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Find frequency of each element in a limited range array in less than O(n) time

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  • Difficulty Level : Medium
  • Last Updated : 12 Sep, 2022
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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 M

Note: Do this without traversing the complete array. i.e. expected time complexity is less than O(n)

Examples: 

Input: arr[] = [1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10]
Output: 
Element 1 occurs 3 times
Element 2 occurs 1 times
Element 3 occurs 2 times
Element 5 occurs 2 times
Element 8 occurs 3 times
Element 9 occurs 2 times
Element 10 occurs 1 times
 

Input: arr[] = [2, 2, 6, 6, 7, 7, 7, 11] 
Output: 
Element 2 occurs 2 times
Element 6 occurs 2 times
Element 7 occurs 3 times
Element 11 occurs 1 times

Frequency of each element in a limited range array using linear search:

To solve the problem follow the below idea:

Traverse the input array and increment the frequency of the element if the current element and the previous element are the same, otherwise reset the frequency and print the element and its frequency

Follow the given steps to solve the problem: 

  • Initialize frequency to 1 and index to 1.
  • Traverse the array from the index position and check if the current element is equal to the previous element.
  • If yes, increment the frequency and index and repeat step 2. Otherwise, print the element and its frequency and repeat step 2.
  • At last(corner case), print the last element and its frequency.

Below is the implementation of the above approach:

C++




// C++ program to count number of occurrences of
// each element in the array in O(n) time and O(1) space
 
#include <bits/stdc++.h>
using namespace std;
 
void findFrequencies(int ele[], int n)
{
    int freq = 1;
    int idx = 1;
    int element = ele[0];
    while (idx < n) {
 
        // check if the current element is equal to
        // previous element.
        if (ele[idx - 1] == ele[idx]) {
            freq++;
            idx++;
        }
        else {
            cout << element << " " << freq << endl;
            element = ele[idx];
            idx++;
 
            // reset the frequency
            freq = 1;
        }
    }
 
    // print the last element and its frequency
    cout << element << " " << freq;
}
 
// Driver code
int main()
{
    cout << "---frequencies in a sorted array----" << endl;
    int arr[]
        = { 10, 20, 30, 30, 30, 40, 50, 50, 50, 50, 70 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    findFrequencies(arr, n);
}
 
// This code is contributed by anushkaseehh

Java




// Java program to count number of occurrences of
// each element in the array in O(n) time and O(1) space
 
import java.io.*;
import java.util.*;
class GFG {
    public static void findFrequencies(int[] ele, int n)
    {
        int freq = 1;
        int idx = 1;
        int element = ele[0];
        while (idx < n) {
            // check if the current element is equal to
            // previous element.
            if (ele[idx - 1] == ele[idx]) {
                freq++;
                idx++;
            }
            else {
                System.out.println(element + " " + freq);
                element = ele[idx];
                idx++;
 
                // reset the frequency
                freq = 1;
            }
        }
 
        // print the last element and its frequency
        System.out.println(element + " " + freq);
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        System.out.println(
            "---frequencies in a sorted array----");
        int[] arr = { 10, 20, 30, 30, 30, 40,
                      50, 50, 50, 50, 70 };
        int n = arr.length;
 
        // Function call
        findFrequencies(arr, n);
    }
}
 
// This code is contributed by Pushpesh raj.

Python3




# python3 program to count number of occurrences of
# each element in the array in O(n) time and O(1) space
 
 
def findFrequencies(ele, n):
 
    freq = 1
    idx = 1
    element = ele[0]
    while (idx < n):
 
        # check if the current element is equal to
        # previous element.
        if (ele[idx - 1] == ele[idx]):
            freq += 1
            idx += 1
 
        else:
            print(element, " ", freq)
            element = ele[idx]
            idx += 1
 
            # reset the frequency
            freq = 1
 
    # print the last element and its frequency
    print(element, " ", freq)
 
 
# Driver code
if __name__ == "__main__":
    print("---frequencies in a sorted array----")
    arr = [10, 20, 30, 30, 30, 40, 50, 50, 50, 50, 70]
    n = len(arr)
 
    # Function call
    findFrequencies(arr, n)
 
# This code is contributed by shivanisinghss2110

C#




// C# program to count number of occurrences of
// each element in the array in O(n) time and O(1) space
using System;
class GFG {
 
    public static void findFrequencies(int[] ele)
    {
        int freq = 1;
        int idx = 1;
        int element = ele[0];
        while (idx < ele.Length) {
 
