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Find the frequency of a number in an array
• Difficulty Level : Easy
• Last Updated : 30 Apr, 2021

Given an array a[] and an element x, find number of occurrences of x in a[].
Examples:

```Input  : a[] = {0, 5, 5, 5, 4}
x = 5
Output : 3

Input  : a[] = {1, 2, 3}
x = 4
Output : 0```

If array is not sorted

The idea is simple, we initialize count as 0. We traverse array in linear fashion. For every element that matches with x, we increment count. Finally, we return count.
Below is the implementation of the approach.

## C++

 `// CPP program to count occurrences of an``// element in an unsorted array``#include``using` `namespace` `std;` `int` `frequency(``int` `a[], ``int` `n, ``int` `x)``{``    ``int` `count = 0;``    ``for` `(``int` `i=0; i < n; i++)``       ``if` `(a[i] == x)``          ``count++;``    ``return` `count;``}` `// Driver program``int` `main() {``    ``int` `a[] = {0, 5, 5, 5, 4};``    ``int` `x = 5;``    ``int` `n = ``sizeof``(a)/``sizeof``(a);``    ``cout << frequency(a, n, x);``    ``return` `0;``}`

## Java

 `// Java program to count``// occurrences of an``// element in an unsorted``// array` `import` `java.io.*;` `class` `GFG {``    ` `    ``static` `int` `frequency(``int` `a[],``    ``int` `n, ``int` `x)``    ``{``        ``int` `count = ``0``;``        ``for` `(``int` `i=``0``; i < n; i++)``        ``if` `(a[i] == x)``            ``count++;``        ``return` `count;``    ``}``    ` `    ``// Driver program``    ``public` `static` `void` `main (String[]``    ``args) {``        ` `        ``int` `a[] = {``0``, ``5``, ``5``, ``5``, ``4``};``        ``int` `x = ``5``;``        ``int` `n = a.length;``        ` `        ``System.out.println(frequency(a, n, x));``    ``}``}` `// This code is contributed``// by Ansu Kumari`

## Python3

 `# Python program to count``# occurrences of an``# element in an unsorted``# array``def` `frequency(a, x):``    ``count ``=` `0``    ` `    ``for` `i ``in` `a:``        ``if` `i ``=``=` `x: count ``+``=` `1``    ``return` `count` `# Driver program``a ``=` `[``0``, ``5``, ``5``, ``5``, ``4``]``x ``=` `5``print``(frequency(a, x))` `# This code is contributed by Ansu Kumari`

## C#

 `// C# program to count``// occurrences of an``// element in an unsorted``// array``using` `System;` `class` `GFG {``    ` `    ``static` `int` `frequency(``int` `[]a,``    ``int` `n, ``int` `x)``    ``{``        ``int` `count = 0;``        ``for` `(``int` `i=0; i < n; i++)``        ``if` `(a[i] == x)``            ``count++;``        ``return` `count;``    ``}``    ` `    ``// Driver program``    ``static` `public` `void` `Main (){``    ` `        ` `        ``int` `[]a = {0, 5, 5, 5, 4};``        ``int` `x = 5;``        ``int` `n = a.Length;``        ` `        ``Console.Write(frequency(a, n, x));``    ``}``}` `// This code is contributed``// by Anuj_67`

## PHP

 ``

## Javascript

 ``

Output:

`3`

Time Complexity : O(n)
Auxiliary Space : O(1)

If array is sorted

We can apply method for both sorted and unsorted. But for sorted array, we can optimize it to work in O(Log n) time using Binary Search. Please refer below article for details.Count number of occurrences (or frequency) in a sorted array.

If there are multiple queries on a single array

We can use hashing to store frequencies of all elements. Then we can answer all queries in O(1) time. Please refer Frequency of each element in an unsorted array for details.

## CPP

 `// CPP program to answer queries for frequencies``// in O(1) time.``#include ``using` `namespace` `std;`` ` `unordered_map<``int``, ``int``> hm;` `void` `countFreq(``int` `a[], ``int` `n)``{``    ``// Insert elements and their``    ``// frequencies in hash map.``    ``for` `(``int` `i=0; i

## Java

 `// Java program to answer``// queries for frequencies``// in O(1) time.` `import` `java.io.*;``import` `java.util.*;` `class` `GFG {``    ` `   ``static` `HashMap hm = ``new` `HashMap();` `   ``static` `void` `countFreq(``int` `a[], ``int` `n)``   ``{``        ``// Insert elements and their``        ``// frequencies in hash map.``        ``for` `(``int` `i=``0``; i

## Python3

 `# Python program to``# answer queries for``# frequencies``# in O(1) time.` `hm ``=` `{}` `def` `countFreq(a):``    ``global` `hm``    ` `    ``# Insert elements and their``    ``# frequencies in hash map.``    ` `    ``for` `i ``in` `a:``        ``if` `i ``in` `hm: hm[i] ``+``=` `1``        ``else``: hm[i] ``=` `1` `# Return frequency``# of x (Assumes that``# countFreq() is``# called before)``def` `query(x):``    ``if` `x ``in` `hm:``        ``return` `hm[x]``    ``return` `0` `# Driver program``a ``=` `[``1``, ``3``, ``2``, ``4``, ``2``, ``1``]``countFreq(a)``print``(query(``2``))``print``(query(``3``))``print``(query(``5``))` `# This code is contributed``# by Ansu Kumari`

## C#

 `// C# program to answer``// queries for frequencies``// in O(1) time.``using` `System;``using` `System.Collections.Generic;``class` `GFG``{``    ` `static` `Dictionary <``int``, ``int``> hm = ``new` `Dictionary<``int``, ``int``>();` `static` `void` `countFreq(``int` `[]a, ``int` `n)``{``    ``// Insert elements and their``    ``// frequencies in hash map.``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `(hm.ContainsKey(a[i]) )``            ``hm[a[i]] = hm[a[i]] + 1;``        ``else` `hm.Add(a[i], 1);``}``    ` `// Return frequency of x (Assumes that``// countFreq() is called before)``static` `int` `query(``int` `x)``{``    ``if` `(hm.ContainsKey(x))``        ``return` `hm[x];``    ``return` `0;``}``    ` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]a = {1, 3, 2, 4, 2, 1};``    ``int` `n = a.Length;``    ``countFreq(a, n);``    ``Console.WriteLine(query(2));``    ``Console.WriteLine(query(3));``    ``Console.WriteLine(query(5));``}``}` `// This code is contributed by 29AjayKumar`

Output :

```2
1
0```

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