Majority Element

Find the majority element in the array. 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 : A[]={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 : A[] = {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.

Naive Approach:

The basic solution is to have two loops and keep track of the maximum count for all different elements. If the maximum count becomes greater than n/2 then break the loops and return the element having the maximum count. If the maximum count doesn’t become more than n/2 then the majority element doesn’t exist.

Illustration:

arr[] = {3, 4, 3, 2, 4, 4, 4, 4}, n = 8

For i = 0:

• count = 0
• Loop over the array, whenever an element is equal to arr[i] (is 3), increment count
• count of arr[i] is 2, which is less than n/2, hence it can’t be majority element.

For i = 1:

• count = 0
• Loop over the array, whenever an element is equal to arr[i] (is 4), increment count
• count of arr[i] is 5, which is greater than n/2 (i.e 4), hence it will be majority element.

Hence, 4 is the majority element.

Follow the steps below to solve the given problem:

• Create a variable to store the max count, count = 0
• Traverse through the array from start to end.
• For every element in the array run another loop to find the count of similar elements in the given array.
• If the count is greater than the max count update the max count and store the index in another variable.
• If the maximum count is greater than half the size of the array, print the element. Else print there is no majority element.

Below is the implementation of the above idea:

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.
}

Python3

# Python3 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 = 1
        # here we compare the element in
        # ith position with i+1th position
        for j in range(i+1, 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

Javascript

<script>

// Javascript program to find Majority
// element in an array

// Function to find Majority element
// in an array
function findMajority(arr, n)
{
    let maxCount = 0;
    let index = -1; // sentinels

    for(let i = 0; i < n; i++)
    {
        let count = 0;
        for(let 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)
        document.write(arr[index]);
    else
        document.write("No Majority Element");
}

// Driver code
let arr = [ 1, 1, 2, 1, 3, 5, 1 ];
let n = arr.length;

// Function calling
findMajority(arr, n);

// This code is contributed by suresh07

</script>

C

#include <stdio.h>

// 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)
        printf("%d\n", arr[index]);

    else
        printf("No Majority Element\n");
}

// 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;
}

// This code is contributed by Vaibhav Saroj.

Output

1

Time Complexity: O(n*n), A nested loop is needed where both the loops traverse the array from start to end.
Auxiliary Space: O(1), No extra space is required.

Majority Element using Binary Search Tree

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.

Illustration:

Follow the steps below to solve the given problem:

• Create a binary search tree, if the same element is entered in the binary search tree the frequency of the node is increased.
• traverse the array and insert the element in the binary search tree.
• If the maximum frequency of any node is greater than half the size of the array, then perform an inorder traversal and find the node with a frequency greater than half
• Else print No majority Element.

Below is the implementation of the above idea:

C++14

// 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 Code
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);
    }

    // Function call
    if (ma > (size / 2))
        inorder(root, size);
    else
        cout << "No majority element\n";
    return 0;
}

Java

// Java program to demonstrate insert
// operation in binary search tree.
import java.io.*;

class Node
{
    int key;
    int c = 0;
    Node left,right;

}
class GFG{

static int ma = 0;

// A utility function to create a
// new BST node
static Node newNode(int item)
{
    Node temp = new 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
static Node insert(Node node, int key)
{

    // 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);
    else if (key > node.key)
        node.right = insert(node.right, key);
    else
        node.c++;

    // Find the max count
    ma = Math.max(ma, node.c);

    // Return the (unchanged) node pointer
    return node;
}

// A utility function to do inorder
// traversal of BST
static void inorder(Node root, int s)
{
    if (root != null)
    {
        inorder(root.left, s);

        if (root.c > (s / 2))
            System.out.println(root.key + "\n");

        inorder(root.right, s);
    }
}

// Driver Code
public static void main(String[] args)
{
    int a[] = { 1, 3, 3, 3, 2 };
    int size = a.length;
    Node root = null;

    for(int i = 0; i < size; i++)
    {
        root = insert(root, a[i]);
    }

    // Function call
    if (ma > (size / 2))
        inorder(root, size);
    else
        System.out.println("No majority element\n");
}
}

// This code is contributed by avanitrachhadiya2155

Python3

# Python3 program to demonstrate insert operation in binary
# search tree.
# class for creating node
class Node():
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
        self.count = 1  # count of number of times data is inserted in tree

