# 1. Introduction

In this post, we'll delve into a solution for the "Majority Element" problem, a common challenge in array processing and voting algorithms. The task is to identify an element in an array that appears more than half the time (i.e., more than ⌊n / 2⌋ times), assuming such an element always exists in the array.

## Problem

Given an array *nums* of size *n*, the objective is to find and return the majority element, which is defined as the element that appears more than ⌊n / 2⌋ times.

# 2. Solution Steps

1. Utilize the Boyer-Moore Voting Algorithm, which is an efficient method to find the majority element.

2. Initialize two variables: *count* to keep track of the frequency, and *candidate* to store the potential majority element.

3. Iterate through the array:

- If *count* is 0, assign the current element as the *candidate*.

- Increment *count* if the current element is the same as *candidate*, otherwise decrement it.

4. Return *candidate* as it is the majority element.

# 3. Code Program

```
public class MajorityElement {
public static void main(String[] args) {
int[] nums = {2,2,1,1,1,2,2};
System.out.println(findMajorityElement(nums)); // Test the function
}
// Function to find the majority element in the array
public static int findMajorityElement(int[] nums) {
int count = 0;
Integer candidate = null;
for (int num : nums) {
if (count == 0) {
candidate = num;
}
count += (num == candidate) ? 1 : -1;
}
return candidate;
}
}
```

### Output:

2

### Explanation:

1. *findMajorityElement*: This function implements the Boyer-Moore Voting Algorithm to find the majority element in *nums*.

2. It initializes *candidate* as *null* and *count* as 0.

3. As the function iterates over *nums*, it tracks the potential majority element by increasing or decreasing the count based on whether the current element matches the candidate.

4. When *count* drops to 0, it selects a new candidate.

5. The algorithm guarantees that the final *candidate* is the majority element as it appears more than ⌊n / 2⌋ times.

6. This approach is highly efficient, requiring only a single pass over the array, resulting in O(n) time complexity and O(1) space complexity.

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