            // check if the current element is equal to
            // previous element.
            if (ele[idx - 1] == ele[idx]) {
                freq++;
                idx++;
            }
            else {
                Console.WriteLine(element + " " + freq);
                element = ele[idx];
                idx++;
 
                // reset the frequency
                freq = 1;
            }
        }
 
        // print the last element and its frequency
        Console.WriteLine(element + " " + freq);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        Console.WriteLine(
            "---frequencies in a sorted array----");
 
        // Function call
        findFrequencies(new int[] { 10, 20, 30, 30, 30, 40,
                                    50, 50, 50, 50, 70 });
    }
}

Javascript




<script>
// JavaScript program to count number of occurrences of
// each element in the array in O(n) time and O(1) space
    function void findFrequencies(ele)
    {
        var freq = 1;
        var idx = 1;
        var element = ele[0];
        while (idx < ele.length)
        {
         
            // check if the current element is equal to
            // previous element.
            if (ele[idx - 1] == ele[idx]) {
                freq++;
                idx++;
            }
            else
            {
                document.write(element + " " + freq);
                element = ele[idx];
                idx++;
                 
                // reset the frequency
                freq = 1;
            }
        }
         
        // print the last element and its frequency
        document.write(element + " " + freq);
    }
     
    // Driver code
        document.write(
            "---frequencies in a sorted array----");
        findFrequencies(new var[] { 10, 20, 30, 30, 30, 40,
                                    50, 50, 50, 50, 70 });
     
// This code is contributed by shivanisinghss2110
</script>

Output

---frequencies in a sorted array----
10 1
20 1
30 3
40 1
50 4
70 1

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

Frequency of each element in a limited range array using Hash-Map:

To solve the problem follow the below idea:

The idea is to traverse the input array and for each distinct element of the array, store its frequency in a HashMap, and finally print the HashMap.

Follow the given steps to solve the problem: 

  • Create a HashMap to map the frequency to the element, i.e to store the element-frequency pair.
  • Traverse the array from start to end.
  • For each element in the array update the frequency, i.e hm[array[i]]++
  • Traverse the HashMap and print the element frequency pair

Below is the implementation of the above approach:

C++




// C++ program to count number of occurrences of
// each element in the array #include <iostream>
#include <bits/stdc++.h>
using namespace std;
 
// It prints number of
// occurrences of each element in the array.
void findFrequency(int arr[], int n)
{
    // HashMap to store frequencies
    unordered_map<int, int> mp;
 
    // traverse the array
    for (int i = 0; i < n; i++) {
        // update the frequency
        mp[arr[i]]++;
    }
 
    // traverse the hashmap
    for (auto i : mp) {
        cout << "Element " << i.first << " occurs "
             << i.second << " times" << endl;
    }
}
 
// Driver code
int main()
{
    int arr[]
        = { 1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    findFrequency(arr, n);
 
    return 0;
}

Java




// Java program to count number
// of occurrences of each
// element in the array
import java.io.*;
import java.util.*;
class GFG {
 
    // It prints number of
    // occurrences of each
    // element in the array.
    static void findFrequency(int[] arr, int n)
    {
        Map<Integer, Integer> mp
            = new HashMap<Integer, Integer>();
 
        // traverse the array
        for (int i = 0; i < n; i++) {
 
            // update the frequency
            if (!mp.containsKey(arr[i]))
                mp.put(arr[i], 0);
 
            mp.put(arr[i], mp.get(arr[i]) + 1);
        }
 
        // traverse the hashmap
        for (Map.Entry<Integer, Integer> kvp :
             mp.entrySet()) {
            System.out.println("Element " + kvp.getKey()
                               + " occurs " + kvp.getValue()
                               + " times");
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        int[] arr
            = { 1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10 };
        int n = arr.length;
 
        // Function call
        findFrequency(arr, n);
    }
}
 
// This code is contributed by avanitrachhadiya2155

Python3




# Python program to count number of occurrences of
# each element in the array #include <iostream>
 
# It prints number of
# occurrences of each element in the array.
 
 
def findFrequency(arr, n):
 
    # HashMap to store frequencies
    mp = {}
 
    # traverse the array
    for i in range(n):
 
        # update the frequency
        if arr[i] not in mp:
            mp[arr[i]] = 0
        mp[arr[i]] += 1
 
    # traverse the hashmap
    for i in mp:
        print("Element", i, "occurs", mp[i], "times")
 
 
# Driver function
if __name__ == "__main__":
    arr = [1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10]
    n = len(arr)
 
    findFrequency(arr, n)
 
