
    int count = 0; 
    for(auto i : m) 
    { 
        if(i.second > size / 2) 
        { 
            count =1; 
            cout << "Majority found :- " << i.first<<endl; 
            break; 
        } 
    } 
    if(count == 0) 
        cout << "No Majority element" << endl; 
} 
  
// Driver code  
int main()  
{  
    int arr[] = {2, 2, 2, 2, 5, 5, 2, 3, 3};  
    int n = sizeof(arr) / sizeof(arr[0]);  
      
    // Function calling  
    findMajority(arr, n);  
  
    return 0;  
}  
  
// This code is contributed by codeMan_d 

Java

import java.util.HashMap; 
  
/* Program for finding out majority element in an array */
   
class MajorityElement  
{ 
    private static void findMajority(int[] arr)  
    { 
        HashMap<Integer,Integer> map = new HashMap<Integer, Integer>(); 
  
        for(int i = 0; i < arr.length; i++) { 
            if (map.containsKey(arr[i])) { 
                    int count = map.get(arr[i]) +1; 
                    if (count > arr.length /2) { 
                        System.out.println("Majority found :- " + arr[i]); 
                        return; 
                    } else
                        map.put(arr[i], count); 
  
            } 
            else
                map.put(arr[i],1); 
            } 
            System.out.println(" No Majority element"); 
    } 
  
   
    /* Driver program to test the above functions */
    public static void main(String[] args)  
    { 
        int a[] = new int[]{2,2,2,2,5,5,2,3,3}; 
          
        findMajority(a); 
    } 
} 
// This code is contributed by  karan malhotra 

Python3

# Python program for finding out majority  
# element in an array  
  
def findMajority(arr, size): 
    m = {} 
    for i in range(size): 
        if arr[i] in m: 
            m[arr[i]] += 1
        else: 
            m[arr[i]] = 1
    count = 0
    for key in m: 
        if m[key] > size / 2: 
            count = 1
            print("Majority found :-",key) 
            break
    if(count == 0): 
        print("No Majority element") 
  
# Driver code  
arr = [2, 2, 2, 2, 5, 5, 2, 3, 3]  
n = len(arr) 
  
# Function calling  
findMajority(arr, n) 
  
# This code is contributed by ankush_953 

C#

// C# Program for finding out majority 
// element in an array  
using System; 
using System.Collections.Generic; 
  
class GFG 
{ 
private static void findMajority(int[] arr) 
{ 
    Dictionary<int,  
               int> map = new Dictionary<int,  
                                         int>(); 
  
    for (int i = 0; i < arr.Length; i++) 
    { 
        if (map.ContainsKey(arr[i])) 
        { 
                int count = map[arr[i]] + 1; 
                if (count > arr.Length / 2) 
                { 
                    Console.WriteLine("Majority found :- " +  
                                                    arr[i]); 
                    return; 
                } 
                else
                { 
                    map[arr[i]] = count; 
                } 
  
        } 
        else
        { 
            map[arr[i]] = 1; 
        } 
    } 
    Console.WriteLine(" No Majority element"); 
} 
  
  
// Driver Code 
public static void Main(string[] args) 
{ 
    int[] a = new int[]{2, 2, 2, 2,  
                        5, 5, 2, 3, 3}; 
  
    findMajority(a); 
} 
} 
  
// This code is contributed by Shrikant13 

Output:

Majority found :- 2

Thanks Ashwani Tanwar, Karan Malhotra for suggesting this.

Complexity Analysis:
- Time Complexity: O(n).
  One traversal of the array is needed, so the time complexity is linear.
- Auxiliary Space : O(n).
  Since a hashmap requires linear space.

METHOD 5

Approach:The idea is to sort the array. Sorting makes similar elements in the array adjacent, so traverse the array and update the count until the present element is similar to the previous one. If the frequency is more than half the size of the array, print the majority element.
Algorithm:
1. Sort the array and create a varibale count and previous ,prev = INT_MIN.
2. Traverse the element from start to end.
3. If the current element is equal to the previous element increase the count.
4. Else set the count to 1.
5. If the count is greater than half the size of array, print the element as majority element and break.
6. If no majority element found, print “No majority element”

Implementation:

// C++ program to find Majority  
// element in an array 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to find Majority element 
// in an array 
// it returns -1 if there is no majority element 
  
int majorityElement(int *arr, int n) 
{ 
    // sort the array in O(nlogn) 
    sort(arr, arr+n); 
      
    int count = 1, max_ele = -1, temp = arr[0], ele, f=0; 
      
    for(int i=1;i<n;i++) 
    { 
        // increases the count if the same element occurs 
        // otherwise starts counting new element 
        if(temp==arr[i]) 
        { 
            count++; 
        } 
        else
        { 
            count = 1; 
            temp = arr[i]; 
        } 
          
        // sets maximum count 
        // and stores maximum occured element so far 
        // if maximum count becomes greater than n/2 
        // it breaks out setting the flag 
        if(max_ele<count) 
        { 
            max_ele = count; 
            ele = arr[i]; 
              
            if(max_ele>(n/2)) 
            { 
                f = 1; 
                break; 
            } 
        } 
    } 
      
    // returns maximum occured element 
    // if there is no such element, returns -1 
    return (f==1 ? ele : -1); 
} 
  
  
// Driver code 
int main() 
{ 
    int arr[] = {1, 1, 2, 1, 3, 5, 1}; 
    int n = sizeof(arr) / sizeof(arr[0]); 
      
    // Function calling  
    cout<<majorityElement(arr, n); 
  
    return 0; 
} 

Output:

Complexity Analysis:
- Time Complexity : O(nlogn).
  Sorting requires O(n log n) time complexity.
- Auxiliary Space : O(1).
  As no extra space is required.

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My Personal Notes

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