Find the frequency of each element in a sorted array

Last Updated : 05 Apr, 2023

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# 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);

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)

Find frequency of each element in given 3D Array

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Find the frequency of each element in a sorted array

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