Find the frequency of a number in an array

Given an array a[] and an element x, find 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



If array is not sorted

The idea is simple, we initialize count as 0. We traverse array in linear fashion. For every element that matches with x, we increment count. Finally we return count.
Below is the implementation of the approach.

C++

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// 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;

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Java

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// 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

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C#

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// 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

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Python3

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# 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

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PHP

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<?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
?>

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Output:

3

Time Complexity : O(n)
Auxiliary Space : O(1)

If array is sorted

We can apply method for both sorted and unsorted. But for sorted array, we can optimize it to work in O(Log n) time using Binary Search. Please refer 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

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// 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;
}

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Java

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// 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

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Python3

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# 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

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Output :

2
1
0

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