# Counting frequencies of array elements

• Difficulty Level : Easy
• Last Updated : 21 Jun, 2022

Given an array which may contain duplicates, print all elements and their frequencies.

Examples:

```Input :  arr[] = {10, 20, 20, 10, 10, 20, 5, 20}
Output : 10 3
20 4
5  1

Input : arr[] = {10, 20, 20}
Output : 10 1
20 2 ```

A simple solution is to run two loops. For every item count number of times, it occurs. To avoid duplicate printing, keep track of processed items.

## C++

 `// CPP program to count frequencies of array items ` `#include ` `using` `namespace` `std; ` ` `  `void` `countFreq(``int` `arr[], ``int` `n) ` `{ ` `    ``// Mark all array elements as not visited ` `    ``vector<``bool``> visited(n, ``false``); ` ` `  `    ``// Traverse through array elements and ` `    ``// count frequencies ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Skip this element if already processed ` `        ``if` `(visited[i] == ``true``) ` `            ``continue``; ` ` `  `        ``// Count frequency ` `        ``int` `count = 1; ` `        ``for` `(``int` `j = i + 1; j < n; j++) { ` `            ``if` `(arr[i] == arr[j]) { ` `                ``visited[j] = ``true``; ` `                ``count++; ` `            ``} ` `        ``} ` `        ``cout << arr[i] << ``" "` `<< count << endl; ` `    ``} ` `} ` ` `  `int` `main() ` `{ ` `    ``int` `arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``countFreq(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to count frequencies of array items ` `import` `java.util.Arrays; ` ` `  `class` `GFG ` `{ ` `public` `static` `void` `countFreq(``int` `arr[], ``int` `n) ` `{ ` `    ``boolean` `visited[] = ``new` `boolean``[n]; ` `     `  `    ``Arrays.fill(visited, ``false``); ` ` `  `    ``// Traverse through array elements and ` `    ``// count frequencies ` `    ``for` `(``int` `i = ``0``; i < n; i++) { ` ` `  `        ``// Skip this element if already processed ` `        ``if` `(visited[i] == ``true``) ` `            ``continue``; ` ` `  `        ``// Count frequency ` `        ``int` `count = ``1``; ` `        ``for` `(``int` `j = i + ``1``; j < n; j++) { ` `            ``if` `(arr[i] == arr[j]) { ` `                ``visited[j] = ``true``; ` `                ``count++; ` `            ``} ` `        ``} ` `        ``System.out.println(arr[i] + ``" "` `+ count); ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String []args) ` `{ ` `    ``int` `arr[] = ``new` `int``[]{ ``10``, ``20``, ``20``, ``10``, ``10``, ``20``, ``5``, ``20` `}; ` `    ``int` `n = arr.length; ` `    ``countFreq(arr, n); ` `} ` `} ` ` `  `// This code contributed by Adarsh_Verma. `

## Python3

 `# Python 3 program to count frequencies ` `# of array items ` `def` `countFreq(arr, n): ` `     `  `    ``# Mark all array elements as not visited ` `    ``visited ``=` `[``False` `for` `i ``in` `range``(n)] ` ` `  `    ``# Traverse through array elements  ` `    ``# and count frequencies ` `    ``for` `i ``in` `range``(n): ` `         `  `        ``# Skip this element if already  ` `        ``# processed ` `        ``if` `(visited[i] ``=``=` `True``): ` `            ``continue` ` `  `        ``# Count frequency ` `        ``count ``=` `1` `        ``for` `j ``in` `range``(i ``+` `1``, n, ``1``): ` `            ``if` `(arr[i] ``=``=` `arr[j]): ` `                ``visited[j] ``=` `True` `                ``count ``+``=` `1` `         `  `        ``print``(arr[i], count) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[``10``, ``20``, ``20``, ``10``, ``10``, ``20``, ``5``, ``20``] ` `    ``n ``=` `len``(arr) ` `    ``countFreq(arr, n) ` `     `  `# This code is contributed by ` `# Shashank_Sharma `

## C#

 `// C# program to count frequencies of array items ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``public` `static` `void` `countFreq(``int` `[]arr, ``int` `n) ` `    ``{ ` `        ``bool` `[]visited = ``new` `bool``[n]; ` `     `  `        ``// Traverse through array elements and ` `        ``// count frequencies ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `     `  `            ``// Skip this element if already processed ` `            ``if` `(visited[i] == ``true``) ` `                ``continue``; ` `     `  `            ``// Count frequency ` `            ``int` `count = 1; ` `            ``for` `(``int` `j = i + 1; j < n; j++)  ` `            ``{ ` `                ``if` `(arr[i] == arr[j])  ` `                ``{ ` `                    ``visited[j] = ``true``; ` `                    ``count++; ` `                ``} ` `            ``} ` `            ``Console.WriteLine(arr[i] + ``" "` `+ count); ` `        ``} ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String []args) ` `    ``{ ` `        ``int` `[]arr = ``new` `int``[]{ 10, 20, 20, 10, 10, 20, 5, 20 }; ` `        ``int` `n = arr.Length; ` `        ``countFreq(arr, n); ` `    ``} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

## Javascript

 ` `

Output

```10 3
20 4
5 1```

Time Complexity : O(n2
Auxiliary Space : O(n)

An efficient solution is to use hashing.