# This code is contributed by shubhamsingh10

C#




// C# program to count number
// of occurrences of each
// element in the array
using System;
using System.Collections.Generic;
class GFG {
 
    // It prints number of
    // occurrences of each
    // element in the array.
    static void findFrequency(int[] arr, int n)
    {
        // HashMap to store frequencies
        Dictionary<int, int> mp
            = new Dictionary<int, int>();
 
        // traverse the array
        for (int i = 0; i < n; i++) {
            // update the frequency
            if (!mp.ContainsKey(arr[i]))
                mp[arr[i]] = 0;
 
            mp[arr[i]]++;
        }
 
        // traverse the hashmap
        foreach(KeyValuePair<int, int> kvp in mp)
            Console.WriteLine("Element " + kvp.Key
                              + " occurs " + kvp.Value
                              + " times");
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr
            = { 1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10 };
        int n = arr.Length;
 
        // Function call
        findFrequency(arr, n);
    }
}
 
// This code is contributed by Chitranayal

Javascript




<script>
 
// Javascript program to count number
// of occurrences of each
// element in the array
 
// It prints number of
// occurrences of each
// element in the array.
function findFrequency(arr, n)
{
    let mp = new Map();
  
    // Traverse the array
    for(let i = 0; i < n; i++)
    {
  
        // Update the frequency
        if (!mp.has(arr[i]))
            mp.set(arr[i],0);
         
        mp.set(arr[i], mp.get(arr[i]) + 1);
    }
  
    // Traverse the hashmap
    for(let [key, value] of mp.entries())
    {
        document.write("Element " + key +
                       " occurs " + value +
                       " times<br>");
    }
}
 
// Driver code
let arr = [ 1, 1, 1, 2, 3, 3, 5,
            5, 8, 8, 8, 9, 9, 10 ];
let n = arr.length;
 
findFrequency(arr, n);
 
// This code is contributed by patel2127
 
</script>

Output

Element 10 occurs 1 times
Element 2 occurs 1 times
Element 9 occurs 2 times
Element 1 occurs 3 times
Element 8 occurs 3 times
Element 3 occurs 2 times
Element 5 occurs 2 times

Time Complexity: O(N), only one traversal of the array is needed.
Auxiliary Space: O(N), to store the elements in the HashMap O(N) extra space is needed.

Frequency of each element in a limited range array using binary search:

The problem can be solved in less than O(n) time if all its elements are sorted, i.e. if similar elements exist in the array then the elements are in a contiguous subarray or it can be said that if the ends of a subarray are the same then all the elements inside the subarray are equal. So the count of that element is the size of the subarray and all the elements of that subarray need not be counted. 
 

Follow the given steps to solve the problem: 

  • Create a HashMap (hm) to store the frequency of elements.
  • Create a recursive function that accepts an array and size.
  • Check if the first element of the array is equal to the last element. If equal then all the elements are the same and update the frequency by hm[array[0]+=size
  • Else divide the array into two equal halves and call the function recursively for both halves.
  • Traverse the hashmap and print the element frequency pair.

Below is the implementation of the above approach:

C++




// C++ program to count number of occurrences of
// each element in the array in less than O(n) time
#include <bits/stdc++.h>
using namespace std;
 
// A recursive function to count number of occurrences
// for each element in the array without traversing
// the whole array
void findFrequencyUtil(int arr[], int low, int high,
                       vector<int>& freq)
{
    // If element at index low is equal to element
    // at index high in the array
    if (arr[low] == arr[high]) {
        // increment the frequency of the element
        // by count of elements between high and low
        freq[arr[low]] += high - low + 1;
    }
    else {
        // Find mid and recurse for left and right
        // subarray
        int mid = (low + high) / 2;
        findFrequencyUtil(arr, low, mid, freq);
        findFrequencyUtil(arr, mid + 1, high, freq);
    }
}
 
// A wrapper over recursive function
// findFrequencyUtil(). It print number of
// occurrences of each element in the array.
void findFrequency(int arr[], int n)
{
    // create a empty vector to store frequencies
    // and initialize it by 0. Size of vector is
    // maximum value (which is last value in sorted
    // array) plus 1.
    vector<int> freq(arr[n - 1] + 1, 0);
 
