Important Note

This is my first blog entry, and I want to acknowledge that some mistakes were initially made. I'm grateful for the community's feedback, which has allowed me to make corrections. I will note areas where I've made corrections below.

Special thanks to @vgtcross and @brownfox2k6 for their input on this article.

Intro

Mex, or the minimum excludant, is a concept that refers to the smallest non-negative integer not present in a given array of non-negative integers. It's a fundamental concept in various algorithmic problems and has applications in areas like game theory and memory management. In this blog post, we'll delve into three different approaches to calculating Mex in an array, each with its unique characteristics and use cases.

1. Quick Mex for Small Array Values

When dealing with small array values, specifically when all values are less than 1 million, a set can be used to calculate Mex. This approach leverages the constant time lookups of a set, making it efficient for small values.

int mex(vector<int>& numberArray) {
  set<int> sett;
  for(int i = 0; i < numberArray.size(); i++) 
    sett.insert(numberArray[i]);
  for(int i = 0; i < 1000001; i++)
    if(!sett.count(i)) return i;
  return -1;
}

This method is simple and effective, but it may not be suitable for larger values due to the limitation of the set's size.

2. Sorting Approach

Another way to calculate Mex is by sorting the array first and then looping through the sorted array. This approach takes O(N log N) time due to the sorting step.

sort(numberArray.begin(), numberArray.end());
int mex = 0;
for(int e : numberArray) {
 if (e == mex) {
  mex++;
 }
}

While this method is more general and can handle larger values, the sorting step adds complexity, making it less efficient for very large arrays.

3. Efficient O(n) Approach

For those looking for an efficient solution, an O(n) approach is available. Special thanks to @vgtcross for his contribution here.

int mex(vector<int> &a) {
    vector<bool> f(a.size() + 1, 0);
    for (int i : a) if (i <= (int) a.size()) f[i] = 1;
    int mex = 0;
    while (f[mex]) ++mex;
    return mex;
}

This method is noteworthy for its efficiency, working well for large arrays. It's based on the principle that the Mex of n integers cannot exceed n, allowing for a quick calculation.

Conclusion

Calculating Mex is a versatile task with multiple approaches, each with its advantages and limitations. Whether you need a quick solution for small values, a general method for larger arrays, or an efficient O(n) approach, understanding these techniques can enhance your problem-solving skills.

Examples of Mex Calculation

Quick Mex for Small Array Values

Array: [3, 1, 4, 0, 2] Mex: 5 Explanation: All integers from 0 to 4 are present in the array, so the Mex is 5.

Sorting Approach

Array: [2, 2, 1, 3, 0] Mex: 4 Explanation: After sorting, the array becomes [0, 1, 2, 2, 3], and the Mex is 4.

Efficient O(n) Approach

Array: [1, 1, 2, 0] Mex: 3 Explanation: Using the efficient O(n) method, the Mex is calculated as 3.

Applications of Mex Calculation

Game Theory: In combinatorial game theory, Mex is used to determine the Grundy number of a game position. It's a key concept in solving impartial games using the Sprague-Grundy theorem.

Memory Management: In computer systems, Mex can be utilized to find the smallest available memory block, optimizing memory allocation and enhancing system performance.

Graph Coloring: In graph algorithms, Mex can be applied to find the smallest unused color when coloring vertices, facilitating efficient graph coloring strategies.

Closing Note

I appreciate this community's feedback, which has helped me correct and improve the content. Mistakes are a part of growth, and I'm committed to growing and improving alongside all of you. I look forward to sharing more insights in the future and will consider all feedback as I strive to become a great writer.

I've expanded on each section to provide more context and detail. If there are specific areas you'd like me to focus on further or any other adjustments needed, please let me know!