## C++

 `// CPP program to count frequencies of array items ` `#include ` `using` `namespace` `std; ` ` `  `void` `countFreq(``int` `arr[], ``int` `n) ` `{ ` `    ``unordered_map<``int``, ``int``> mp; ` ` `  `    ``// Traverse through array elements and ` `    ``// count frequencies ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``mp[arr[i]]++; ` ` `  `    ``// Traverse through map and print frequencies ` `    ``for` `(``auto` `x : mp) ` `        ``cout << x.first << ``" "` `<< x.second << endl; ` `} ` ` `  `int` `main() ` `{ ` `    ``int` `arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``countFreq(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to count frequencies of array items ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `    ``static` `void` `countFreq(``int` `arr[], ``int` `n) ` `    ``{ ` `        ``Map mp = ``new` `HashMap<>(); ` ` `  `        ``// Traverse through array elements and ` `        ``// count frequencies ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``if` `(mp.containsKey(arr[i]))  ` `            ``{ ` `                ``mp.put(arr[i], mp.get(arr[i]) + ``1``); ` `            ``}  ` `            ``else` `            ``{ ` `                ``mp.put(arr[i], ``1``); ` `            ``} ` `        ``} ` `        ``// Traverse through map and print frequencies ` `        ``for` `(Map.Entry entry : mp.entrySet()) ` `        ``{ ` `            ``System.out.println(entry.getKey() + ``" "` `+ entry.getValue()); ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[])  ` `    ``{ ` `        ``int` `arr[] = {``10``, ``20``, ``20``, ``10``, ``10``, ``20``, ``5``, ``20``}; ` `        ``int` `n = arr.length; ` `        ``countFreq(arr, n); ` `    ``} ` `} ` ` `  `// This code contributed by Rajput-Ji `

## Python3

 `# Python3 program to count frequencies  ` `# of array items ` `def` `countFreq(arr, n): ` ` `  `    ``mp ``=` `dict``() ` ` `  `    ``# Traverse through array elements  ` `    ``# and count frequencies ` `    ``for` `i ``in` `range``(n): ` `        ``if` `arr[i] ``in` `mp.keys(): ` `            ``mp[arr[i]] ``+``=` `1` `        ``else``: ` `            ``mp[arr[i]] ``=` `1` `             `  `    ``# Traverse through map and print  ` `    ``# frequencies ` `    ``for` `x ``in` `mp: ` `        ``print``(x, ``" "``, mp[x]) ` ` `  `# Driver code ` `arr ``=` `[``10``, ``20``, ``20``, ``10``, ``10``, ``20``, ``5``, ``20` `] ` `n ``=` `len``(arr) ` `countFreq(arr, n) ` ` `  `# This code is contributed by  ` `# Mohit kumar 29 `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic;  ` ` `  `class` `GFG ` `{ ` ` `  `    ``static` `void` `countFreq(``int` `[]arr, ``int` `n) ` `    ``{ ` `        ``Dictionary<``int``, ``int``> mp = ``new` `Dictionary<``int``,``int``>(); ` ` `  `        ``// Traverse through array elements and ` `        ``// count frequencies ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``if` `(mp.ContainsKey(arr[i]))  ` `            ``{ ` `                ``var` `val = mp[arr[i]]; ` `                ``mp.Remove(arr[i]); ` `                ``mp.Add(arr[i], val + 1);  ` `            ``}  ` `            ``else` `            ``{ ` `                ``mp.Add(arr[i], 1); ` `            ``} ` `        ``} ` `         `  `        ``// Traverse through map and print frequencies ` `        ``foreach``(KeyValuePair<``int``, ``int``> entry ``in` `mp) ` `        ``{ ` `            ``Console.WriteLine(entry.Key + ``" "` `+ entry.Value); ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String []args)  ` `    ``{ ` `        ``int` `[]arr = {10, 20, 20, 10, 10, 20, 5, 20}; ` `        ``int` `n = arr.Length; ` `        ``countFreq(arr, n); ` `    ``} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