# class for binary search tree
# it initialises tree with None root
# insert function inserts node as per BST rule
# and also checks for majority element
# if no majority element is found yet, it returns None


class BST():
    def __init__(self):
        self.root = None

    def insert(self, data, n):
        out = None
        if (self.root == None):
            self.root = Node(data)
        else:
            out = self.insertNode(self.root, data, n)
        return out

    def insertNode(self, currentNode, data, n):
        if (currentNode.data == data):
            currentNode.count += 1
            if (currentNode.count > n//2):
                return currentNode.data
            else:
                return None
        elif (currentNode.data < data):
            if (currentNode.right):
                self.insertNode(currentNode.right, data, n)
            else:
                currentNode.right = Node(data)
        elif (currentNode.data > data):
            if (currentNode.left):
                self.insertNode(currentNode.left, data, n)
            else:
                currentNode.left = Node(data)


# Driver code
# declaring an array
arr = [3, 2, 3]
n = len(arr)

# declaring None tree
tree = BST()
flag = 0
for i in range(n):
    out = tree.insert(arr[i], n)
    if (out != None):
        print(arr[i])
        flag = 1
        break
if (flag == 0):
    print("No Majority Element")

C#

// C# program to demonstrate insert
// operation in binary search tree.
using System;

public class Node
{
    public int key;
    public int c = 0;
    public Node left,right;

}

class GFG{

static int ma = 0;

// A utility function to create a
// new BST node
static Node newNode(int item)
{
    Node temp = new 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
static Node insert(Node node, int key)
{

    // 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);
    else if (key > node.key)
        node.right = insert(node.right, key);
    else
        node.c++;

    // Find the max count
    ma = Math.Max(ma, node.c);

    // Return the (unchanged) node pointer
    return node;
}

// A utility function to do inorder
// traversal of BST
static void inorder(Node root, int s)
{
    if (root != null)
    {
        inorder(root.left, s);

        if (root.c > (s / 2))
            Console.WriteLine(root.key + "\n");

        inorder(root.right, s);
    }
}

// Driver Code
static public void Main()
{
    int[] a = { 1, 3, 3, 3, 2 };
    int size = a.Length;
    Node root = null;

    for(int i = 0; i < size; i++)
    {
        root = insert(root, a[i]);
    }

    // Function call
    if (ma > (size / 2))
        inorder(root, size);
    else
        Console.WriteLine("No majority element\n");
}
}

// This code is contributed by rag2127

Javascript

<script>
// javascript program to demonstrate insert
// operation in binary search tree.
class Node {
    constructor(){
    this.key = 0;
    this.c = 0;
    this.left = null,
    this.right = null;
    }

}
    var ma = 0;

    // A utility function to create a
    // new BST node
    function newNode(item)
    {
        var temp = new 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
    function insert(node , key) {

        // 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);
        else if (key > node.key)
            node.right = insert(node.right, key);
        else
            node.c++;

        // Find the max count
        ma = Math.max(ma, node.c);

        // Return the (unchanged) node pointer
        return node;
    }

    // A utility function to do inorder
    // traversal of BST
    function inorder(root , s) {
        if (root != null) {
            inorder(root.left, s);

            if (root.c > (s / 2))
                document.write(root.key + "\n");

            inorder(root.right, s);
        }
    }

    // Driver Code

        var a = [ 1, 3, 3, 3, 2 ];
        var size = a.length;
        var root = null;

        for (i = 0; i < size; i++) {
            root = insert(root, a[i]);
        }

        // Function call
        if (ma > (size / 2))
            inorder(root, size);
        else
            document.write("No majority element\n");

// This code is contributed by gauravrajput1
</script>

C

#include <stdio.h>
#include <stdlib.h>

struct node {
    int key;
    int c;
    struct node* left;
    struct node* right;
};

struct node* newNode(int item) {
    struct node* temp = (struct node*)malloc(sizeof(struct node));
    temp->key = item;
    temp->c = 1;
    temp->left = NULL;
    temp->right = NULL;
    return temp;
}

struct node* insert(struct node* node, int key, int* ma) {
    if (node == NULL) {
        if (*ma == 0)
            *ma = 1;

        return newNode(key);
    }

    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++;

    *ma = (*ma > node->c) ? *ma : node->c;

    return node;
}

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);
    }
}

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
        printf("No majority element\n");

    return 0;
}

// This code is contributed by Vaibhav Saroj.

