C++ Calculating the Mode of a Sorted Array

Question

I have to write a C++ code that finds the median and mode of an array. I'm told that it's much easier to find the mode of an array AFTER the numbers have been sorted. I sorted the function but still cannot find the mode.

 int counter = 0;
    for (int pass = 0; pass < size - 1; pass++)
        for (int count = pass + 1; count < size; count++) {
            if (array [count] == array [pass])
                counter++;
            cout << "The mode is: " << counter << endl; 

Answer 1

This is how I did it, my solution will take a sorted vector as input. It has O(n) time complexity and can work with the case where there are more than 1 "mode" number in the vector.

void findMode(vector<double> data) {

double biggestMode = 1;
vector<double> mode, numbers;
numbers.push_back(data.at(0));
mode.push_back(1);
int count = 0;
for (int i = 1; i < data.size(); i++) {
    if (data.at(i) == numbers.at(count)) {
        mode.at(count)++;
    }
    else {
        if (biggestMode < mode.at(count)) {
            biggestMode = mode.at(count);
        }
        count++;
        mode.push_back(1);
        numbers.push_back(data.at(i));
    }
}

for (int i = 0; i < mode.size(); i++) {
    if (mode.at(i) == biggestMode)
        cout << numbers.at(i) << " ";
}
cout << endl;

}

Answer 2

This code uses "map" to find out the MODE from the given array. It assumes the array is already sorted.

int findMode(int * arr, int arraySize)
{
    map<int, int> modeMap;
    for (int i = 0; i < arraySize; ++i) {
        ++modeMap[arr[i]];
    }

    auto x = std::max_element(modeMap.begin(), modeMap.end(),
        [](const pair<int, int>& a, const pair<int, int>& b) {
        return a.second < b.second; });

    return x->first;
}

Answer 3

There is an old adage that states "If you put 10 programmers in a room and give them the same program to code you will get 12 different results", hence my version of answering your question. It may not be as fast (I'm planning on testing it's speed versus some of the other suggestions) but I feel it is easy to understand.

#include <iostream>

using namespace std;

int main ()
{
    short z[10];
    short maxCount = 0, curCount = 0, cur = 0, most = 0;

    for (int i = 0; i < 10; i++)
        {
         cout << "Enter a number: " << endl;
         cin >> z[i];
        }

    for (int i = 0; i < 10; i++)
        {
         cur = z[i];
            for (int a = i; a < 10; a++)
                {
                 if (cur == z[a])
                    {
                     curCount++;
                     cur = z[a];
                    }
                if (curCount > maxCount)
                   {
                    maxCount = curCount;
                    most = z[a];
                   }
            }
            curCount = 0;
        }

    cout << "the mode is : " << maxCount << ", the number is: " << most << endl;
}

Answer 4

Here is the code snippet:

int number = array[0];
int mode = number;
int count = 1;
int countMode = 1;

for (int i=1; i<size; i++)
{
    if (array[i] == number) 
    {
        count++;
    }
    else
    {
        if (count > countMode) 
        {
            countMode = count;
            mode = number;
        }
        count = 1;
        number = array[i];
    }
}

cout << "mode : " << mode << endl;

Answer 5

One way is that you can use Run Length encoding. In Run Length encoding, representation would be like; (Item, Its frequency).

While doing so, keep track of the maximum frequency and Item. This will give you the mode once you complete the Run Length.

for example:

 1 1  2 2 2 3 3 4 5

It run length encoding would be

 {1, 2}, {2, 3}, {3, 2}, {4, 1}, {5, 1}

It needs O(n) space.

Answer 6

I did it this way:

    int main()
{ 
    int mode,modecount2,modecount1;
    bool is_nomode=false;
    vector<int> numbers = { 15,43,25,25,25,25,16,14,93,93,58,14,55,55,55,64,14,43,14,25,15,56,78,13,15,29,14,14,16 };
    sort(numbers);

    //If you uncomment the following part, you can see the sorted list of above numbers
    //for (int i = 0; i < numbers.size(); ++i) std::cout << numbers[i] << '\n';
    //keep_window_open();

    mode = numbers[0];
    modecount1 = 0;
    modecount2 = 1; //Obviously any number exists at least once!
    for (int i = 1; i < numbers.size(); ++i) {
        if(numbers[i]==numbers[i-1]) ++modecount2;
        else {
            if (modecount2 > modecount1) {
                mode = numbers[i - 1];
                modecount1 = modecount2;
            }
            else if (i != 1 && modecount2 == modecount1) { std::cout << "No mode!\n"; is_nomode = true; break; }
            modecount2 = 1;
        }
    }
    if(!is_nomode) std::cout << "Mode of these numbers is: " << mode << std::endl;
    keep_window_open();

Also you can add another 25 to the list of numbers and see what happens if two numbers have the same occurrence! I hope it helps.