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.
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 the majority element doesn’t exist.
- Algorithm:
- 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 the 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++
#include <bits/stdc++.h>
using namespace std;
void findMajority(int arr[], int n)
{
int maxCount = 0;
int index = -1;
for (int i = 0; i < n; i++) {
int count = 0;
for (int j = 0; j < n; j++) {
if (arr[i] == arr[j])
count++;
}
if (count > maxCount) {
maxCount = count;
index = i;
}
}
if (maxCount > n / 2)
cout << arr[index] << endl;
else
cout << "No Majority Element" << endl;
}
int main()
{
int arr[] = { 1, 1, 2, 1, 3, 5, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
findMajority(arr, n);
return 0;
}
|
Java
import java.io.*;
class GFG {
static void findMajority(int arr[], int n)
{
int maxCount = 0;
int index = -1;
for (int i = 0; i < n; i++) {
int count = 0;
for (int j = 0; j < n; j++) {
if (arr[i] == arr[j])
count++;
}
if (count > maxCount) {
maxCount = count;
index = i;
}
}
if (maxCount > n / 2)
System.out.println(arr[index]);
else
System.out.println("No Majority Element");
}
public static void main(String[] args)
{
int arr[] = { 1, 1, 2, 1, 3, 5, 1 };
int n = arr.length;
findMajority(arr, n);
}
}
|
Python3
def findMajority(arr, n):
maxCount = 0
index = -1
for i in range(n):
count = 0
for j in range(n):
if(arr[i] == arr[j]):
count += 1
if(count > maxCount):
maxCount = count
index = i
if (maxCount > n//2):
print(arr[index])
else:
print("No Majority Element")
if __name__ == "__main__":
arr = [1, 1, 2, 1, 3, 5, 1]
n = len(arr)
findMajority(arr, n)
|
C#
using System;
public class GFG {
static void findMajority(int[] arr, int n)
{
int maxCount = 0;
int index = -1;
for (int i = 0; i < n; i++) {
int count = 0;
for (int j = 0; j < n; j++) {
if (arr[i] == arr[j])
count++;
}
if (count > maxCount) {
maxCount = count;
index = i;
}
}
if (maxCount > n / 2)
Console.WriteLine(arr[index]);
else
Console.WriteLine("No Majority Element");
}
static public void Main()
{
int[] arr = { 1, 1, 2, 1, 3, 5, 1 };
int n = arr.Length;
findMajority(arr, n);
}
}
|
PHP
<?php
function findMajority($arr, $n)
{
$maxCount = 0;
$index = -1;
for($i = 0; $i < $n; $i++)
{
$count = 0;
for($j = 0; $j < $n; $j++)
{
if($arr[$i] == $arr[$j])
$count++;
}
if($count > $maxCount)
{
$maxCount = $count;
$index = $i;
}
}
if ($maxCount > $n/2)
echo $arr[$index] . "\n";
else
echo "No Majority Element" . "\n";
}
$arr = array(1, 1, 2, 1, 3, 5, 1);
$n = sizeof($arr);
findMajority($arr, $n);
|
Javascript
<script>
function findMajority(arr, n)
{
let maxCount = 0;
let index = -1;
for(let i = 0; i < n; i++)
{
let count = 0;
for(let j = 0; j < n; j++)
{
if (arr[i] == arr[j])
count++;
}
if (count > maxCount)
{
maxCount = count;
index = i;
}
}
if (maxCount > n / 2)
document.write(arr[index]);
else
document.write("No Majority Element");
}
let arr = [ 1, 1, 2, 1, 3, 5, 1 ];
let n = arr.length;
findMajority(arr, n);
</script>
|
Complexity 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(n²).
- 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:
- Create a binary search tree, if 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 the half the size of the array, then perform a inorder traversal and find the node with frequency greater than half
- Else print No majority Element.