Output

3 

Time Complexity: If a Binary Search Tree is used then time complexity will be O(n²). 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 the tree.

Majority Element Using Moore’s Voting Algorithm:

This is a two-step process:

• The first step gives the element that may be 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 the majority element. This step is necessary as there might be no majority element. 

Illustration:

arr[] = {3, 4, 3, 2, 4, 4, 4, 4}, n = 8

maj_index = 0, count = 1

At i = 1: arr[maj_index] != arr[i]

• count = count – 1 = 1 – 1 = 0
• now count == 0 then:
  • maj_index = i = 1
  • count = count + 1 = 0 + 1 = 1

At i = 2: arr[maj_index] != arr[i]

• count = count – 1 = 1 – 1 = 0
• now count == 0 then:
  • maj_index = i = 2
  • count = count + 1 = 0 + 1 = 1

At i = 3: arr[maj_index] != arr[i]

• count = count – 1 = 1 – 1 = 0
• now count == 0 then:
  • maj_index = i = 3
  • count = count + 1 = 0 + 1 = 1

At i = 4: arr[maj_index] != arr[i]

• count = count – 1 = 1 – 1 = 0
• now count == 0 then:
  • maj_index = i = 4
  • count = count + 1 = 0 + 1 = 1

At i = 5: arr[maj_index] == arr[i]

• count = count + 1 = 1 + 1 = 2

At i = 6: arr[maj_index] == arr[i]

• count = count + 1 = 2 + 1 = 3

At i = 7: arr[maj_index] == arr[i]

• count = count + 1 = 3 + 1 = 4

Therefore, the arr[maj_index] may be the possible candidate for majority element.

Now, Again traverse the array and check whether arr[maj_index] is the majority element or not.

arr[maj_index] is 4

4 occurs 5 times in the array therefore 4 is our majority element.

Follow the steps below to solve the given problem:

• Loop through each element and maintains a count of the majority element, and a majority index, maj_index
• If the next element is the same then increment the count if the next element is not the same then decrement the count.
• if the count reaches 0 then change the maj_index to the current element and set the count again to 1.
• Now again traverse through the array and find the count of the majority element found.
• If the count is greater than half the size of the array, print the element
• Else print that there is no majority element

Below is the implementation of the above idea: 

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 code */
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 code */
int main()
{
    int a[] = { 1, 3, 3, 1, 2 };
    int size = (sizeof(a)) / sizeof(a[0]);

    // Function call
    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 code */
    public static void main(String[] args)
    {
        MajorityElement majorelement
            = new MajorityElement();
        int a[] = new int[] { 1, 3, 3, 1, 2 };

        // Function call
        int size = a.length;
        majorelement.printMajority(a, size);
    }
}

// This code has been contributed by Mayank Jaiswal

Python3

# 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 code
A = [1, 3, 3, 1, 2]

# Function call
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;

        // Function call
        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
?>

Javascript

<script>
    // Javascript Program for finding out majority element in an array

    /* Function to print Majority Element */
    function printMajority(a, size)
    {
        /* Find the candidate for Majority*/
        let cand = findCandidate(a, size);

        /* Print the candidate if it is Majority*/
        if (isMajority(a, size, cand))
            document.write(" " + cand + " ");
        else
            document.write("No Majority Element");
    }

    /* Function to find the candidate for Majority */
    function findCandidate(a, size)
    {
        let maj_index = 0, count = 1;
        let 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
    function isMajority(a, size, cand)
    {
        let i, count = 0;
        for (i = 0; i < size; i++) {
            if (a[i] == cand)
                count++;
        }
        if (count > parseInt(size / 2, 10))
            return true;
        else
            return false;
    }

    let a = [ 1, 3, 3, 1, 2 ];
    let size = a.length;

    // Function call
    printMajority(a, size);

// This code is contributed by rameshtravel07.
</script>

Output

No Majority Element

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.

Majority Element Using Hashing:

In Hashtable(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). 

Illustration:

arr[] = {3, 4, 3, 2, 4, 4, 4, 4}, n = 8

Create a hashtable for the array

3 -> 2
4 -> 5
2 -> 1

Now traverse the hashtable

• Count for 3 is 2, which is less than n/2 (4) therefore it can’t be the majority element.
• Count for 4 is 5, which is greater than n/2 (4) therefore 4 is the majority element.

