# Majority Element – Part 1

**Objective**: Given an array of integer write an algorithm to find the majority element in it (if exist).

**Majority Element**: If an element appears more than n/2 times in array where n is the size of the array.

**Example**:

int [] arrA = {1,3,5,5,5,5,4,1,5}; Output: Element appearing more than n/2 times: 5 int []arrA = {1,2,3,4}; Output: No element appearing more than n/2 times

Click here to read O(n) approach-Boyer–Moore majority vote algorithm

**Approach 1: Brute Force**

Use nested for loops and count each element and if count>n/2 for each element.

Time Complexity: O(n^2)

**Code**:

**Approach 2: Hash Map**

- Store the count of each element in Hash map.
- Iterate through hash map and check if any element has count >n/2

Time Complexity: O(n), Space Complexity: O(n)

**Code**:

**Approach 3: Sorting**

- Sort array, this will bring all the same elements together.
- Iterate through array and check if any element has count >n/2

Time Complexity: O(nlogn), Space Complexity: O(1)

**Code**:

**Output**:

Element appearing more than n/2 times: 5

In Nested Loop approach, break the loop after found=true;