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#
// 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>
OutputFrequency 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);
}
}
Output10 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)
Similar Reads
Find the Frequency of Each Element in a Vector in C++
The frequency of an element is number of times it occurred in the vector. In this article, we will learn different methods to find the frequency of each element in a vector in C++.The simplest method to find the frequency of each element in a vector is by iterating the vector and storing the count o
3 min read
Find frequency of each element in given 3D Array
Given a 3D array of size N*M*P consisting only of English alphabet characters, the task is to print the frequency of all the elements in increasing order. If the frequency is the same, print them in lexicographic ordering. Examples: Input: N = 3, M = 4, P = 5, matrix = { { {a, b, c, d, e}, {f, g, h,
9 min read
Find frequency of each element in a limited range array in less than O(n) time
Given a sorted array arr[] of positive integers, the task is to find the frequency for each element in the array. Assume all elements in the array are less than some constant MNote: Do this without traversing the complete array. i.e. expected time complexity is less than O(n)Examples: Input: arr[] =
10 min read
Find the array element having maximum frequency of the digit K
Given an array arr[] of size N and an integer K, the task is to find an array element that contains the digit K a maximum number of times. If more than one solutions exist, then print any one of them. Otherwise, print -1. Examples: Input: arr[] = {3, 77, 343, 456}, K = 3 Output: 343 Explanation: Fre
15 min read
Cumulative frequency of count of each element in an unsorted array
Given an unsorted array. The task is to calculate the cumulative frequency of each element of the array using a count array.Examples: Input : arr[] = [1, 2, 2, 1, 3, 4]Output :1->2 2->4 3->5 4->6Input : arr[] = [1, 1, 1, 2, 2, 2]Output :1->3 2->6 A simple solution is to use two nes
9 min read
Find array elements with frequencies in range [l , r]
Given an array of integers, find the elements from the array whose frequency lies in the range [l, r]. Examples: Input : arr[] = { 1, 2, 3, 3, 2, 2, 5 } l = 2, r = 3 Output : 2 3 3 2 2 Approach : Take a hash map, which will store the frequency of all the elements in the array.Now, traverse once agai
9 min read
Find element with highest frequency in given nested Array
Given an array arr[] of N integers. The task is to create a frequency array freq[] of the given array arr[] and find the maximum element of the frequency array. If two elements have the same frequency in the array freq[], then return the element which has a smaller value. Examples: Input: arr[] = {1
8 min read
Find the Kth occurrence of an element in a sorted Array
Given a sorted array arr[] of size N, an integer X, and a positive integer K, the task is to find the index of Kth occurrence of X in the given array. Examples: Input: N = 10, arr[] = [1, 2, 3, 3, 4, 5, 5, 5, 5, 5], X = 5, K = 2Output: Starting index of the array is '0' Second occurrence of 5 is at
15+ min read
Frequency of an element in an array
Given an array, a[], and an element x, find a number of occurrences of x in a[].Examples: Input : a[] = {0, 5, 5, 5, 4} x = 5Output : 3Input : a[] = {1, 2, 3} x = 4Output : 0Unsorted ArrayThe idea is simple, we initialize count as 0. We traverse the array in a linear fashion. For every element that
9 min read
Find element in a sorted array whose frequency is greater than or equal to n/2.
Given a sorted array of length n, find the number in array that appears more than or equal to n/2 times. It is given that such element always exists. Examples: Input : 2 3 3 4 Output : 3 Input : 3 4 5 5 5 Output : 5 Input : 1 1 1 2 3 Output : 1 To find that number, we traverse the array and check th
3 min read