Frequency of an element in an array
Last Updated :
17 Jan, 2025
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 = 5
Output : 3
Input : a[] = {1, 2, 3}
x = 4
Output : 0
Unsorted Array
The idea is simple, we initialize count as 0. We traverse the array in a linear fashion. For every element that matches with x, we increment count. Finally, we return count.
Below is the implementation of the approach.
C++
// CPP program to count occurrences of an
// element in an unsorted array
#include<iostream>
using namespace std;
int frequency(int a[], int n, int x)
{
int count = 0;
for (int i=0; i < n; i++)
if (a[i] == x)
count++;
return count;
}
// Driver program
int main() {
int a[] = {0, 5, 5, 5, 4};
int x = 5;
int n = sizeof(a)/sizeof(a[0]);
cout << frequency(a, n, x);
return 0;
}
// Java program to count
// occurrences of an
// element in an unsorted
// array
import java.io.*;
class GFG {
static int frequency(int a[],
int n, int x)
{
int count = 0;
for (int i=0; i < n; i++)
if (a[i] == x)
count++;
return count;
}
// Driver program
public static void main (String[]
args) {
int a[] = {0, 5, 5, 5, 4};
int x = 5;
int n = a.length;
System.out.println(frequency(a, n, x));
}
}
// This code is contributed
// by Ansu Kumari
# Python program to count
# occurrences of an
# element in an unsorted
# array
def frequency(a, x):
count = 0
for i in a:
if i == x: count += 1
return count
# Driver program
a = [0, 5, 5, 5, 4]
x = 5
print(frequency(a, x))
# This code is contributed by Ansu Kumari
// C# program to count
// occurrences of an
// element in an unsorted
// array
using System;
class GFG {
static int frequency(int []a,
int n, int x)
{
int count = 0;
for (int i=0; i < n; i++)
if (a[i] == x)
count++;
return count;
}
// Driver program
static public void Main (){
int []a = {0, 5, 5, 5, 4};
int x = 5;
int n = a.Length;
Console.Write(frequency(a, n, x));
}
}
// This code is contributed
// by Anuj_67
<script>
// C# program to count occurrences of an element in an unsorted array
function frequency(a, n, x)
{
let count = 0;
for (let i=0; i < n; i++)
if (a[i] == x)
count++;
return count;
}
let a = [0, 5, 5, 5, 4];
let x = 5;
let n = a.length;
document.write(frequency(a, n, x));
</script>
<?php
// PHP program to count occurrences of an
// element in an unsorted array
function frequency($a, $n, $x)
{
$count = 0;
for ($i = 0; $i < $n; $i++)
if ($a[$i] == $x)
$count++;
return $count;
}
// Driver Code
$a = array (0, 5, 5, 5, 4);
$x = 5;
$n = sizeof($a);
echo frequency($a, $n, $x);
// This code is contributed by ajit
?>
Time Complexity: O(n)
Auxiliary Space: O(1)
Sorted Array
We can apply methods for both sorted and unsorted. But for a sorted array, we can optimize it to work in O(Log n) time using Binary Search. Please refer to below article for details.Count number of occurrences (or frequency) in a sorted array.
If there are multiple queries on a single array
We can use hashing to store frequencies of all elements. Then we can answer all queries in O(1) time. Please refer Frequency of each element in an unsorted array for details.
CPP
// CPP program to answer queries for frequencies
// in O(1) time.
#include <bits/stdc++.h>
using namespace std;
unordered_map<int, int> hm;
void countFreq(int a[], int n)
{
// Insert elements and their
// frequencies in hash map.
for (int i=0; i<n; i++)
hm[a[i]]++;
}
// Return frequency of x (Assumes that
// countFreq() is called before)
int query(int x)
{
return hm[x];
}
// Driver program
int main()
{
int a[] = {1, 3, 2, 4, 2, 1};
int n = sizeof(a)/sizeof(a[0]);
countFreq(a, n);
cout << query(2) << endl;
cout << query(3) << endl;
cout << query(5) << endl;
return 0;
}
// Java program to answer
// queries for frequencies
// in O(1) time.
import java.io.*;
import java.util.*;
class GFG {
static HashMap <Integer, Integer> hm = new HashMap<Integer, Integer>();
static void countFreq(int a[], int n)
{
// Insert elements and their
// frequencies in hash map.
for (int i=0; i<n; i++)
if (hm.containsKey(a[i]) )
hm.put(a[i], hm.get(a[i]) + 1);
else hm.put(a[i] , 1);
}
// Return frequency of x (Assumes that
// countFreq() is called before)
static int query(int x)
{
if (hm.containsKey(x))
return hm.get(x);
return 0;
}
// Driver program
public static void main (String[] args) {
int a[] = {1, 3, 2, 4, 2, 1};
int n = a.length;
countFreq(a, n);
System.out.println(query(2));
System.out.println(query(3));
System.out.println(query(5));
}
}
// This code is contributed by Ansu Kumari
# Python program to
# answer queries for
# frequencies
# in O(1) time.
hm = {}
def countFreq(a):
global hm
# Insert elements and their
# frequencies in hash map.
for i in a:
if i in hm: hm[i] += 1
else: hm[i] = 1
# Return frequency
# of x (Assumes that
# countFreq() is
# called before)
def query(x):
if x in hm:
return hm[x]
return 0
# Driver program
a = [1, 3, 2, 4, 2, 1]
countFreq(a)
print(query(2))
print(query(3))
print(query(5))
# This code is contributed
# by Ansu Kumari
// C# program to answer
// queries for frequencies
// in O(1) time.
