Frequency of an element in an array (original) (raw)

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 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

// 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

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#

// 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

JavaScript

PHP

n,n, n,x) { $count = 0; for ($i = 0; i<i < i<n; $i++) if ($a[$i] == $x) $count++; return $count; } // Driver Code $a = array (0, 5, 5, 5, 4); $x = 5; n=sizeof(n = sizeof(n=sizeof(a); echo frequency($a, n,n, 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

// 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

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#

// 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

JavaScript

**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 using namespace std; #include #include int main() { int digit; digit = 7; // Specify the digit as a input to find the // occurrence of digit vector 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; }

Java

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 v = Arrays.asList(7, 2, 3, 1, 7, 6, 7, 1, 3, 7); List partitioned = v.stream().filter(x -> x == digit).collect(Collectors.toList()); int count = partitioned.size(); System.out.println(count); } }

Python

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)

C#

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 v = new List { 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); } }

JavaScript

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)