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Find the frequency of each element in a sorted array

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Given a sorted array, arr[] consisting of N integers, the task is to find the frequencies of each array element.

Examples: 

Input: arr[] = {1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10} 
Output: Frequency of 1 is: 3
              Frequency of 2 is: 1
              Frequency of 3 is: 2
              Frequency of 5 is: 2
              Frequency of 8 is: 3
              Frequency of 9 is: 2
              Frequency of 10 is: 1

Input: arr[] = {2, 2, 6, 6, 7, 7, 7, 11} 
Output:  Frequency of 2 is: 2
               Frequency of 6 is: 2
               Frequency of 7 is: 3
               Frequency of 11 is: 1

 

Naive Approach: The simplest approach is to traverse the array and keep the count of every element encountered in a HashMap and then, in the end, print the frequencies of every element by traversing the HashMap. This approach is already implemented here.

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

Efficient Approach: The above approach can be optimized in terms of space used based on the fact that, in a sorted array, the same elements occur consecutively, so the idea is to maintain a variable to keep track of the frequency of elements while traversing the array. Follow the steps below to solve the problem:

  • Initialize a variable, say freq as 1 to store the frequency of elements.
  • Iterate in the range [1, N-1] using the variable i and perform the following steps:
    • If the value of arr[i] is equal to arr[i-1], increment freq by 1.
    • Else print value the frequency of arr[i-1] obtained in freq and then update freq to 1.
  • Finally, after the above step, print the frequency of the last distinct element of the array as freq.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
    // Function to print the frequency
    // of each element of the sorted array
     void printFreq(vector<int> &arr, int N)
    {
        
        // Stores the frequency of an element
        int freq = 1;
        
       // Traverse the array arr[]
        for (int i = 1; i < N; i++)
        {
           
            // If the current element is equal
            // to the previous element
            if (arr[i] == arr[i - 1])
            {
               
                // Increment the freq by 1
                freq++;
            }
           
        // Otherwise,
            else {
                cout<<"Frequency of "<<arr[i - 1]<< " is: " << freq<<endl;
                // Update freq
                freq = 1;
            }
        }
        
        // Print the frequency of the last element
       cout<<"Frequency of "<<arr[N - 1]<< " is: " << freq<<endl;
       }
 
    // Driver Code
    int main()
    {
       
        // Given Input
        vector<int> arr
            = { 1, 1, 1, 2, 3, 3, 5, 5,
                8, 8, 8, 9, 9, 10 };
        int N = arr.size();
       
        // Function Call
        printFreq(arr, N);
    return 0;
    }   
 
// This code is contributed by codersaty


Java




// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
    // Function to print the frequency
    // of each element of the sorted array
    static void printFreq(int arr[], int N)
    {
 
        // Stores the frequency of an element
        int freq = 1;
 
        // Traverse the array arr[]
        for (int i = 1; i < N; i++) {
            // If the current element is equal
            // to the previous element
            if (arr[i] == arr[i - 1]) {
                // Increment the freq by 1
                freq++;
            }
 
            // Otherwise,
            else {
                System.out.println("Frequency of "
                                   + arr[i - 1]
                                   + " is: " + freq);
                // Update freq
                freq = 1;
            }
        }
 
        // Print the frequency of the last element
        System.out.println("Frequency of "
                           + arr[N - 1]
                           + " is: " + freq);
    }
 
    // Driver Code
    public static void main(String args[])
    {
        // Given Input
        int arr[]
            = { 1, 1, 1, 2, 3, 3, 5, 5,
                8, 8, 8, 9, 9, 10 };
        int N = arr.length;
 
        // Function Call
        printFreq(arr, N);
    }
}


Python3




# Python3 program for the above approach
 
# Function to print the frequency
# of each element of the sorted array
def printFreq(arr, N):
   
    # Stores the frequency of an element
    freq = 1
 
    # Traverse the array arr[]
    for i in range(1, N, 1):
       
        # If the current element is equal
        # to the previous element
        if (arr[i] == arr[i - 1]):
           
            # Increment the freq by 1
            freq += 1
 
        # Otherwise,
        else:
            print("Frequency of",arr[i - 1],"is:",freq)
                # Update freq
            freq = 1
 