Below is the implementation of the above idea:
C++14
#include <bits/stdc++.h>
using namespace std;
struct node {
int key;
int c = 0;
struct node *left, *right;
};
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;
}
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 = max(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
cout << "No majority element\n";
return 0;
}
|
Java
import java.io.*;
class Node
{
int key;
int c = 0;
Node left,right;
}
class GFG{
static int ma = 0;
static Node newNode(int item)
{
Node temp = new Node();
temp.key = item;
temp.c = 1;
temp.left = temp.right = null;
return temp;
}
static Node insert(Node node, int key)
{
if (node == null)
{
if (ma == 0)
ma = 1;
return newNode(key);
}
if (key < node.key)
node.left = insert(node.left, key);
else if (key > node.key)
node.right = insert(node.right, key);
else
node.c++;
ma = Math.max(ma, node.c);
return node;
}
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);
}
}
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]);
}
if (ma > (size / 2))
inorder(root, size);
else
System.out.println("No majority element\n");
}
}
|
Python3
class Node():
def __init__(self, data):
self.data = data
self.left = None
self.right = None
self.count = 1
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)
arr = [3, 2, 3]
n = len(arr)
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#
using System;
public class Node
{
public int key;
public int c = 0;
public Node left,right;
}
class GFG{
static int ma = 0;
static Node newNode(int item)
{
Node temp = new Node();
temp.key = item;
temp.c = 1;
temp.left = temp.right = null;
return temp;
}
static Node insert(Node node, int key)
{
if (node == null)
{
if (ma == 0)
ma = 1;
return newNode(key);
}
if (key < node.key)
node.left = insert(node.left, key);
else if (key > node.key)
node.right = insert(node.right, key);
else
node.c++;
ma = Math.Max(ma, node.c);
return node;
}
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);
}
}
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]);
}
if (ma > (size / 2))
inorder(root, size);
else
Console.WriteLine("No majority element\n");
}
}
|
Javascript
<script>
class Node {
constructor(){
this.key = 0;
this.c = 0;
this.left = null,
this.right = null;
}
}
var ma = 0;
function newNode(item)
{
var temp = new Node();
temp.key = item;
temp.c = 1;
temp.left = temp.right = null;
return temp;
}
function insert(node , key) {
if (node == null) {
if (ma == 0)
ma = 1;
return newNode(key);
}
if (key < node.key)
node.left = insert(node.left, key);
else if (key > node.key)
node.right = insert(node.right, key);
else
node.c++;
ma = Math.max(ma, node.c);
return node;
}
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);
}
}
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]);
}
if (ma > (size / 2))
inorder(root, size);
else
document.write("No majority element\n");
</script>
|
Complexity Analysis:
- 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 tree.
METHOD 3 (Using Moore’s Voting Algorithm):
- 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.
- Algorithm:
- Loop through each element and maintains a count of majority element, and a majority index, maj_index
- If the next element is same then increment the count if the next element is not same then decrement the count.
- if the count reaches 0 then changes 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 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 above idea:
C++
#include <bits/stdc++.h>
using namespace std;
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];
}
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;
}
void printMajority(int a[], int size)
{
int cand = findCandidate(a, size);
if (isMajority(a, size, cand))
cout << " " << cand << " ";
else
cout << "No Majority Element";
}
int main()
{
int a[] = { 1, 3, 3, 1, 2 };
int size = (sizeof(a)) / sizeof(a[0]);
printMajority(a, size);
return 0;
}
|
C
#include <stdio.h>
#define bool int
int findCandidate(int*, int);
bool isMajority(int*, int, int);
void printMajority(int a[], int size)
{
int cand = findCandidate(a, size);
if (isMajority(a, size, cand))
printf(" %d ", cand);
else
printf("No Majority Element");
}
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];
}
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;
}
int main()
{
int a[] = { 1, 3, 3, 1, 2 };
int size = (sizeof(a)) / sizeof(a[0]);
printMajority(a, size);
getchar();
return 0;
}
|
Java
class MajorityElement {
void printMajority(int a[], int size)
{
int cand = findCandidate(a, size);
if (isMajority(a, size, cand))
System.out.println(" " + cand + " ");
else
System.out.println("No Majority Element");
}
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];
}
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;
}
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);
}
}
|
Python3
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]
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
def printMajority(A):
cand = findCandidate(A)
if isMajority(A, cand) == True:
print(cand)
else:
print("No Majority Element")