Hence, 4 is the majority element.

Follow the steps below to solve the given problem:

• Create a hashmap to store a key-value pair, i.e. element-frequency pair.
• Traverse the array from start to end.
• For every element in the array, insert the element in the hashmap if the element does not exist as a key, else fetch the value of the key ( array[i] ), and increase the value by 1
• If the count is greater than half then print the majority element and break.
• If no majority element is found print “No Majority element”

Below is the implementation of the above idea:

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

# Python3 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
    is_majority_present = False
    for key in m:
        if m[key] > size / 2:
            is_majority_present = True
            print("Majority found :-",key)
            break
    if not is_majority_present:
        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

Javascript

<script>

// Javascript program for the above approach

    function findMajority(arr)
    {
        let map = new Map();

        for(let i = 0; i < arr.length; i++) {
            if (map.has(arr[i])) {
                    let count = map.get(arr[i]) +1;
                    if (count > arr.length /2) {
                        document.write("Majority found :- " + arr[i]);
                        return;
                    } else
                        map.set(arr[i], count);

            }
            else
                map.set(arr[i],1);
            }
            document.write(" No Majority element");
    }

// Driver Code
    let a = [ 2,2,2,2,5,5,2,3,3 ];

    findMajority(a);

// This code is contributed by splevel62.
</script>

C

#include <stdio.h>
#include <stdlib.h>

void findMajority(int arr[], int size)
{
    int i, j;
    int count = 0;
    for(i = 0; i < size; i++)
    {
        int current = arr[i];
        int freq = 0;
        for(j = 0; j < size; j++)
        {
            if(arr[j] == current)
            {
                freq++;
            }
        }
        if(freq > size / 2)
        {
            count = 1;
            printf("Majority found :- %d\n", current);
            break;
        }
    }
    if(count == 0)
    {
        printf("No Majority element\n");
    }
}

// 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;
}

Output

Majority found :- 2

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.

Majority Element Using Sorting:

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.

Illustration:

arr[] = {3, 4, 3, 2, 4, 4, 4, 4}, n = 8

Array after sorting => arr[] = {2, 3, 3, 4, 4, 4, 4, 4}, count = 1

At i = 1:

• arr[i] != arr[i – 1] => arr[1] != arr[0]
• count is not greater than n/2, therefore reinitialise count with, count = 1

At i = 2:

• arr[i] == arr[i – 1] => arr[2] == arr[1] = 3
• count = count + 1 = 1 + 1 = 2

At i = 3

• arr[i] != arr[i – 1] => arr[3] != arr[2]
• count is not greater than n/2, therefore reinitialise count with, count = 1

At i = 4

• arr[i] == arr[i – 1] => arr[4] == arr[3] = 4
• count = count + 1 = 1 + 1 = 2

At i = 5

• arr[i] == arr[i – 1] => arr[5] == arr[4] = 4
• count = count + 1 = 2 + 1 = 3

At i = 6

• arr[i] == arr[i – 1] => arr[6] == arr[5] = 4
• count = count + 1 = 3 + 1 = 4

At i = 7

• arr[i] == arr[i – 1] => arr[7] == arr[6] = 4
• count = count + 1 = 4 + 1 = 5
• Therefore, the count of 4 is now greater than n/2.

Hence, 4 is the majority element.

Follow the steps below to solve the given problem:

• Sort the array and create a variable count and previous, prev = INT_MIN.
• Traverse the element from start to end.
• If the current element is equal to the previous element increase the count.
• Else set the count to 1.
• If the count is greater than half the size of the array, print the element as the majority element and break.
• If no majority element is found, print “No majority element”

Below is the implementation of the above idea: 

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
// it returns -1 if there is no majority element

int majorityElement(int *arr, int n)
{
    if (n == 1) return arr[0];

    int cnt = 1;
      // sort the array, o(nlogn)
    sort(arr, arr + n);
    for (int i = 1; i <= n; i++){
        if (arr[i - 1] == arr[i]){
            cnt++;
        }
        else{
            if (cnt > n / 2){
                return arr[i - 1];
            }
            cnt = 1;
        }
    }
    // if no majority element, return -1
    return -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;
}

