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).

Write a function which takes an array and emits the majority element (if it exists), otherwise prints NONE as follows:

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.

Time Complexity: O(n*n). Auxiliary Space : O(1).

METHOD 2 (Using Binary Search Tree)
Thanks to Sachin Midha for suggesting this solution. Node of the Binary Search Tree (used in this approach) will be as follows.

structtree

{

intelement;

intcount;

}BST;

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)

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METHOD 3 (Using Moore’s Voting Algorithm)
This is a two step process.
1. Get an element occurring most of the time in the array. This phase will make sure that if there is a majority element then it will return that only.
2. Check if the element obtained from above 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 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 next element is same then increments the count, if next element is not same then decrements the count, and if the count reaches 0 then changes the maj_index to the current element and sets count to 1.
First Phase algorithm gives us a candidate element. In 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.

Same as a[maj_index] => count = 2
2, 2, 3, 5, 2, 2, 6

Different from a[maj_index] => count = 1
2, 2, 3, 5, 2, 2, 6

Different from a[maj_index] => count = 0
Since count = 0, change candidate for majority element to 5 => maj_index = 3, count = 1
2, 2, 3, 5, 2, 2, 6

Different from a[maj_index] => count = 0
Since count = 0, change candidate for majority element to 2 => maj_index = 4
2, 2, 3, 5, 2, 2, 6

Same as a[maj_index] => count = 2
2, 2, 3, 5, 2, 2, 6

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

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 "NONE"

Implementation of method 3:

C

Java

/* Program for finding out majority element in an array */

# include<stdio.h>

# define bool int

intfindCandidate(int*, int);

boolisMajority(int*, int, int);

/* Function to print Majority Element */

voidprintMajority(inta[], intsize)

{

/* Find the candidate for Majority*/

intcand = 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 */

intfindCandidate(inta[], intsize)

{

intmaj_index = 0, count = 1;

inti;

for(i = 1; i < size; i++)

{

if(a[maj_index] == a[i])

count++;

else

count--;

if(count == 0)

{

maj_index = i;

count = 1;

}

}

returna[maj_index];

}

/* Function to check if the candidate occurs more than n/2 times */