# 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)

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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.

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:**

- C
- Java

`/* 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;` `}` |

Output:

No Majority Element

**Time Complexity:** O(n)

**Auxiliary Space :** O(1)