Counting frequencies of array elements
Last Updated : 03 Oct, 2023
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 1Input : 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.
Implementation:
C++ // CPP program to count frequencies of array items #include <bits/stdc++.h> 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 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. # 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# 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 <script> // JavaScript program to count frequencies of array items function countFreq(arr, n) { let visited = Array.from({length: n}, (_, i) => false); // Traverse through array elements and // count frequencies for (let i = 0; i < n; i++) { // Skip this element if already processed if (visited[i] == true) continue; // Count frequency let count = 1; for (let j = i + 1; j < n; j++) { if (arr[i] == arr[j]) { visited[j] = true; count++; } } document.write(arr[i] + " " + count + "<br/>"); } } // Driver Code let arr = [ 10, 20, 20, 10, 10, 20, 5, 20 ]; let n = arr.length; countFreq(arr, n); </script> Complexity Analysis:
- Time Complexity : O(n2)
- Auxiliary Space : O(n)
An efficient solution is to use hashing.
Implementation:
C++ // CPP program to count frequencies of array items #include <bits/stdc++.h> 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 program to count frequencies of array items import java.util.*; class GFG { static void countFreq(int arr[], int n) { Map<Integer, Integer> 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<Integer, Integer> 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 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# 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 */ <script> // JavaScript program to count // frequencies of array items function countFreq(arr, n) { const map ={} // Traverse through array elements and // count frequencies for (let i = 0; i < n; i++) { if(map[arr[i]){ map[arr[i]]+=1 } else{ map[arr[i]] =1 } } console.log(map) } var arr = [10, 20, 20, 10, 10, 20, 5, 20]; var n = arr.length; countFreq(arr, n); </script> Complexity Analysis:
- Time Complexity : O(n)
- Auxiliary Space : O(n)
In the above efficient solution, how to print elements in same order as they appear in input?
Implementation:
C++ // CPP program to count frequencies of array items #include <bits/stdc++.h> 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 program to count frequencies of array items import java.util.*; class GFG { static void countFreq(int arr[], int n) { Map<Integer, Integer> 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 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# 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 <script> // Javascript program to count frequencies of array items function countFreq(arr, n) { var mp = new Map(); // Traverse through array elements and // count frequencies for (var i = 0; i < n; i++) { if(mp.has(arr[i])) mp.set(arr[i], mp.get(arr[i])+1) else mp.set(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 (var i = 0; i < n; i++) { if (mp.get(arr[i]) != -1) { document.write( arr[i] + " " + mp.get(arr[i]) + "<br>"); mp.set(arr[i], -1); } } } var arr = [10, 20, 20, 10, 10, 20, 5, 20]; var n = arr.length; countFreq(arr, n); // This code is contributed by rrrtnx. </script> Complexity Analysis:
- Time Complexity : O(n)
- Auxiliary Space : O(n)
This problem can be solved in Java using Hashmap. Below is the program.
Implementation:
C++ // C++ program to count frequencies of // integers in array using Hashmap #include <bits/stdc++.h> 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<size;i++) { freqMap[arr[i]]++; } // Printing the freqMap for (auto it : freqMap) { cout<<it.first<<" "<<it.second<<endl; } } int main() { int arr[] = {10, 20, 20, 10, 10, 20, 5, 20}; int size = sizeof(arr)/sizeof(arr[0]); frequencyNumber(arr,size); } // This code is contributed by shinjanpatra. // 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<Integer, Integer> freqMap = new HashMap<Integer, Integer>(); for (int i=0;i<size;i++) { if (freqMap.containsKey(arr[i])) { // If number is present in freqMap, // incrementing it's count by 1 freqMap.put(arr[i], freqMap.get(arr[i]) + 1); } else { // If integer is not present in freqMap, // putting this integer to freqMap with 1 as it's value freqMap.put(arr[i], 1); } } // Printing the freqMap for (Map.Entry entry : freqMap.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 size = arr.length; frequencyNumber(arr,size); } } # 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# 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 <script> // Javascript program to count frequencies of // integers in array using Hashmap function frequencyNumber(arr,size) { // Creating a HashMap containing integer // as a key and occurrences as a value let freqMap = new Map(); for (let i=0;i<size;i++) { if (freqMap.has(arr[i])) { // If number is present in freqMap, // incrementing it's count by 1 freqMap.set(arr[i], freqMap.get(arr[i]) + 1); } else { // If integer is not present in freqMap, // putting this integer to freqMap with 1 as it's value freqMap.set(arr[i], 1); } } // Printing the freqMap for (let [key, value] of freqMap.entries()) { document.write(key + " " + value+"<br>"); } } // Driver Code let arr=[10, 20, 20, 10, 10, 20, 5, 20]; let size = arr.length; frequencyNumber(arr,size); // This code is contributed by patel2127 </script> Complexity Analysis:
- Time Complexity: O(n) since using a single loop to track frequency
- Auxiliary Space: O(n) for hashmap.