A = [1, 3, 3, 1, 2]
printMajority(A)
|
C#
using System;
class GFG {
static void printMajority(int[] a, int size)
{
int cand = findCandidate(a, size);
if (isMajority(a, size, cand))
Console.Write(" " + cand + " ");
else
Console.Write("No Majority Element");
}
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];
}
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;
}
public static void Main()
{
int[] a = { 1, 3, 3, 1, 2 };
int size = a.Length;
printMajority(a, size);
}
}
|
PHP
<?php
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 isMajority($a, $size, $cand)
{
$count = 0;
for ($i = 0; $i < $size; $i++)
if ($a[$i] == $cand)
$count++;
if ($count > $size / 2)
return 1;
else
return 0;
}
function printMajority($a, $size)
{
$cand = findCandidate($a, $size);
if (isMajority($a, $size, $cand))
echo " ", $cand, " ";
else
echo "No Majority Element";
}
$a = array(1, 3, 3, 1, 2);
$size = sizeof($a);
printMajority($a, $size);
?>
|
Javascript
<script>
function printMajority(a, size)
{
let cand = findCandidate(a, size);
if (isMajority(a, size, cand))
document.write(" " + cand + " ");
else
document.write("No Majority Element");
}
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 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;
printMajority(a, size);
</script>
|
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:
- 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 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 above idea:
C++
#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;
}
int main()
{
int arr[] = {2, 2, 2, 2, 5, 5, 2, 3, 3};
int n = sizeof(arr) / sizeof(arr[0]);
findMajority(arr, n);
return 0;
}
|
Java
import java.util.HashMap;
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");
}
public static void main(String[] args)
{
int a[] = new int[]{2,2,2,2,5,5,2,3,3};
findMajority(a);
}
}
|
Python3
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")
arr = [2, 2, 2, 2, 5, 5, 2, 3, 3]
n = len(arr)
findMajority(arr, n)
|
C#
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");
}
public static void Main(string[] args)
{
int[] a = new int[]{2, 2, 2, 2,
5, 5, 2, 3, 3};
findMajority(a);
}
}
|
Javascript
<script>
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");
}
let a = [ 2,2,2,2,5,5,2,3,3 ];
findMajority(a);
</script>
|
Output
Majority found :- 2
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.
Thanks, Ashwani Tanwar, Karan Malhotra for suggesting this.
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:
- 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 array, print the element as majority element and break.
- If no majority element found, print “No majority element”
Below is the implementation of the above idea:
C++
#include <bits/stdc++.h>
using namespace std;
int majorityElement(int *arr, int n)
{
if (n == 1) return arr[0];
int cnt = 1;
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;
}
}
return -1;
}
int main()
{
int arr[] = {1, 1, 2, 1, 3, 5, 1};
int n = sizeof(arr) / sizeof(arr[0]);
cout<<majorityElement(arr, n);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG{
public static int majorityElement(int[] arr, int n)
{
Arrays.sort(arr);
int count = 1, max_ele = -1,
temp = arr[0], ele = 0,
f = 0;
for(int i = 1; i < n; i++)
{
if (temp == arr[i])
{
count++;
}
else
{
count = 1;
temp = arr[i];
}
if (max_ele < count)
{
max_ele = count;
ele = arr[i];
if (max_ele > (n / 2))
{
f = 1;
break;
}
}
}
return (f == 1 ? ele : -1);
}
public static void main(String[] args)
{
int arr[] = { 1, 1, 2, 1, 3, 5, 1 };
int n = 7;
System.out.println(majorityElement(arr, n));
}
}
|
Python3
def majorityElement(arr, n) :
arr.sort()
count, max_ele, temp, f = 1, -1, arr[0], 0
for i in range(1, n) :
if(temp == arr[i]) :
count += 1
else :
count = 1
temp = arr[i]
if(max_ele < count) :
max_ele = count
ele = arr[i]
if(max_ele > (n//2)) :
f = 1
break
if f == 1 :
return ele
else :
return -1
arr = [1, 1, 2, 1, 3, 5, 1]
n = len(arr)
print(majorityElement(arr, n))
|
C#
using System;
class GFG
{
public static int majorityElement(int[] arr, int n)
{
Array.Sort(arr);
int count = 1, max_ele = -1,
temp = arr[0], ele = 0,
f = 0;
for(int i = 1; i < n; i++)
{
if (temp == arr[i])
{
count++;
}
else
{
count = 1;
temp = arr[i];
}
if (max_ele < count)
{
max_ele = count;
ele = arr[i];
if (max_ele > (n / 2))
{
f = 1;
break;
}
}
}
return (f == 1 ? ele : -1);
}
public static void Main(String[] args)
{
int []arr = { 1, 1, 2, 1, 3, 5, 1 };
int n = 7;
Console.WriteLine(majorityElement(arr, n));
}
}
|
Javascript
<script>
function majorityElement(arr, n)
{
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++)
{
if (temp == arr[i])
{
count++;
}
else
{
count = 1;
temp = arr[i];
}
if (max_ele < count)
{
max_ele = count;
ele = arr[i];
if (max_ele > parseInt(n / 2, 10))
{
f = 1;
break;
}
}
}
return (f == 1 ? ele : -1);
}
let arr = [ 1, 1, 2, 1, 3, 5, 1 ];
let n = 7;
document.write(majorityElement(arr, n));
</script>
|
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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