Find the frequency of each element in a sorted array
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: 1Input: 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> |
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); } } |
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)
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