Median and Mode using Counting Sort

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 + b)/2.
Median = (2 + 3)/2 = 2.5

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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;     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);     printModeMedian(a, n);     return 0; }

Java

 import java.util.Arrays;    // Java Program for Mode and  // Median using Counting  // Sort technique  class GFG  { // function that sort input array a[] and  // calculate mode and median using counting  // sort.         static void printModeMedian(int a[], int n)      {         // The output array b[] will          // have sorted array          int[] b = new int[n];            // variable to store max of          // input array which will          // to have size of count array          int max = Arrays.stream(a).max().getAsInt();            // 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[] = new int[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;         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 = (float) ((b[(n - 1) / 2]                     + b[(n / 2)]) / 2.0);         }            // Output the result          System.out.println("median = " + median);         System.out.println("mode = " + mode);     }    // Driver program  public static void main(String[] args) {     int a[] = {1, 4, 1, 2, 7, 1, 2, 5, 3, 6};     int n = a.length;     printModeMedian(a, n);    } }    // This code is contributed by 29AjayKumar

C#

 // C# Program for Mode and  // Median using Counting  // Sort technique  using System; using System.Linq;    class GFG  {     // function that sort input array a[] and      // calculate mode and median using counting      // sort.      static void printModeMedian(int []a, int n)      {         // The output array b[] will          // have sorted array          int[] b = new int[n];            // variable to store max of          // input array which will          // to have size of count array          int max = a.Max();            // 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 = new int[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;         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 = (float) ((b[(n - 1) / 2] +                                 b[(n / 2)]) / 2.0);         }            // Output the result          Console.WriteLine("median = " + median);         Console.WriteLine("mode = " + mode);     }    // Driver Code public static void Main(String[] args) {     int []a = {1, 4, 1, 2, 7, 1, 2, 5, 3, 6};     int n = a.Length;     printModeMedian(a, n); } }    // This code is contributed by Rajput-Ji

PHP

 \$k)         {             \$k = \$count[\$i];             \$mode = \$i;         }     }         // Update count[] array with sum     for (\$i = 1; \$i < \$t; \$i++)         \$count[\$i] = \$count[\$i] + \$count[\$i - 1];        // Sorted output array b[]     // to calculate median     for (\$i = 0; \$i < \$n; \$i++)     {         \$b[\$count[\$a[\$i]] - 1] = \$a[\$i];         \$count[\$a[\$i]]--;     }            // Median according to odd and even      // array size respectively.     \$median;     if (\$n % 2 != 0)         \$median = \$b[\$n / 2];     else         \$median = (\$b[(\$n - 1) / 2] +                     \$b[(\$n / 2)]) / 2.0;            // Output the result      echo "median = ", \$median, "\n" ;     echo "mode = " , \$mode; }    // Driver Code \$a = array( 1, 4, 1, 2, 7,              1, 2, 5, 3, 6 ); \$n = sizeof(\$a); printModeMedian(\$a, \$n);    // This code is contributed by jit_t ?>

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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