Another Efficient Solution (Space optimization): we can find frequency of array elements using Binary search function . First we will sort the array for binary search . Our frequency of element will be '(last occ - first occ)+1' of a element in a array .
Below is the implementation of the above approach:
C++ // C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; //Function to find frequency of elements in the array void countFreq(int arr[], int n) { sort(arr,arr+n);//sort array for binary search for(int i = 0 ; i < n ;i++) { //index of first and last occ of arr[i] int first_index = lower_bound(arr,arr+n,arr[i])- arr; int last_index = upper_bound(arr,arr+n,arr[i])- arr-1; i=last_index; int fre=last_index-first_index+1;//finding frequency cout << arr[i] <<" "<<fre <<endl;//printing frequency } } // Drive code int main() { int arr[] = { 10, 20, 20, 10, 10, 20, 5, 20 }; int n = sizeof(arr) / sizeof(arr[0]); // Function call countFreq(arr, n); return 0; } // This Code is contributed by nikhilsainiofficial546 // Java program to count frequencies of // integers in array using Hashmap import java.io.*; import java.util.*; class OccurenceOfNumberInArray { public static void countFreq(int[] arr, int n) { Arrays.sort(arr); //sort array for binary search for (int i = 0; i < n; i++) { //index of first and last occ of arr[i] int first_index = Arrays.binarySearch(arr, arr[i]); int last_index = Arrays.binarySearch(arr, arr[i]); while (first_index > 0 && arr[first_index - 1] == arr[i]) { first_index--; } while (last_index < n - 1 && arr[last_index + 1] == arr[i]) { last_index++; } i = last_index; int fre = last_index - first_index + 1;//finding frequency System.out.println(arr[i] + " " + fre);//printing frequency } } // Driver Code public static void main(String[] args) { int arr[] = {10, 20, 20, 10, 10, 20, 5, 20}; int size = arr.length; countFreq(arr,size); } } from bisect import bisect_left, bisect_right # Function to find frequency of elements in the array def countFreq(arr, n): arr.sort() # sort array for binary search i = 0 while i < n: # index of first and last occ of arr[i] first_index = bisect_left(arr, arr[i]) last_index = bisect_right(arr, arr[i]) - 1 i = last_index + 1 fre = last_index - first_index + 1 # finding frequency print(arr[i - 1], fre) # printing frequency # Driver code arr = [10, 20, 20, 10, 10, 20, 5, 20] n = len(arr) countFreq(arr, n)
using System; public class GFG { // Function to find frequency of elements in the array public static void countFreq(int[] arr, int n) { Array.Sort(arr); // sort array for binary search for (int i = 0; i < n; i++) { // index of first and last occurrence of arr[i] int first_index = Array.IndexOf(arr, arr[i]); int last_index = Array.LastIndexOf(arr, arr[i]); int fre = last_index - first_index + 1; // finding frequency Console.WriteLine(arr[i] + " " + fre); // printing frequency i = last_index; } } // Driver code static public void Main () { int[] arr = { 10, 20, 20, 10, 10, 20, 5, 20 }; int n = arr.Length; // Function call countFreq(arr, n); } } // Javascript implementation of the above approach //Function to find frequency of elements in the array function countFreq(arr,n) { arr.sort((a, b) => a - b); //sort array for binary search let i = 0; while (i < n) { //index of first and last occ of arr[i] const first_index = arr.indexOf(arr[i]); const last_index = arr.lastIndexOf(arr[i]); i = last_index; const fre = last_index - first_index + 1; //finding frequency console.log(arr[i], fre); //printing frequency i++; } } // Driver code const arr = [ 10, 20, 20, 10, 10, 20, 5, 20 ]; countFreq(arr,arr.length); Time Complexity: O(n*log2n) , where O(log2n) time for binary search function .
Auxiliary Space: O(1)
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