using System;
using System.Collections.Generic;
class GFG
{
static Dictionary <int, int> hm = new Dictionary<int, int>();
static void countFreq(int []a, int n)
{
// Insert elements and their
// frequencies in hash map.
for (int i = 0; i < n; i++)
if (hm.ContainsKey(a[i]) )
hm[a[i]] = hm[a[i]] + 1;
else hm.Add(a[i], 1);
}
// Return frequency of x (Assumes that
// countFreq() is called before)
static int query(int x)
{
if (hm.ContainsKey(x))
return hm[x];
return 0;
}
// Driver code
public static void Main(String[] args)
{
int []a = {1, 3, 2, 4, 2, 1};
int n = a.Length;
countFreq(a, n);
Console.WriteLine(query(2));
Console.WriteLine(query(3));
Console.WriteLine(query(5));
}
}
// This code is contributed by 29AjayKumar
<script>
// JavaScript program to answer
// queries for frequencies
// in O(1) time.
var hm = new Map();
function countFreq(a, n)
{
// Insert elements and their
// frequencies in hash map.
for (var i=0; i<n; i++)
{
if(hm.has(a[i]))
{
hm.set(a[i], hm.get(a[i])+1);
}
else
{
hm.set(a[i], 1);
}
}
}
// Return frequency of x (Assumes that
// countFreq() is called before)
function query(x)
{
if(hm.has(x))
{
return hm.get(x);
}
return 0;
}
// Driver program
var a = [1, 3, 2, 4, 2, 1];
var n = a.length;
countFreq(a, n);
document.write( query(2) + "<br>");
document.write( query(3) + "<br>");
document.write( query(5) + "<br>");
</script>
Time complexity: O(n) where n is the number of elements in the given array.
Auxiliary space: O(n) because using extra space for unordered_map.
Using Library Method
In this approach I am using Language functions only with below conditions.
Conditions :
- To find the counts of digit we can't use count_if() and count() functions.
- Use STL and lambda functions only.
Examples :
Input : v = {7,2,3,1,7,6,7,1,3,7} digit = 7
Output : 4
Input : v = {7,2,3,1,7,6,7,1,3,7} digit = 10
Output : 0
Explanation :
To find the occurrence of a digit with these conditions follow the below steps,
1. Use partition(start, end, condition) function to get all the digits and return the pointer of the last digit.
2. Use the distance(start , end) to get the distance from vector starting point to the last digit pointer which partition() function returns.
So, distance() function returns the occurrence of the digit and we can print it.
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
#include <algorithm>
#include <vector>
int main()
{
int digit;
digit = 7; // Specify the digit as a input to find the
// occurrence of digit
vector<int> v{ 7, 2, 3, 1, 7, 6, 7, 1, 3, 7 };
auto itr
= partition(v.begin(), v.end(),
[&digit](int x) { return x == digit; });
int count = distance(v.begin(), itr);
cout << count << endl;
return 0;
}
import java.util.*;
import java.util.stream.*;
class Main {
public static void main(String[] args) {
int digit;
digit = 7; // Specify the digit as a input to find the
// occurrence of digit
List<Integer> v = Arrays.asList(7, 2, 3, 1, 7, 6, 7, 1, 3, 7);
List<Integer> partitioned = v.stream().filter(x -> x == digit).collect(Collectors.toList());
int count = partitioned.size();
System.out.println(count);
}
}
# Specify the digit as an input to find the occurrence of digit
digit = 7
v = [7, 2, 3, 1, 7, 6, 7, 1, 3, 7]
# Use the filter function to find all occurrences of the digit in the list
itr = filter(lambda x: x == digit, v)
# Count the number of occurrences
count = len(list(itr))
print(count)
using System;
using System.Collections.Generic;
using System.Linq;
public class GFG
{
static public void Main()
{
int digit;
digit = 7; // Specify the digit as a input to find the
// occurrence of digit
List<int> v = new List<int> { 7, 2, 3, 1, 7, 6, 7, 1, 3, 7 };
var partitioned = v.Where(x => x == digit).ToList();
int count = partitioned.Count();
Console.WriteLine(count);
}
}
let digit;
digit = 7; // Specify the digit as a input to find the
// occurrence of digit
let v = [7, 2, 3, 1, 7, 6, 7, 1, 3, 7];
let itr = v.filter(x => x === digit);
let count = itr.length;
console.log(count);
Time Complexity: O(N)
Auxiliary Space: O(1)
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