Majority Element

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:

       I/P : 3 3 4 2 4 4 2 4 4
       O/P : 4 

       I/P : 3 3 4 2 4 4 2 4
       O/P : NONE

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.

struct tree {   int element;   int count; }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)

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.

Example:
A[] = 2, 2, 3, 5, 2, 2, 6
Initialize:
maj_index = 0, count = 1 –> candidate ‘2?
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
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:

/* 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};     printMajority(a, 5);     getchar();     return 0; }

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

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).