# 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[4] + b[5])/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[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; ` `} `

## 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[``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 = (``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[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 = (``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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