First element occurring k times in an array

Given an array of n integers. The task is to find the first element that occurs k number of times. If no element occurs k times the print -1. The distribution of integer elements could be in any range.
Examples:

Input: {1, 7, 4, 3, 4, 8, 7}, k = 2 
Output: 7 
Explanation: Both 7 and 4 occur 2 times. But 7 is the first that occurs 2 times. 

Input: {4, 1, 6, 1, 6, 4}, k = 1 
Output: -1

Naive Approach: The idea is to use two nested loops. one for the selection of element and other for counting the number of time the selected element occurs in the given array.

Below is the implementation of the above approach:

C++

// C++ implementation to find first
// element occurring k times
#include <bits/stdc++.h>
using namespace std;

// Function to find the first element
// occurring k number of times
int firstElement(int arr[], int n, int k)
{
    // This loop is used for selection
    //  of elements
    for (int i = 0; i < n; i++) {
        // Count how many time selected element
        // occurs
        int count = 0;
        for (int j = 0; j < n; j++) {
            if (arr[i] == arr[j])
                count++;
        }

        // Check, if it occurs k times or not
        if (count == k)
            return arr[i];
    }

    return -1;
}

// Driver Code
int main()
{
    int arr[] = { 1, 7, 4, 3, 4, 8, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 2;
    cout << firstElement(arr, n, k);
    return 0;
}

Java

public class GFG {
    // Java implementation to find first
    // element occurring k times

    // Function to find the first element
    // occurring k number of times
    public static int firstElement(int[] arr, int n, int k)
    {
        // This loop is used for selection
        // of elements
        for (int i = 0; i < n; i++) {
            // Count how many time selected element
            // occurs
            int count = 0;
            for (int j = 0; j < n; j++) {
                if (arr[i] == arr[j]) {
                    count++;
                }
            }

            // Check, if it occurs k times or not
            if (count == k) {
                return arr[i];
            }
        }

        return -1;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 1, 7, 4, 3, 4, 8, 7 };
        int n = arr.length;
        int k = 2;
        System.out.print(firstElement(arr, n, k));
    }
}

// This code is contributed by Aarti_Rathi

Python3

# Python3 implementation to
# find first element
# occurring k times

# function to find the
# first element occurring
# k number of times
def firstElement(arr, n, k):

    # dictionary to count
    # occurrences of
    # each element
    for i in arr:
      count=0
      for j in arr:
        if i==j:
          count=count+1
      if count == k:
        return i

    # no element occurs k times
    return -1

# Driver Code
if __name__=="__main__":

    arr = [1, 7, 4, 3, 4, 8, 7];
    n = len(arr)
    k = 2
    print(firstElement(arr, n, k))

# This code is contributed by Arpit Jain

C#

// C# implementation to find first
// element occurring k times
using System;

public class GFG {

    // Function to find the first element
    // occurring k number of times
    public static int firstElement(int[] arr, int n, int k)
    {
        // This loop is used for selection
        // of elements
        for (int i = 0; i < n; i++) {
            // Count how many time selected element
            // occurs
            int count = 0;
            for (int j = 0; j < n; j++) {
                if (arr[i] == arr[j]) {
                    count++;
                }
            }

            // Check, if it occurs k times or not
            if (count == k) {
                return arr[i];
            }
        }

        return -1;
    }

    // Driver Code
    public static void Main(String[] args)
    {
        int[] arr = { 1, 7, 4, 3, 4, 8, 7 };
        int n = arr.Length;
        int k = 2;
        Console.Write(firstElement(arr, n, k));
    }
}

// This code is contributed by Abhijeet Kumar(abhijeet19403)

Javascript

class GFG
{
    // javascript implementation to find first
    // element occurring k times
    // Function to find the first element
    // occurring k number of times
    static firstElement(arr, n, k)
    {

        // This loop is used for selection
        // of elements
        for (var i = 0; i < n; i++)
        {

            // Count how many time selected element
            // occurs
            var count = 0;
            for (var j=0; j < n; j++)
            {
                if (arr[i] == arr[j])
                {
                    count++;
                }
            }

            // Check, if it occurs k times or not
            if (count == k)
            {
                return arr[i];
            }
        }
        return -1;
    }

    // Driver Code
    static main(args)
    {
        var arr = [1, 7, 4, 3, 4, 8, 7];
        var n = arr.length;
        var k = 2;
        console.log(GFG.firstElement(arr, n, k));
    }
}
GFG.main([]);

// This code is contributed by aadityaburujwale.

Output

7

Time complexity: O(n²).
Auxiliary space: O(1) as it is using constant space for variables

Efficient Approach: Use unordered_map for hashing as the range is not known. Steps:  

1. Traverse the array of elements from left to right.
2. While traversing increment their count in the hash table.
3. Again traverse the array from left to right and check which element has a count equal to k. Print that element and stop.
4. If no element has a count equal to k, print -1.