    // Fill the vector with frequency
    findFrequencyUtil(arr, 0, n - 1, freq);
 
    // Print the frequencies
    for (int i = 0; i <= arr[n - 1]; i++)
        if (freq[i] != 0)
            cout << "Element " << i << " occurs " << freq[i]
                 << " times" << endl;
}
 
// Driver code
int main()
{
    int arr[]
        = { 1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    findFrequency(arr, n);
 
    return 0;
}

Java




// Java program to count number of occurrences of
// each element in the array in less than O(n) time
import java.util.*;
 
class GFG {
 
    // A recursive function to count number of occurrences
    // for each element in the array without traversing
    // the whole array
    static void findFrequencyUtil(int arr[], int low,
                                  int high, int[] freq)
    {
        // If element at index low is equal to element
        // at index high in the array
        if (arr[low] == arr[high]) {
            // increment the frequency of the element
            // by count of elements between high and low
            freq[arr[low]] += high - low + 1;
        }
        else {
            // Find mid and recurse for left and right
            // subarray
            int mid = (low + high) / 2;
            findFrequencyUtil(arr, low, mid, freq);
            findFrequencyUtil(arr, mid + 1, high, freq);
        }
    }
 
    // A wrapper over recursive function
    // findFrequencyUtil(). It print number of
    // occurrences of each element in the array.
    static void findFrequency(int arr[], int n)
    {
        // create a empty vector to store frequencies
        // and initialize it by 0. Size of vector is
        // maximum value (which is last value in sorted
        // array) plus 1.
        int[] freq = new int[arr[n - 1] + 1];
 
        // Fill the vector with frequency
        findFrequencyUtil(arr, 0, n - 1, freq);
 
        // Print the frequencies
        for (int i = 0; i <= arr[n - 1]; i++)
            if (freq[i] != 0)
                System.out.println("Element " + i
                                   + " occurs " + freq[i]
                                   + " times");
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[]
            = { 1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10 };
        int n = arr.length;
 
        // Function call
        findFrequency(arr, n);
    }
}
 
// This code is contributed by 29AjayKumar

Python3




# Python 3 program to count number of occurrences of
# each element in the array in less than O(n) time
 
# A recursive function to count number of occurrences
# for each element in the array without traversing
# the whole array
 
 
def findFrequencyUtil(arr, low, high, freq):
 
    # If element at index low is equal to element
    # at index high in the array
    if (arr[low] == arr[high]):
 
        # increment the frequency of the element
        # by count of elements between high and low
        freq[arr[low]] += high - low + 1
 
    else:
 
        # Find mid and recurse for left
        # and right subarray
        mid = int((low + high) / 2)
        findFrequencyUtil(arr, low, mid, freq)
        findFrequencyUtil(arr, mid + 1, high, freq)
 
# A wrapper over recursive function
# findFrequencyUtil(). It print number of
# occurrences of each element in the array.
 
 
def findFrequency(arr, n):
 
    # create a empty vector to store frequencies
    # and initialize it by 0. Size of vector is
    # maximum value (which is last value in sorted
    # array) plus 1.
    freq = [0 for i in range(n - 1 + 1)]
 
    # Fill the vector with frequency
    findFrequencyUtil(arr, 0, n - 1, freq)
 
    # Print the frequencies
    for i in range(0, arr[n - 1] + 1, 1):
        if (freq[i] != 0):
            print("Element", i, "occurs",
                  freq[i], "times")
 
 
# Driver Code
if __name__ == '__main__':
    arr = [1, 1, 1, 2, 3, 3, 5,
           5, 8, 8, 8, 9, 9, 10]
    n = len(arr)
 
    # Function call
    findFrequency(arr, n)
 
# This code is contributed by
# Surendra_Gangwar

C#




// C# program to count number of occurrences of
// each element in the array in less than O(n) time
using System;
 
class GFG {
 
    // A recursive function to count number of occurrences
    // for each element in the array without traversing
    // the whole array
    static void findFrequencyUtil(int[] arr, int low,
                                  int high, int[] freq)
    {
        // If element at index low is equal to element
        // at index high in the array
        if (arr[low] == arr[high]) {
            // increment the frequency of the element
            // by count of elements between high and low
            freq[arr[low]] += high - low + 1;
        }
        else {
            // Find mid and recurse for left and right
            // subarray
            int mid = (low + high) / 2;
            findFrequencyUtil(arr, low, mid, freq);
            findFrequencyUtil(arr, mid + 1, high, freq);
        }
    }
 
