Related Articles

# Frequency of each element of an array of small ranged values

• Last Updated : 13 Sep, 2021

Given an array where elements in small range. The maximum element in the array does not go beyond size of array. Find frequencies of elements.
Examples:

```Input : arr[] = {3, 1, 2, 3, 4, 5, 4}
Output: 1-->1
2-->1
3-->2
4-->2
5-->1

Input : arr[] = {1, 2, 2, 1, 2}
Output: 1-->2
2-->3

Input : arr[] = {1, 2, 4}
Output: 1-->1
2-->1
4-->1```

A simple solution is to use two nested loops. For each element (from 1 to n where n is size of array), count how many times it appears. Time complexity of this solution is O(n*n)
A better solution is to use sorting. First sort the array, after sorting, linearly traverse the array and count occurrences of each element. Time complexity of this solution is O(n Log n)
An efficient solution is to use hashing. We insert every element in a hash table and increment frequency. Time complexity of this solution is O(n). Please see Frequency Measuring Techniques for Competitive Programming for implementation.
An efficient solution for limited range
The hashing based solution is fast, but requires hash function computations, etc. If we know that range is small, we use direct address table where we create an array of size equal to maximum value and use array elements as index.
Following is the implementation for above explanation:

## C++

 `// CPP program to find frequencies of elements in``// limited range array.``#include ``using` `namespace` `std;` `void` `frequencyOfEach(``int``* arr, ``int` `n)``{``    ``// finding maximum element in array``    ``int` `max = *max_element(arr, arr + n);` `    ``// make hash array of size equal to maximum``    ``// element in array``    ``int` `hash[max + 1] = { 0 };` `    ``/* Counting frequency of each element of array``       ``and storing it in hash*/``    ``for` `(``int` `i = 0; i < n; i++) {``        ``hash[arr[i]]++;``    ``}` `    ``// printing frequency of elements``    ``for` `(``int` `i = 0; i <= max; i++) {` `        ``/* If hash[i] has stored any value``           ``i.e element has occurred atleast``           ``once in array */``        ``if` `(hash[i] != 0)``            ``cout << i << ``"-->"` `<< hash[i] << ``"\n"``;``    ``}``}` `int` `main()``{``    ``int` `arr[] = { 5, 2, 2, 3, 5, 1, 1, 5, 3, 4 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``frequencyOfEach(arr, n);``    ``return` `0;``}`

## Java

 `// Java program to find frequencies of elements in``// limited range array.``import` `java.util.*;` `class` `solution``{` `static` `void` `frequencyOfEach(``int` `[]arr, ``int` `n)``{``    ``int` `max = Integer.MIN_VALUE;``    ``// finding maximum element in array``    ``for` `(``int` `i = ``0``;imax)``        ``max = arr[i];``}``    `  `    ``// make hash array of size equal to maximum``    ``// element in array``    ``int` `[]hash = ``new` `int``[max + ``1``];``    ``Arrays.fill(hash,``0``);` `    ``/* Counting frequency of each element of array``    ``and storing it in hash*/``    ``for` `(``int` `i = ``0``; i < n; i++) {``        ``hash[arr[i]]++;``    ``}` `    ``// printing frequency of elements``    ``for` `(``int` `i = ``0``; i <= max; i++) {` `        ``/* If hash[i] has stored any value``        ``i.e element has occurred atleast``        ``once in array */``        ``if` `(hash[i] != ``0``)``            ``System.out.println(i+``"-->"``+hash[i]);``    ``}``}` `public` `static` `void` `main(String args[])``{``    ``int` `[]arr = { ``5``, ``2``, ``2``, ``3``, ``5``, ``1``, ``1``, ``5``, ``3``, ``4` `};``    ``int` `n = arr.length;``    ``frequencyOfEach(arr, n);``    ` `}``}``//This code is contributed by Surendra_Gangwar`

## Python3

 `# Python 3 program to find frequencies``# of elements in limited range array.``def` `frequencyOfEach(arr, n) :``    ` `    ``# finding maximum element in array``    ``max_element ``=` `max``(arr)` `    ``# make hash array of size equal``    ``# to maximum element in array``    ``hash` `=` `[``0``] ``*` `(max_element ``+` `1``)` `    ``# Counting frequency of each element``    ``# of array and storing it in hash``    ``for` `i ``in` `range``(n) :``        ``hash``[arr[i]] ``+``=` `1``    ` `    ``# printing frequency of elements``    ``for` `i ``in` `range``(max_element ``+` `1``) :` `        ``# If hash[i] has stored any value``        ``# i.e element has occurred atleast``        ``# once in array``        ``if` `(``hash``[i] !``=` `0``) :``            ``print``(i, ``"-->"``, ``hash``[i])``    ` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:` `    ``arr ``=` `[ ``5``, ``2``, ``2``, ``3``, ``5``,``            ``1``, ``1``, ``5``, ``3``, ``4` `]``    ``n ``=` `len``(arr)``    ``frequencyOfEach(arr, n);` `# This code is contributed by Ryuga`

## C#

 `// C# program to find frequencies of elements in``// limited range array.``using` `System;`` ` `public` `class` `solution{` `    ``static` `void` `frequencyOfEach(``int` `[]arr, ``int` `n)``    ``{``        ``int` `max = ``int``.MinValue;``        ``// finding maximum element in array``        ``for` `(``int` `i = 0;imax)``            ``max = arr[i];``    ``}`  `        ``// make hash array of size equal to maximum``        ``// element in array``        ``int` `[]hash = ``new` `int``[max + 1];` `        ``/* Counting frequency of each element of array``        ``and storing it in hash*/``        ``for` `(``int` `i = 0; i < n; i++) {``            ``hash[arr[i]]++;``        ``}` `        ``// printing frequency of elements``        ``for` `(``int` `i = 0; i <= max; i++) {` `            ``/* If hash[i] has stored any value``            ``i.e element has occurred atleast``            ``once in array */``            ``if` `(hash[i] != 0)``                ``Console.WriteLine (i+``"-->"``+hash[i]);``        ``}``    ``}` `    ``public` `static` `void` `Main()``    ``{``        ``int` `[]arr = { 5, 2, 2, 3, 5, 1, 1, 5, 3, 4 };``        ``int` `n = arr.Length;``        ``frequencyOfEach(arr, n);` `    ``}``}``// This code is contributed by PrinciRaj1992`

## PHP

 `"` `. ``\$hash``[``\$i``] . ``"\n"``;``    ``}``}` `// Driver Code``\$arr` `= ``array``(5, 2, 2, 3, 5, 1, 1, 5, 3, 4 );``\$n` `= sizeof(``\$arr``);``frequencyOfEach(``\$arr``, ``\$n``);` `// This code is contributed by ita_c``?>`

## Javascript

 ``
Output:
```1-->2
2-->2
3-->2
4-->1
5-->3```

Time Complexity: O(max_value)
Auxiliary Space: O(max_value)

My Personal Notes arrow_drop_up