# Mastering LeetCode's Majority Element Problem: Strategies for Every Engineer

## Unlock the secret to acing the Majority Element problem on LeetCode with sorting, hash maps, and the Boyer–Moore algorithm.

Welcome to another installment in the software engineering interview tutorial series. Today, we're diving deep into a popular LeetCode problem: finding the Majority Element in an array (LeetCode 169).

This problem is a fantastic opportunity to explore different problem-solving strategies, from simple to sophisticated, each with its unique advantages and computational complexities.

Whether you're an experienced engineer brushing up on your skills or new to coding interviews, understanding these approaches will significantly bolster your problem-solving arsenal.

## Introduction to the Majority Element Problem

The Majority Element problem asks you to find an element in an array that appears more than `⌊n / 2⌋`

times, where `n`

is the array's size.

For instance, in the array `[3,2,3]`

, the majority element is `3`

, and in `[2,2,1,1,1,2,2]`

, it's `2`

. The beauty of this problem lies in its guarantee: the majority element always exists within the array.

Given an array

`nums`

of size`n`

, returnthe majority element.The majority element is the element that appears more than

`⌊n / 2⌋`

times. You may assume that the majority element always exists in the array.

Example 1:`Input: nums = [3,2,3] Output: 3`

Example 2:`Input: nums = [2,2,1,1,1,2,2] Output: 2`

Constraints:

`n == nums.length`

`1 <= n <= 5 * 10^4`

`-10^9 <= nums[i] <= 10^9`

## Naive Sorting Solution

A straightforward approach to solve this problem involves sorting the array. Once sorted, the element in the middle of the array ((n/2) position) must be the majority element, as it will occupy more than half of the array's positions.

**Big O Notation Analysis**: This method has a time complexity of `O(n log n)`

due to the sorting process, with `O(1)`

space complexity if the sort is done in-place.

### Sorting Solution in Python

```
def findMajorityElement(nums):
# Sort the array
nums.sort()
# The majority element is at the middle position after sorting
return nums[len(nums) // 2]
```

## Hash Map Solution

Another method involves using a hash map to count occurrences of each element. This technique allows us to track and identify the element that surpasses the `n/2`

occurrence threshold.

**Big O Notation Analysis**: Counting elements using a hash map results in `O(n)`

time complexity, with `O(n)`

space complexity to store the counts.

### Hash Map Solution in Python

```
def findMajorityElement(nums):
counts = {}
for num in nums:
if num in counts:
counts[num] += 1
else:
counts[num] = 1
if counts[num] > len(nums) // 2:
return num
```

## Boyer–Moore Majority Vote Algorithm Solution

The Boyer–Moore Majority Vote Algorithm is an ingenious solution that effectively finds the majority element with a linear time complexity and constant space usage. It operates on the principle that the majority element's count can offset all other elements' counts combined.

**Big O Notation Analysis**: This algorithm boasts an impressive `O(n)`

time complexity with `O(1)`

space complexity.

### Step-by-step explanation of the Boyer-Moore Voting Algorithm

**Initialize two variables**:**candidate**: This will eventually hold the majority element.**count**: This is used to track the "strength" of the current candidate. It increases when we see an instance of the candidate and decreases when we see anything else.

**Identify a Candidate**:Iterate through each element in the array.

If

`count`

is 0, we set the current element as our candidate.Then, for each element, if it is the same as our current candidate, we increment

`count`

. If it's different, we decrement`count`

.

### Moore Solution in Python

```
def findMajorityElement(nums):
count = 0
candidate = None
for num in nums:
if count == 0:
candidate = num
count += (1 if num == candidate else -1)
return candidate
```

### Moore Solution in TypeScript

```
function majorityElement(nums: number[]): number {
let count = 0;
let candidate = 0;
for (let i = 0; i < nums.length; i++) {
if (count === 0) {
candidate = nums[i];
}
if (nums[i] === candidate) {
count += 1;
} else {
count -= 1;
}
}
return candidate;
};
```

### Moore Solution in Java

```
public int majorityElement(int[] nums) {
int count = 0;
int candidate = 0;
for (int i = 0; i < nums.length; i++) {
if (count == 0) {
candidate = nums[i];
}
if (nums[i] == candidate) {
count += 1;
} else {
count -=1;
}
}
return candidate;
}
```

## Conclusion

Each solution to the Majority Element problem offers different insights into algorithmic design and complexity analysis. Starting from the simple sorting method to the efficient Boyer–Moore Voting Algorithm, these approaches equip you with versatile strategies for tackling array manipulation and frequency counting problems.

Understanding the underlying principles and trade-offs of each method is crucial for software engineering interviews and beyond. Happy coding, and may your journey through coding challenges be enlightening and empowering!

I hope this post helps illuminate the various paths you can take to solve the Majority Element problem and enhance your problem-solving skills for your coding interviews. If you have questions or need further clarifications, feel free to reach out. Thank you for joining me on this learning adventure!