    # Print the frequency of the last element
    print("Frequency of",arr[N - 1],"is:",freq)
 
# Driver Code
if __name__ == '__main__':
   
    # Given Input
    arr = [1, 1, 1, 2, 3, 3, 5, 5,8, 8, 8, 9, 9, 10]
    N = len(arr)
 
    # Function Call
    printFreq(arr, N)
     
    # This code is contributed by ipg2016107.


C#




// C# program for the above approach
using System;
 
public class GFG{
     
// Function to print the frequency
// of each element of the sorted array
static void printFreq(int[] arr, int N)
{
 
    // Stores the frequency of an element
    int freq = 1;
 
    // Traverse the array arr[]
    for (int i = 1; i < N; i++)
    {
 
      // If the current element is equal
      // to the previous element
      if (arr[i] == arr[i - 1])
      {
 
        // Increment the freq by 1
        freq++;
      }
 
      // Otherwise,
      else {
        Console.WriteLine("Frequency of " + arr[i - 1] + " is: " + freq);
        // Update freq
        freq = 1;
      }
    }
 
    // Print the frequency of the last element
    Console.WriteLine("Frequency of " + arr[N - 1] + " is: " + freq);
}
 
// Driver Code
static public void Main (){
 
    // Given Input
    int[] arr = { 1, 1, 1, 2, 3, 3, 5, 5,
                  8, 8, 8, 9, 9, 10 };
    int N = arr.Length;
       
    // Function Call
    printFreq(arr, N);
}
}
 
// This code is contributed by Dharanendra L V.


Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to print the frequency
// of each element of the sorted array
function printFreq(arr, N)
{
     
    // Stores the frequency of an element
    let freq = 1;
 
    // Traverse the array arr[]
    for(let i = 1; i < N; i++)
    {
         
        // If the current element is equal
        // to the previous element
        if (arr[i] == arr[i - 1])
        {
             
            // Increment the freq by 1
            freq++;
        }
 
        // Otherwise,
        else
        {
            document.write("Frequency of " +
                           parseInt(arr[i - 1]) + " is: " +
                           parseInt(freq) + "<br>");
                            
            // Update freq
            freq = 1;
        }
    }
 
    // Print the frequency of the last element
    document.write("Frequency of " +
                   parseInt(arr[N - 1]) + " is: " +
                   parseInt(freq) + "<br>");
}
 
// Driver Code
 
// Given Input
let arr = [ 1, 1, 1, 2, 3, 3, 5, 5,
            8, 8, 8, 9, 9, 10 ];
let N = arr.length;
 
// Function Call
printFreq(arr, N);
 
// This code is contributed by Potta Lokesh
 
</script>


Output

Frequency of 1 is: 3
Frequency of 2 is: 1
Frequency of 3 is: 2
Frequency of 5 is: 2
Frequency of 8 is: 3
Frequency of 9 is: 2
Frequency of 10 is: 1

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

Another approach :- using unordered_map

   Initialize a hashmap to store the frequency of each element.
   Traverse the array and insert each element into the hashmap. If an element already exists in the hashmap, increment its frequency.
   Print the frequency of each element in the hashmap.

