Check if an array has a majority element
Last Updated :
01 Aug, 2022
Given an array, the task is to find if the input array contains a majority element or not. An element is
Examples:
Input : arr[] = {2, 3, 9, 2, 2}
Output : Yes
A majority element 2 is present in arr[]
Input : arr[] = {1, 8, 9, 2, 5}
Output : No
A simple solution is to traverse through the array. For every element, count its occurrences. If the count of occurrence of any element becomes n/2, we return true.
An efficient solution is to use hashing. We count occurrences of all elements. If count becomes n/2 or more return true.
Below is the implementation of the approach.
C++
// Hashing based C++ program to find if there
// is a majority element in input array.
#include <bits/stdc++.h>
using namespace std;
// Returns true if there is a majority element
// in a[]
bool isMajority(int a[], int n)
{
// Insert all elements in a hash table
unordered_map<int, int> mp;
for (int i = 0; i < n; i++)
mp[a[i]]++;
// Check if frequency of any element is
// n/2 or more.
for (auto x : mp)
if (x.second >= n/2)
return true;
return false;
}
// Driver code
int main()
{
int a[] = { 2, 3, 9, 2, 2 };
int n = sizeof(a) / sizeof(a[0]);
if (isMajority(a, n))
cout << "Yes";
else
cout << "No";
return 0;
}
// Hashing based Java program
// to find if there is a
// majority element in input array.
import java.util.*;
import java.lang.*;
import java.io.*;
class Gfg
{
// Returns true if there is a
// majority element in a[]
static boolean isMajority(int a[], int n)
{
// Insert all elements
// in a hash table
HashMap <Integer,Integer> mp = new
HashMap<Integer,Integer>();
for (int i = 0; i < n; i++)
if (mp.containsKey(a[i]))
mp.put(a[i], mp.get(a[i]) + 1);
else mp.put(a[i] , 1);
// Check if frequency of any
// element is n/2 or more.
for (Map.Entry<Integer, Integer> x : mp.entrySet())
if (x.getValue() >= n/2)
return true;
return false;
}
// Driver code
public static void main (String[] args)
{
int a[] = { 2, 3, 9, 2, 2 };
int n = a.length;
if (isMajority(a, n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Ansu Kumari
# Hashing based Python
# program to find if
# there is a majority
# element in input array.
# Returns true if there
# is a majority element
# in a[]
def isMajority(a):
# Insert all elements
# in a hash table
mp = {}
for i in a:
if i in mp: mp[i] += 1
else: mp[i] = 1
# Check if frequency
# of any element is
# n/2 or more.
for x in mp:
if mp[x] >= len(a)//2:
return True
return False
# Driver code
a = [ 2, 3, 9, 2, 2 ]
print("Yes" if isMajority(a) else "No")
#This code is contributed by Ansu Kumari
// Hashing based C# program
// to find if there is a
// majority element in input array.
using System;
using System.Collections.Generic;
class GFG
{
// Returns true if there is a
// majority element in a[]
static Boolean isMajority(int []a, int n)
{
// Insert all elements
// in a hash table
Dictionary<int,
int> mp = new Dictionary<int,
int>();
for (int i = 0; i < n; i++)
{
if(mp.ContainsKey(a[i]))
{
var val = mp[a[i]];
mp.Remove(a[i]);
mp.Add(a[i], val + 1);
}
else
{
mp.Add(a[i], 1);
}
}
// Check if frequency of any
// element is n/2 or more.
foreach(KeyValuePair<int, int> x in mp)
if (x.Value >= n / 2)
return true;
return false;
}
// Driver code
public static void Main (String[] args)
{
int []a = { 2, 3, 9, 2, 2 };
int n = a.Length;
if (isMajority(a, n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by PrinciRaj1992
<script>
// Hashing based Javascript program to find if there
// is a majority element in input array.
// Returns true if there is a majority element
// in a[]
function isMajority(a, n)
{
// Insert all elements in a hash table
let mp = new Map();
for (let i = 0; i < n; i++){
if(mp.has(a[i])){
mp.set(a[i], mp.get(a[i]) + 1)
}else{
mp.set(a[i], 1)
}
}
// Check if frequency of any element is
// n/2 or more.
for (let x of mp)
if (x[1] >= n/2)
return true;
return false;
}
// Driver code
let a = [ 2, 3, 9, 2, 2 ];
let n = a.length;
if (isMajority(a, n))
document.write("Yes");
else
document.write("No");
// This code is contributed by saurabh_jaiswal.
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
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