Java

// Java program to find Majority
// element in an array
import java.io.*;
import java.util.*;

class GFG{

// Function to find Majority element
// in an array it returns -1 if there
// is no majority element
public static int majorityElement(int[] arr, int n)
{

    // Sort the array in O(nlogn)
    Arrays.sort(arr);

    int count = 1, max_ele = -1,
         temp = arr[0], ele = 0,
            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 occurred 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 occurred element
    // if there is no such element, returns -1
    return (f == 1 ? ele : -1);
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 1, 2, 1, 3, 5, 1 };
    int n = 7;

    System.out.println(majorityElement(arr, n));
}
}

// This code is contributed by RohitOberoi

Python3

# Python3 program to find Majority
# element in an array

# Function to find Majority element
# in an array
# it returns -1 if there is no majority element
def majorityElement(arr, n) :

    # sort the array in O(nlogn)
    arr.sort()
    count, max_ele, temp, f = 1, -1, arr[0], 0
    for i in range(1, n) :

        # increases the count if the same element occurs
        # otherwise starts counting new element
        if(temp == arr[i]) :
            count += 1
        else :
            count = 1
            temp = arr[i]

        # sets maximum count
        # and stores maximum occurred 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 occurred element
    # if there is no such element, returns -1
    if f == 1 :
        return ele
    else :
        return -1

# Driver code
arr = [1, 1, 2, 1, 3, 5, 1]
n = len(arr)

# Function calling
print(majorityElement(arr, n))

# This code is contributed by divyeshrabadiya07

C#

// C# program to find Majority
// element in an array
using System;
class GFG
{

// Function to find Majority element
// in an array it returns -1 if there
// is no majority element
public static int majorityElement(int[] arr, int n)
{

    // Sort the array in O(nlogn)
    Array.Sort(arr);

    int count = 1, max_ele = -1,
         temp = arr[0], ele = 0,
            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 occurred 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 occurred element
    // if there is no such element, returns -1
    return (f == 1 ? ele : -1);
}

// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 1, 2, 1, 3, 5, 1 };
    int n = 7;
    Console.WriteLine(majorityElement(arr, n));
}
}

// This code is contributed by aashish1995

Javascript

<script>

    // Javascript program to find Majority
    // element in an array

    // Function to find Majority element
    // in an array it returns -1 if there
    // is no majority element
    function majorityElement(arr, n)
    {

        // Sort the array in O(nlogn)
        arr.sort(function(a, b){return a - b});

        let count = 1, max_ele = -1,
             temp = arr[0], ele = 0,
                f = 0;

        for(let 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 occurred 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 > parseInt(n / 2, 10))
                {
                    f = 1;
                    break;
                }
            }
        }

        // Returns maximum occurred element
        // if there is no such element, returns -1
        return (f == 1 ? ele : -1);
    }

    let arr = [ 1, 1, 2, 1, 3, 5, 1 ];
    let n = 7;
    document.write(majorityElement(arr, n));

</script>

Output

1

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

Approach using Recursion: The idea is to use the divide and conquer strategy to identify the majority element in an array using a recursive approach.

• Define a function “countOccurrences”  which takes an array “arr”, its size “n”, and an integer “num” as input parameters. This function counts the number of occurrences of the given integer in the array and returns the count. And the initial count is 0.
• Then we’ll define a recursive function “findMajorityUtil” that takes an array, its lower “low” and upper “high” indices as input parameters. This function has the following steps:
  • Check if the array has only one element (base case). If yes, return that element.
  • Divide the array into two equal parts, left and right by finding the mid index.
  • Recursively find the majority element in the left part and right part.
  •  If both the left and right parts have the same majority elements, then return that element.
  • Count the total number of occurrences of the left and right majority elements in the entire array.
  • Return the element that occurs more than n/2 times.
• Define a function “findMajority” which takes an array “arr” and its size “n” as input parameters. This function calls the “findMajorityUtil” function with the array and its indices and prints the majority element if it exists; otherwise, it will print “No Majority Element”.
• The “main” function will initialize an array “arr” and size “n”. It will call the “findMajority” function with these parameters.

