Given an **n** sized unsorted array, find median and mode using **counting sort** technique. Thia can be useful when array elements are in limited range.

**Examples:**

Input : array a[] = {1, 1, 1, 2, 7, 1} Output : Mode = 1 Median = 1.5 Input : array a[] = {9, 9, 9, 9, 9} Output : Mode = 9 Median = 9

**Prerequisites:** Count Sort, Median of Array, Mode (Most frequent element in array)

1. Auxiliary(count) array before summing its previous counts, c[]:

Index: 0 1 2 3 4 5 6 7 8 9 10

count: 0 4 1 0 0 0 0 1 0 0 0

2. Mode = index with maximum value of count.

Mode = 1(for above example)

3. count array is modified similarly as it is done while performing count sort.

Index: 0 1 2 3 4 5 6 7 8 9 10

count: 0 3 5 6 7 8 9 10 10 10 10

4. output array is calculated normally as in count sort, b[]:

output array b[] = {1, 1, 1, 2, 2, 3, 4, 5, 6, 7}

5. If size of array b[] is odd, Median = b[n/2]

Else Median = (b[(n-1)/2] + b[n/2])/2

6. For above example size of b[] is even hence, Median = (b[4] + b[5])/2.

Median = (2 + 3)/2 = 2.5

Basic Approach to be followed :

Assuming size of input array is **n**:

**Step1:** Take the count array before summing its previous counts into next index.

**Step2:** The index with maximum value stored in it is the mode of given data.

**Step3:** In case there are more than one indexes with maximum value in it, all are results for mode so we can take any.

**Step4:** Store the value at that index in a separate variable called mode.

**Step5:** Continue with the normal processing of the count sort.

**Step6:** In the resultant(sorted) array, if n is odd then median = middle-most element of the

sorted array, And if n is even the median = average of two middle-most elements of the sorted array.

**Step7:** Store the result in a separate variable called median.

Following is the implementation of problem discussed above:

## C++

// C++ Program for Mode and // Median using Counting // Sort technique #include <bits/stdc++.h> using namespace std; // function that sort input array a[] and // calculate mode and median using counting // sort. void printModeMedian(int a[], int n) { // The output array b[] will // have sorted array int b[n]; // variable to store max of // input array which will // to have size of count array int max = *max_element(a, a+n); // auxiliary(count) array to // store count. Initialize // count array as 0. Size // of count array will be // equal to (max + 1). int t = max + 1; int count[t]; for (int i = 0; i < t; i++) count[i] = 0; // Store count of each element // of input array for (int i = 0; i < n; i++) count[a[i]]++; // mode is the index with maximum count int mode = 0; int k = count[0]; for (int i = 1; i < t; i++) { if (count[i] > k) { k = count[i]; mode = i; } } // Update count[] array with sum for (int i = 1; i < t; i++) count[i] = count[i] + count[i-1]; // Sorted output array b[] // to calculate median for (int i = 0; i < n; i++) { b[count[a[i]]-1] = a[i]; count[a[i]]--; } // Median according to odd and even // array size respectively. float median; if (n % 2 != 0) median = b[n/2]; else median = (b[(n-1)/2] + b[(n/2)])/2.0; // Output the result cout << "median = " << median << endl; cout << "mode = " << mode; } // Driver program int main() { int a[] = { 1, 4, 1, 2, 7, 1, 2, 5, 3, 6 }; int n = sizeof(a)/sizeof(a[0]); printModeMedian(a, n); return 0; }

**Output:**

median = 2.5 mode = 1

**Time Complexity** = O(N + P), where N is the time for input array and P is time for count array.

**Space Complexity** = O(P), where P is the size of auxiliary array.

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