## Javascript

 ``

Output

```5 1
10 3
20 4```

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

In above efficient solution, how to print elements in same order as they appear in input?

## C++

 `// CPP program to count frequencies of array items ` `#include ` `using` `namespace` `std; ` ` `  `void` `countFreq(``int` `arr[], ``int` `n) ` `{ ` `    ``unordered_map<``int``, ``int``> mp; ` ` `  `    ``// Traverse through array elements and ` `    ``// count frequencies ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``mp[arr[i]]++; ` ` `  `    ``// To print elements according to first ` `    ``// occurrence, traverse array one more time ` `    ``// print frequencies of elements and mark ` `    ``// frequencies as -1 so that same element ` `    ``// is not printed multiple times. ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `      ``if` `(mp[arr[i]] != -1) ` `      ``{ ` `          ``cout << arr[i] << ``" "` `<< mp[arr[i]] << endl; ` `          ``mp[arr[i]] = -1; ` `      ``} ` `    ``} ` `} ` ` `  `int` `main() ` `{ ` `    ``int` `arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``countFreq(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to count frequencies of array items  ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``static` `void` `countFreq(``int` `arr[], ``int` `n)  ` `    ``{ ` `        ``Map mp = ``new` `HashMap<>(); ` ` `  `        ``// Traverse through array elements and  ` `        ``// count frequencies  ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``mp.put(arr[i], mp.get(arr[i]) == ``null` `? ``1` `: mp.get(arr[i]) + ``1``); ` `        ``} ` ` `  `        ``// To print elements according to first  ` `        ``// occurrence, traverse array one more time  ` `        ``// print frequencies of elements and mark  ` `        ``// frequencies as -1 so that same element  ` `        ``// is not printed multiple times.  ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``{ ` `            ``if` `(mp.get(arr[i]) != -``1``)  ` `            ``{ ` `                ``System.out.println(arr[i] + ``" "` `+ mp.get(arr[i])); ` `                ``mp.put(arr[i], -``1``); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `arr[] = {``10``, ``20``, ``20``, ``10``, ``10``, ``20``, ``5``, ``20``}; ` `        ``int` `n = arr.length; ` `        ``countFreq(arr, n); ` `    ``} ` `} ` ` `  `// This code contributed by Rajput-Ji `

## Python3

 `# Python3 program to count frequencies of array items ` `def` `countFreq(arr, n): ` `     `  `    ``mp ``=` `{} ` `     `  `    ``# Traverse through array elements and ` `    ``# count frequencies ` `    ``for` `i ``in` `range``(n): ` `        ``if` `arr[i] ``not` `in` `mp: ` `            ``mp[arr[i]] ``=` `0` `        ``mp[arr[i]] ``+``=` `1` `         `  `    ``# To print elements according to first ` `    ``# occurrence, traverse array one more time ` `    ``# print frequencies of elements and mark ` `    ``# frequencies as -1 so that same element ` `    ``# is not printed multiple times. ` `    ``for` `i ``in` `range``(n): ` `        ``if` `(mp[arr[i]] !``=` `-``1``): ` `            ``print``(arr[i],mp[arr[i]]) ` `        ``mp[arr[i]] ``=` `-``1` ` `  `# Driver code ` ` `  `arr ``=` `[``10``, ``20``, ``20``, ``10``, ``10``, ``20``, ``5``, ``20``] ` `n ``=` `len``(arr) ` `countFreq(arr, n) ` ` `  `# This code is contributed by shubhamsingh10 `

## C#

 `// C# program to count frequencies of array items  ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG  ` `{  ` ` `  `    ``static` `void` `countFreq(``int` `[]arr, ``int` `n)  ` `    ``{  ` `        ``Dictionary<``int``,``int``> mp = ``new` `Dictionary<``int``,``int``>();  ` ` `  `        ``// Traverse through array elements and  ` `        ``// count frequencies  ` `         `  `        ``for` `(``int` `i = 0 ; i < n; i++) ` `        ``{ ` `            ``if``(mp.ContainsKey(arr[i])) ` `            ``{ ` `                ``var` `val = mp[arr[i]]; ` `                ``mp.Remove(arr[i]); ` `                ``mp.Add(arr[i], val + 1);  ` `            ``} ` `            ``else` `            ``{ ` `                ``mp.Add(arr[i], 1); ` `            ``} ` `        ``} ` ` `  `        ``// To print elements according to first  ` `        ``// occurrence, traverse array one more time  ` `        ``// print frequencies of elements and mark  ` `        ``// frequencies as -1 so that same element  ` `        ``// is not printed multiple times.  ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{  ` `            ``if` `(mp.ContainsKey(arr[i]) && mp[arr[i]] != -1)  ` `            ``{  ` `                ``Console.WriteLine(arr[i] + ``" "` `+ mp[arr[i]]);  ` `                ``mp.Remove(arr[i]); ` `                ``mp.Add(arr[i], -1);  ` `            ``}  ` `        ``}  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{  ` `        ``int` `[]arr = {10, 20, 20, 10, 10, 20, 5, 20};  ` `        ``int` `n = arr.Length;  ` `        ``countFreq(arr, n);  ` `    ``}  ` `}  ` ` `  `// This code is contributed by Princi Singh `

## Javascript

 ``

Output

```10 3
20 4
5 1```

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

This problem can be solved in Java using Hashmap. Below is the program.

