## Introduction

The “majority element in a list” problem is a common problem in computer science and data analysis. In this problem, you are given a list or array of elements, and you need to determine if there exists an element in the list that appears more than half of the total number of elements. If such an element exists, it is called the “majority element.”

The problem can be formally stated as follows:

**Input**: A list or array of elements of length ‘n’.

**Output**: Determine if there is a majority element in the list, and if so, return the majority element. A majority element is an element that appears more than n/2 times in the list.

For example:

- In the list
`[2, 2, 1, 1, 2, 2, 3]`

, the majority element is`2`

because it appears 4 times, which is more than half of the list’s length. - In the list
`[3, 3, 4, 2, 4, 4, 2, 4, 4]`

, the majority element is`4`

because it appears 5 times, which is more than half of the list’s length.

Solving this problem efficiently can be important in various applications, such as data analysis, voting systems, and algorithmic design. One common algorithm to solve this problem efficiently is the Boyer-Moore Majority Vote algorithm.

## Code Implementation

To find the majority element in a list in Python, you can use the Boyer-Moore Majority Vote algorithm. This algorithm allows you to find the element that appears more than half the length of the list (i.e., the majority element) with a time complexity of O(n). Here’s how the algorithm works, step by step:

```
def find_majority_element(nums):
candidate = None # The current potential majority candidate
count = 0 # Count of occurrences of the current candidate
# Step 1: Finding a potential candidate
for num in nums:
if count == 0:
candidate = num
count = 1
elif num == candidate:
count += 1
else:
count -= 1
# Step 2: Verifying the candidate
count = 0 # Resetting the count for the second pass
for num in nums:
if num == candidate:
count += 1
# Step 3: Checking if the candidate is the majority element
if count > len(nums) // 2:
return candidate
else:
return None # No majority element
# Example usage
nums = [2, 2, 1, 1, 2, 2, 3]
result = find_majority_element(nums)
if result is not None:
print("Majority element:", result)
else:
print("No majority element")
```

Now, let’s break down the steps:

**Finding a potential candidate**: In the first loop, we iterate through the list. If the`count`

is 0, we assume the current element as the potential candidate and set the`count`

to 1. If the current element is the same as the candidate, we increment the`count`

. If the current element is different, we decrement the`count`

. This step helps us narrow down the potential majority candidate.**Verifying the candidate**: After the first pass, we’ve identified a potential candidate. In the second loop, we count the occurrences of this candidate to make sure it appears more than half the length of the list.**Checking if the candidate is the majority element**: If the candidate’s count is indeed greater than half the length of the list, we return it as the majority element. Otherwise, we return`None`

to indicate that there’s no majority element.

In the provided example usage, the majority element is `2`

because it appears more than half the number of times in the list `[2, 2, 1, 1, 2, 2, 3]`

.

## Space Complexity

The space complexity of the Boyer-Moore Majority Vote algorithm implementation you provided is O(1), which means it uses constant space regardless of the size of the input list.

The algorithm uses a constant amount of extra space to store the `candidate`

and `count`

variables, regardless of the length of the input list. These variables are used to keep track of the potential majority candidate and its count, and their memory consumption does not increase with the size of the input list. Therefore, the space complexity remains constant, making this algorithm very memory-efficient.

## Time Complexity

The time complexity of the Boyer-Moore Majority Vote algorithm implemented in the provided code is O(n), where ‘n’ is the length of the input list (`nums`

).

Here’s why the time complexity is O(n):

**First Pass (Finding a Potential Candidate):**In the first loop, we iterate through the input list`nums`

once. In each iteration, we perform constant time operations (assignments and comparisons). Therefore, the first pass takes O(n) time.**Second Pass (Verifying the Candidate):**In the second loop, we again iterate through the input list`nums`

once to count the occurrences of the candidate element. This is also a linear operation that takes O(n) time.

Overall, the two passes combined take O(n) time. The algorithm is efficient and scales linearly with the size of the input list, making it an effective way to find the majority element.