Majority Element

easyFREEArrays & HashingNetflix DSA

Given an array nums, return the majority element (appears more than n/2 times). The majority element always exists.

Example

Input

3,2,3

Output

Reasoning

nums is a list of integers. Here it is 3, 2, 3, so there are $n = 3$ values.
The majority element is the value whose count is strictly greater than $\frac{n}{2} = \frac{3}{2} = 1.5$ .
Tally the values: 3 appears $2$ times and 2 appears $1$ time. Since $2 > 1.5$ , 3 qualifies, while $1 < 1.5$ rules out 2.
The output is that single value, 3, returned as an integer, not its count.
Notice that at most one value can ever clear the bar: if two different values each appeared more than $\frac{n}{2}$ times, together they would need more than $n$ slots. That is why the answer is always unique, and why "more than half" is a stronger condition than simply "most frequent".

Constraints

1 <= len(nums) <= 5 * 10^4
-10^9 <= nums[i] <= 10^9

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Concept primer

Background Knowledge

The Majority Element problem is a classic example of a problem that can be solved using Arrays & Hashing techniques. In this problem, we are given an array nums and asked to find the majority element, which is the element that appears more than $n/2$ times, where $n$ is the length of the array. To understand this problem, we need to have a basic understanding of arrays and hashing. An array is a collection of elements of the same data type stored in contiguous memory locations, while hashing is a technique used to store and retrieve data efficiently using a hash function.

The key concept in this problem is the idea of a majority element, which is an element that appears more than half of the time in the array. This means that if we have an array of length $n$ , the majority element will appear at least $n/2 + 1$ times. To find the majority element, we can use various techniques such as counting, hashing, or voting. The voting technique is particularly useful in this problem, as it allows us to find the majority element in a single pass through the array.

In terms of theory, the Majority Element problem is related to the concept of frequency analysis, which is the study of the distribution of elements in a dataset. In this problem, we are interested in finding the element with the highest frequency, which is the majority element. The problem also has applications in data analysis and machine learning, where finding the majority element can be useful in identifying patterns and trends in data.

Algorithm/Approach

The general approach to solving the Majority Element problem is to use a voting algorithm, which works by essentially maintaining a counter for the majority element. The algorithm iterates through the array, incrementing the counter when it encounters the majority element and decrementing it when it encounters a different element. The element that is left at the end of the algorithm is the majority element. This approach is based on the idea that the majority element will always be the one that is left after all other elements have been cancelled out.

Step-by-Step Strategy

To implement the solution, we can follow these steps:

Initialize a count variable to 0 and a candidate variable to the first element of the array.
Iterate through the array, starting from the second element.
For each element, check if it is equal to the candidate. If it is, increment the count. If it is not, decrement the count.
If the count becomes 0, set the candidate to the current element and reset the count to 1.
After iterating through the entire array, the candidate will be the majority element.

Common Pitfalls

Some common pitfalls to watch out for when implementing the solution include:

Not initializing the count and candidate variables correctly.
Not handling the case where the count becomes 0 correctly.
Not iterating through the entire array.

Time & Space Complexity

The expected time complexity of the solution is $O(n)$ , where $n$ is the length of the array, since we only need to iterate through the array once. The expected space complexity is $O(1)$ , since we only need to use a constant amount of space to store the count and candidate variables.

solution.py

Python 3.13

Test results