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count the occurrences
int countOccurrences(int arr[], int n, int num) {
    int count = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] == num)
            count++;
    }
    return count;
}

// Function to find the majority element
// using recursion
int findMajorityUtil(int arr[], int low, int high) {

    // Base case: single element array
    if (low == high)
        return arr[low];

    // Divide the array into left
  // and right halves
    int mid = (low + high) / 2;
    int leftMajority = findMajorityUtil(arr, low, mid);
    int rightMajority = findMajorityUtil(arr, mid+1, high);

    // If left and right halves have the
  // same majority element
    if (leftMajority == rightMajority)
        return leftMajority;

    // Count the occurrences of the
  // majority element in entire array
    int leftCount = countOccurrences(arr, high-low+1, leftMajority);
    int rightCount = countOccurrences(arr, high-low+1, rightMajority);

    // Return the element that occurs
  // more than n/2 times
    if (leftCount > (high-low+1) / 2)
        return leftMajority;
    if (rightCount > (high-low+1) / 2)
        return rightMajority;

    // No majority element
    return -1;
}

// Function to find the majority element
void findMajority(int arr[], int n) {
    int majority = findMajorityUtil(arr, 0, n-1);
    if (majority != -1)
        cout << majority << endl;
    else
        cout << "No Majority Element" << endl;
}

// Driver Code
int main() {
    int arr[] = {1, 3, 3, 3, 2};
    int n = sizeof(arr) / sizeof(arr[0]);

    findMajority(arr, n);

    return 0;
}

C

#include <stdio.h>

int countOccurrences(int arr[], int n, int num) {
    int count = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] == num)
            count++;
    }
    return count;
}

int findMajorityUtil(int arr[], int low, int high) {
    // Base case: single element array
    if (low == high)
        return arr[low];

    // Divide the array into left
  // and right halves
    int mid = (low + high) / 2;
    int leftMajority = findMajorityUtil(arr, low, mid);
    int rightMajority = findMajorityUtil(arr, mid+1, high);

    // If left and right halves have the
  //same majority element
    if (leftMajority == rightMajority)
        return leftMajority;

    // Count the occurrences of the majority
  // element in the entire array
    int leftCount = countOccurrences(arr, high-low+1, leftMajority);
    int rightCount = countOccurrences(arr, high-low+1, rightMajority);

    // Return the element that occurs
  // more than n/2 times
    if (leftCount > (high-low+1) / 2)
        return leftMajority;
    if (rightCount > (high-low+1) / 2)
        return rightMajority;

    // No majority element
    return -1;
}

void findMajority(int arr[], int n) {
    int majority = findMajorityUtil(arr, 0, n-1);
    if (majority != -1)
        printf("%d\n", majority);
    else
        printf("No Majority Element\n");
}

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

    findMajority(arr, n);

    return 0;
}

Java

import java.util.*;

public class MajorityElement {

    // Function to count the occurrences
    public static int countOccurrences(int[] arr, int n, int num) {
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (arr[i] == num)
                count++;
        }
        return count;
    }

    // Function to find the majority element using recursion
    public static int findMajorityUtil(int[] arr, int low, int high) {

        // Base case: single element array
        if (low == high)
            return arr[low];

        // Divide the array into left and right halves
        int mid = (low + high) / 2;
        int leftMajority = findMajorityUtil(arr, low, mid);
        int rightMajority = findMajorityUtil(arr, mid + 1, high);

        // If left and right halves have the same majority element
        if (leftMajority == rightMajority)
            return leftMajority;

        // Count the occurrences of the majority element in entire array
        int leftCount = countOccurrences(arr, high - low + 1, leftMajority);
        int rightCount = countOccurrences(arr, high - low + 1, rightMajority);

        // Return the element that occurs more than n/2 times
        if (leftCount > (high - low + 1) / 2)
            return leftMajority;
        if (rightCount > (high - low + 1) / 2)
            return rightMajority;

        // No majority element
        return -1;
    }

    // Function to find the majority element
    public static void findMajority(int[] arr, int n) {
        int majority = findMajorityUtil(arr, 0, n - 1);
        if (majority != -1)
            System.out.println(majority);
        else
            System.out.println("No Majority Element");
    }

    // Driver Code
    public static void main(String[] args) {
        int[] arr = {1, 3, 3, 3, 2};
        int n = arr.length;

        findMajority(arr, n);
    }
}
// This code is contributed by Prajwal Kandekar

Output

3

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