Here’s the implementation of the function frequency_sorted_array_3() using this approach:

C++




#include <iostream>
#include <unordered_map>
#include <vector>
 
using namespace std;
 
// Function to calculate the frequency of each element in a
// sorted array using the hashmap approach
void frequency_sorted_array_3(vector<int> arr)
{
    int n = arr.size();
 
    // Create a hashmap to store the frequency of each
    // element
    unordered_map<int, int> freq;
 
    // Traverse the array and insert each element into the
    // hashmap
    for (int i = 0; i < n; i++) {
        freq[arr[i]]++;
    }
 
    // Print the frequency of each element in the hashmap
    for (auto it : freq) {
        cout << it.first << " occurs " << it.second
             << " times" << endl;
    }
}
 
int main()
{
    vector<int> arr
        = { 1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10 };
 
    frequency_sorted_array_3(arr);
 
    return 0;
}


Java




import java.util.*;
 
public class Main {
 
    // Function to calculate the frequency of
    // each element in a sorted array using the hashmap
    // approach
    static void frequencySortedArray(List<Integer> arr)
    {
        int n = arr.size();
 
        // Create a hashmap to store the frequency of each
        // element
        Map<Integer, Integer> freq = new LinkedHashMap<>();
 
        // Traverse the array and insert each element into
        // the hashmap
        for (int i = 0; i < n; i++) {
            freq.put(arr.get(i),
                     freq.getOrDefault(arr.get(i), 0) + 1);
        }
 
        // Print the frequency of each element in the
        // hashmap
        for (Map.Entry<Integer, Integer> entry :
             freq.entrySet()) {
            System.out.println(entry.getKey() + " occurs "
                               + entry.getValue()
                               + " times");
        }
    }
 
    public static void main(String[] args)
    {
        List<Integer> arr = Arrays.asList(
            1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10);
 
        frequencySortedArray(arr);
    }
}


Python3




def frequency_sorted_array_3(arr):
    n = len(arr)
 
    # Create a dictionary to store the frequency of each element
    freq = {}
 
    # Traverse the array and insert each element into the dictionary
    for i in range(n):
        if arr[i] in freq:
            freq[arr[i]] += 1
        else:
            freq[arr[i]] = 1
 
    # Print the frequency of each element in the dictionary
    for key in freq:
        print(f"{key} occurs {freq[key]} times")
 
 
arr = [1, 1, 1, 2, 3, 3, 5, 5,
       8, 8, 8, 9, 9, 10]
frequency_sorted_array_3(arr)


Javascript




function frequency_sorted_array_3(arr) {
    let n = arr.length;
 
    // Create a hashmap to store the frequency of each element
    let freq = {};
 
    // Traverse the array and insert each element into the hashmap
    for (let i = 0; i < n; i++) {
        if (freq[arr[i]]) {
            freq[arr[i]]++;
        } else {
            freq[arr[i]] = 1;
        }
    }
 
    // Print the frequency of each element in the hashmap
    for (let key in freq) {
        console.log(key + " occurs " + freq[key] + " times");
    }
}
 
let arr = [1, 1, 1, 2, 3, 3, 5, 5,
                8, 8, 8, 9, 9, 10];
frequency_sorted_array_3(arr);


C#




using System;
using System.Collections.Generic;
 
public class Program {
    // Function to calculate the frequency of
    // each element in a sorted array using the hashmap
    // approach
    static void FrequencySortedArray(List<int> arr)
    {
        int n = arr.Count;
 
        // Create a dictionary to store the frequency of
        // each element
        Dictionary<int, int> freq
            = new Dictionary<int, int>();
 
        // Traverse the array and insert each element into
        // the dictionary
        for (int i = 0; i < n; i++) {
            if (freq.ContainsKey(arr[i]))
                freq[arr[i]]++;
            else
                freq.Add(arr[i], 1);
        }
 
        // Print the frequency of each element in the
        // dictionary
        foreach(KeyValuePair<int, int> entry in freq)
        {
            Console.WriteLine(entry.Key + " occurs "
                              + entry.Value + " times");
        }
    }
 
    public static void Main(string[] args)
    {
        List<int> arr = new List<int>() {
            1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10
        };
 
        FrequencySortedArray(arr);
    }
}


Output

10 occurs 1 times
9 occurs 2 times
8 occurs 3 times
5 occurs 2 times
3 occurs 2 times
2 occurs 1 times
1 occurs 3 times

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



Last Updated : 05 Apr, 2023
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