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
Input : {3, 3, 4, 2, 4, 4, 2, 4}
Output : No Majority Element
METHOD 1 (Basic)
The basic solution is to have two loops and keep track of 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 maximum count doesn’t become more than n/2 then majority element doesn’t exist.
Below is the implementation of the above approach :
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;
}
Output :
1
Time Complexity : O(n*n).
Auxiliary Space : O(1).
METHOD 2 (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 count of a node becomes more than n/2 then return.
The method works well for the cases where n/2+1 occurrences of the majority element is present in the starting of the array, for example {1, 1, 1, 1, 1, 2, 3, 4}.
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 O(nlogn)
Auxiliary Space : O(n)
METHOD 3 (Using Moore’s Voting Algorithm)
This is a two step process.
- The first step gives the element that may be 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 above step is majority element.This step is necessary as we are not always sure that element return by first step is majority element.
1. Finding a Candidate :
The algorithm for first phase that works in O(n) is known as Moore’s Voting Algorithm. Basic idea of the algorithm is that if we cancel out each occurrence of an element e with all the other elements that are different from e then e will exist till end if it is a majority element.
findCandidate(a[], size)
1. Initialize index and count of majority element
maj_index = 0, count = 1
2. Loop for i = 1 to size – 1
(a) If a[maj_index] == a[i]
count++
(b) Else
count--;
(c) If count == 0
maj_index = i;
count = 1
3. Return a[maj_index]
Above algorithm loops through each element and maintains a count of a[maj_index]. If the next element is same then increment the count, if the next element is not same then decrement the count, and if the count reaches 0 then changes the maj_index to the current element and set the count again to 1. So, the first phase of the algorithm gives us a candidate element.
In the second phase we need to check if the candidate is really a majority element. Second phase is simple and can be easily done in O(n). We just need to check if count of the candidate element is greater than n/2.
Example :
Let the array be A[] = 2, 2, 3, 5, 2, 2, 6
- Initialize maj_index = 0, count = 1
- Next element is 2, which is same as a[maj_index] => count = 2
- Next element is 3, which is different from a[maj_index] => count = 1
- Next element is 5, which is different from a[maj_index] => count = 0
Since count = 0, change candidate for majority element to 5 => maj_index = 3, count = 1 - Next element is 2, which is different from a[maj_index] => count = 0
Since count = 0, change candidate for majority element to 2 => maj_index = 4 - Next element is 2, which is same as a[maj_index] => count = 2
- Next element is 6, which is different from a[maj_index] => count = 1
- Finally candidate for majority element is 2.
First step uses Moore’s Voting Algorithm to get a candidate for majority element.
2. Check if the element obtained in step 1 is majority element or not :
printMajority (a[], size)
1. Find the candidate for majority
2. If candidate is majority. i.e., appears more than n/2 times.
Print the candidate
3. Else
Print "No Majority Element"
Below is the implementation of the above approach :
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
Output:
No Majority Element
Time Complexity: O(n)
Auxiliary Space : O(1)
METHOD 4 (Using Hashmap) :This method is somewhat similar to Moore voting algorithm in terms of time complexity, but in this case there is no need of 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 count is greater than half of array length, we are just returning that key(majority element).
Time Complexity : O(n)
Auxiliary Space : O(n)
Below is the implementation.
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
Output:
Majority found :- 2
Thanks Ashwani Tanwar, Karan Malhotra for suggesting this.
Now give a try to below question
Given an array of 2n elements of which n elements are same and the remaining n elements are all different. Write a C program to find out the value which is present n times in the array. There is no restriction on the elements in the array. They are random (In particular they not sequential).
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