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++ 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 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 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. |
<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:
#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;
} |
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);
}
} |
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) |
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);
}
} |
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)