## C++

 `// C++ program to count frequencies of ` `// integers in array using Hashmap ` `#include ` `using` `namespace` `std; ` ` `  `void` `frequencyNumber(``int` `arr[],``int` `size) ` `{ ` `   `  `  ``// Creating a HashMap containing integer ` `  ``// as a key and occurrences as a value ` `  ``unordered_map<``int``,``int``>freqMap; ` ` `  `  ``for` `(``int` `i=0;i

## Java

 `// Java program to count frequencies of ` `// integers in array using Hashmap ` `import` `java.io.*; ` `import` `java.util.*; ` `class` `OccurenceOfNumberInArray { ` `    ``static` `void` `frequencyNumber(``int` `arr[], ``int` `size) ` `    ``{ ` `        ``// Creating a HashMap containing integer ` `        ``// as a key and occurrences as a value ` `        ``HashMap freqMap ` `            ``= ``new` `HashMap(); ` ` `  `        ``for` `(``int` `i=``0``;i

## Python3

 `# Python program to count frequencies of ` `# integers in array using Hashmap ` ` `  `def` `frequencyNumber(arr,size): ` `    ``# Creating a HashMap containing integer ` `        ``# as a key and occurrences as a value ` `        ``freqMap ``=` `{} ` `  `  `        ``for` `i ``in` `range``(size): ` `            ``if` `(arr[i] ``in` `freqMap): ` `  `  `                ``# If number is present in freqMap, ` `                ``# incrementing it's count by 1 ` `                ``freqMap[arr[i]] ``=` `freqMap[arr[i]] ``+` `1` `            ``else``: ` `  `  `                ``# If integer is not present in freqMap, ` `                ``# putting this integer to freqMap with 1 as it's value ` `                ``freqMap[arr[i]] ``=` `1` `  `  `        ``# Printing the freqMap ` `        ``for` `key, value ``in` `freqMap.items(): ` `            ``print``(f``"{key} {value}"``) ` ` `  `# Driver Code ` `arr ``=` `[``10``, ``20``, ``20``, ``10``, ``10``, ``20``, ``5``, ``20``] ` `size ``=` `len``(arr) ` `frequencyNumber(arr,size) ` ` `  `# This code is contributed by shinjanpatra`

## C#

 `// C# program to count frequencies of ` `// integers in array using Hashmap ` `using` `System; ` `using` `System.Collections.Generic;  ` `class` `GFG ` `{ ` `  ``static` `void` `frequencyNumber(``int` `[]arr,``int` `size) ` `  ``{ ` ` `  `    ``// Creating a Dictionary containing integer ` `    ``// as a key and occurrences as a value ` `    ``Dictionary<``int``, ``int``> freqMap = ``new` `Dictionary<``int``,``int``>(); ` ` `  `    ``for``(``int` `i = 0; i < size; i++){ ` `      ``if` `(freqMap.ContainsKey(arr[i]))  ` `      ``{ ` `        ``var` `val = freqMap[arr[i]]; ` `        ``freqMap.Remove(arr[i]); ` `        ``freqMap.Add(arr[i], val + 1);  ` `      ``}  ` `      ``else` `      ``{ ` `        ``freqMap.Add(arr[i], 1); ` `      ``} ` `    ``} ` ` `  `    ``// Printing the freqMap ` `    ``foreach``(KeyValuePair<``int``, ``int``> entry ``in` `freqMap) ` `    ``{ ` `      ``Console.WriteLine(entry.Key + ``" "` `+ entry.Value); ` `    ``} ` `  ``} ` ` `  `  ``public` `static` `void` `Main(String []args)  ` `  ``{ ` `    ``int` `[]arr = {10, 20, 20, 10, 10, 20, 5, 20}; ` `    ``int` `size = arr.Length; ` `    ``frequencyNumber(arr,size); ` `  ``} ` `} ` `// This code is contributed by Taranpreet`

## Javascript

 ` `

Output

```5 1
10 3
20 4```

Time Complexity: O(n) since using a single loop to track frequency
Auxiliary Space: O(n) for hashmap

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