    // A wrapper over recursive function
    // findFrequencyUtil(). It print number of
    // occurrences of each element in the array.
    static void findFrequency(int[] arr, int n)
    {
        // create a empty vector to store frequencies
        // and initialize it by 0. Size of vector is
        // maximum value (which is last value in sorted
        // array) plus 1.
        int[] freq = new int[arr[n - 1] + 1];
 
        // Fill the vector with frequency
        findFrequencyUtil(arr, 0, n - 1, freq);
 
        // Print the frequencies
        for (int i = 0; i <= arr[n - 1]; i++)
            if (freq[i] != 0)
                Console.WriteLine("Element " + i
                                  + " occurs " + freq[i]
                                  + " times");
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int[] arr
            = { 1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10 };
        int n = arr.Length;
 
        // Function call
        findFrequency(arr, n);
    }
}
 
// This code is contributed by Princi Singh

Javascript




<script>
// Javascript program to count number of occurrences of
// each element in the array in less than O(n) time
     
    // A recursive function to count number of occurrences
    // for each element in the array without traversing
    // the whole array
    function findFrequencyUtil(arr, low, high, freq)
    {
     
        // If element at index low is equal to element
        // at index high in the array
        if (arr[low] == arr[high])
        {
         
            // increment the frequency of the element
            // by count of elements between high and low
            freq[arr[low]] += high - low + 1;
        }
        else
        {
         
            // Find mid and recurse for left and right
            // subarray
            let mid = Math.floor((low + high) / 2);
            findFrequencyUtil(arr, low, mid, freq);
            findFrequencyUtil(arr, mid + 1, high, freq);
        }
    }
     
    // A wrapper over recursive function
    // findFrequencyUtil(). It print number of
    // occurrences of each element in the array.
    function findFrequency(arr, n)
    {
     
        // create a empty vector to store frequencies
        // and initialize it by 0. Size of vector is
        // maximum value (which is last value in sorted
        // array) plus 1.
        let freq = new Array(arr[n - 1] + 1);
         
        for(let i = 0; i < arr[n - 1] + 1; i++)
        {
            freq[i] = 0;
        }
         
        // Fill the vector with frequency
        findFrequencyUtil(arr, 0, n - 1, freq);
  
        // Print the frequencies
        for (let i = 0; i <= arr[n - 1]; i++)
            if (freq[i] != 0)
                document.write("Element " + i + " occurs " + freq[i] + " times<br>");
    }
     
    // Driver Code
    let arr = [1, 1, 1, 2, 3, 3, 5,
                      5, 8, 8, 8, 9, 9, 10 ];
    let n = arr.length;
    findFrequency(arr, n);
     
    // This code is contributed by rag2127.
</script>

Output

Element 1 occurs 3 times
Element 2 occurs 1 times
Element 3 occurs 2 times
Element 5 occurs 2 times
Element 8 occurs 3 times
Element 9 occurs 2 times
Element 10 occurs 1 times

Time Complexity: O(m log N). Where m is the number of distinct elements in the array of size N. Since m <= M (a constant) (elements are in a limited range), the time complexity of this solution is O(log N)
Auxiliary Space: O(N). To store the elements in the HashMap O(n) extra space is needed.

Frequency of each element in a limited range array using the input array as a Hash-Map:

In this method, we use the same array as the hash map by modifying its content:

Dry run of this approach:

Input: arr = { 1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10 };

Step 1: Subtract 1 from each element of the array
arr  = {0 ,0 ,0 ,1 ,2 ,2 ,4 ,4 ,7 ,7 ,7 ,8 ,8 ,9 }

Step 2: Add n to the index at which the current array element points.
for example :- 
when i=0, arr[arr[0]%n] = 0 adding n to the arr[0], arr[0] =  14;
when i=1, arr[arr[1]%n] = 14 adding n to arr[0] ,arr[0] = 28;
Similarly finding the modified array in the same way we will get array as  
arr = {42 ,14 ,28 ,1 ,30, 2, 4, 46, 35, 21, 7, 8, 8, 9}

Step 3: Now in step 2 if you have noticed we added the n value to the index at which a particular element points to. So if we have more than one time have a element that point to the same index then in that case the division of the modified number with the n gives us the frequency of the number. 
for example 
at i=0; arr[0] =42; arr[0] / n = 3 it means that 0  appeared three times in the modified array as you can see in the arr of step 1.
at i=1; arr[1] =14; arr[1]/14 = 1 it means that 1 appeared once in the modified array as you can see in the arr of step 1 .
and similarly for other values we can calculate.

Below is the implementation of the above approach:

C++




// C++ program to count number of occurrences of
// each element in the array #include <iostream>
#include <bits/stdc++.h>
using namespace std;
 
// It prints number of occurrences of each element in the
// array.
void findFrequency(int input[], int n)
{
    for (int i = 0; i < n; i++)
        input[i]--;
 
    for (int i = 0; i < n; i++)
        input[input[i] % n] += n;
 
    for (int i = 0; i < n; i++) {
        if (input[i] / n)
            cout << "Element " << (i + 1) << " occurs "
                 << input[i] / n << " times" << endl;
        // Change the element back to original value
        input[i] = input[i] % n + 1;
    }
}
 
// Driver code
int main()
{
    int arr[]
        = { 1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    findFrequency(arr, n);
 
    return 0;
}
 
// This code is contributed by aditya kumar(adiyakumar129)

Java




// Java program to count number of occurrences of each
// element in the array
import java.io.*;
import java.util.*;
class GFG {
 
    // It prints number of occurrences of each element in
    // the array.
    static void findFrequency(int[] input, int n)
    {
        for (int i = 0; i < n; i++)
            input[i]--;
 
        for (int i = 0; i < n; i++)
            input[input[i] % n] += n;
 
        for (int i = 0; i < n; i++) {
            if ((input[i] / n) != 0)
                System.out.println(
                    "Element " + (i + 1) + " occurs "
                    + input[i] / n + " times");
            // Change the element back to original value
            input[i] = input[i] % n + 1;
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int[] arr
            = { 1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10 };
        int n = arr.length;
 
        // Function call
        findFrequency(arr, n);
    }
}
 
// This code is contributed by aditya kumar(adiyakumar129)

Python3




# Javascript program to count number of occurrences of
# each element in the array
 
# It prints number of
# occurrences of each element in the array.
 
 
def findFrequency(input, n):
    for i in range(n):
        input[i] -= 1
 
    for i in range(n):
        input[input[i] % n] += n
 
    for i in range(n):
        if input[i] // n:
            print("Element", i + 1, "occurs", input[i] // n, "times")
 
        # change element back to original value
        input[i] = input[i] % n + 1
 
 
# Driver code
if __name__ == "__main__":
    arr = [1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10]
    n = len(arr)
 
    # Function call
    findFrequency(arr, n)
 
# This code is contributed by phasing17

C#




// C# program to count number of occurrences of each element
// in the array
using System;
 
public class GFG {
 
    // It prints number of occurrences of each element in
    // the array.
    static void findFrequency(int[] input, int n)
    {
        for (int i = 0; i < n; i++)
            input[i]--;
 
        for (int i = 0; i < n; i++)
            input[input[i] % n] += n;
 
        for (int i = 0; i < n; i++) {
            if ((input[i] / n) != 0)
                Console.WriteLine(
                    "Element " + (i + 1) + " occurs "
                    + input[i] / n + " times");
            // Change the element back to original value
            input[i] = input[i] % n + 1;
        }
    }
 
    // Driver function
    public static void Main(String[] args)
    {
        int[] arr
            = { 1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10 };
        int n = arr.Length;
 
        // Function call
        findFrequency(arr, n);
    }
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
// Javascript program to count number of occurrences of
// each element in the array
 
// It prints number of
// occurrences of each element in the array.
function findFrequency(input, n)
{
    for (let i = 0; i < n; i++)
        input[i]--;
 
    for (let i = 0; i < n; i++)
        input[input[i] % n] += n;
 
    console.log(input)
    for (let i = 0; i < n; i++) {
        if (Math.floor(input[i] / n))
            document.write("Element " + (i + 1) +
            " occurs " + Math.floor(input[i] / n) + " times <br>");
             
        // Change the element back to original value
        input[i] = input[i] % n + 1;
    }
}
 
// Driver function
let arr = [1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10];
let n = arr.length;
 
findFrequency(arr, n);
 
// This code is contributed by Saurabh Jaiswal
</script>

Output

Element 1 occurs 3 times
Element 2 occurs 1 times
Element 3 occurs 2 times
Element 5 occurs 2 times
Element 8 occurs 3 times
Element 9 occurs 2 times
Element 10 occurs 1 times

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

https://youtu.be/B2hI-QPoisk 
This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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