Below is a dry run of the above approach: 

Below is the implementation of the above approach: 

C++

// C++ implementation to find first
// element occurring k times
#include <bits/stdc++.h>

using namespace std;

// function to find the first element
// occurring k number of times
int firstElement(int arr[], int n, int k)
{
    // unordered_map to count
    // occurrences of each element
    unordered_map<int, int> count_map;
    for (int i=0; i<n; i++)
        count_map[arr[i]]++;

    for (int i=0; i<n; i++)

        // if count of element == k ,then
        // it is the required first element
        if (count_map[arr[i]] == k)
            return arr[i];

    // no element occurs k times
    return -1;
}

// Driver program to test above
int main()
{
    int arr[] = {1, 7, 4, 3, 4, 8, 7};
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 2;
    cout << firstElement(arr, n, k);
    return 0;
}

Java

import java.util.HashMap;

// Java implementation to find first
// element occurring k times
class GFG {

// function to find the first element
// occurring k number of times
    static int firstElement(int arr[], int n, int k) {
        // unordered_map to count
        // occurrences of each element

        HashMap<Integer, Integer> count_map = new HashMap<>();
        for (int i = 0; i < n; i++) {
            int a = 0;
            if(count_map.get(arr[i])!=null){
                a = count_map.get(arr[i]);
            }

            count_map.put(arr[i], a+1);
        }
        //count_map[arr[i]]++;

        for (int i = 0; i < n; i++) // if count of element == k ,then
        // it is the required first element
        {
            if (count_map.get(arr[i]) == k) {
                return arr[i];
            }
        }

        // no element occurs k times
        return -1;
    }

// Driver program to test above
    public static void main(String[] args) {
        int arr[] = {1, 7, 4, 3, 4, 8, 7};
        int n = arr.length;
        int k = 2;
        System.out.println(firstElement(arr, n, k));
    }
}

//this code contributed by Rajput-Ji

Python3

# Python3 implementation to
# find first element
# occurring k times

# function to find the
# first element occurring
# k number of times
def firstElement(arr, n, k):

    # dictionary to count
    # occurrences of
    # each element
    count_map = {};
    for i in range(0, n):
        if(arr[i] in count_map.keys()):
            count_map[arr[i]] += 1
        else:
            count_map[arr[i]] = 1
        i += 1

    for i in range(0, n):

        # if count of element == k ,
        # then it is the required
        # first element
        if (count_map[arr[i]] == k):
            return arr[i]
        i += 1

    # no element occurs k times
    return -1

# Driver Code
if __name__=="__main__":

    arr = [1, 7, 4, 3, 4, 8, 7];
    n = len(arr)
    k = 2
    print(firstElement(arr, n, k))

# This code is contributed
# by Abhishek Sharma

C#

// C# implementation to find first
// element occurring k times
using System;
using System.Collections.Generic;

class GFG
{

    // function to find the first element
    // occurring k number of times
    static int firstElement(int []arr, int n, int k)
    {
        // unordered_map to count
        // occurrences of each element

        Dictionary<int, int> count_map = new Dictionary<int,int>();
        for (int i = 0; i < n; i++)
        {
            int a = 0;
            if(count_map.ContainsKey(arr[i]))
            {
                a = count_map[arr[i]];
                count_map.Remove(arr[i]);
                count_map.Add(arr[i], a+1);
            }
            else
                count_map.Add(arr[i], 1);
        }
        //count_map[arr[i]]++;

        for (int i = 0; i < n; i++) // if count of element == k ,then
        // it is the required first element
        {
            if (count_map[arr[i]] == k)
            {
                return arr[i];
            }
        }

        // no element occurs k times
        return -1;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {1, 7, 4, 3, 4, 8, 7};
        int n = arr.Length;
        int k = 2;
        Console.WriteLine(firstElement(arr, n, k));
    }
}

// This code has been contributed by 29AjayKumar

Javascript

<script>

// JavaScript implementation to find first
// element occurring k times

// function to find the first element
// occurring k number of times
function firstElement(arr, n, k)
{
    // unordered_map to count
    // occurrences of each element
    count_map = new Map()
    for (let i=0; i<n; i++)
        count_map[arr[i]] = 0;
    for (let i=0; i<n; i++)
        count_map[arr[i]]++;

    for (let i=0; i<n; i++)

        // if count of element == k ,then
        // it is the required first element
        if (count_map[arr[i]] == k)
            return arr[i];

    // no element occurs k times
    return -1;
}

// Driver program to test above

let arr = [1, 7, 4, 3, 4, 8, 7];
let n = arr.length;
let k = 2;
document.write(firstElement(arr, n, k));

<script>

Output

7

Time Complexity: O(n)
Auxiliary Space: O(n) because we are using an auxiliary array of size n to store the count

Method 3:Using Built-in Python functions:

• Count the frequencies of every element using Counter function
• Traverse in frequency dictionary
• Checks which element has a count equal to k. Print that element and stop.
• If no element has a count equal to k, print -1.

C++

// C++ implementation to find first
// element occurring k times
#include <bits/stdc++.h>
using namespace std;

// function to find the first element
// occurring k number of times
int firstElement(int arr[], int n, int k)
{

  // unordered_map to count
  // occurrences of each element
  map<int, int> count_map;
  for (int i = 0; i < n; i++) {

    count_map[arr[i]]++;
  }
  // count_map[arr[i]]++;

  for (int i = 0; i < n;
       i++) // if count of element == k ,then
    // it is the required first element
  {
    if (count_map.find(arr[i]) != count_map.end()) {
      if (count_map[arr[i]] == k)
        return arr[i];
    }
  }

  // no element occurs k times
  return -1;
}

// Driver code
int main()
{
  int arr[] = { 1, 7, 4, 3, 4, 8, 7 };
  int n = sizeof(arr) / sizeof(arr[0]);
  int k = 2;
  cout << (firstElement(arr, n, k));
  return 0;
}

// This code is contributed by Rajput-Ji

Java

// java implementation to find first
// element occurring k times
import java.util.*;
class GFG
{

  // function to find the first element
  // occurring k number of times
  static int firstElement(int []arr, int n, int k)
  {

    // unordered_map to count
    // occurrences of each element
    HashMap<Integer,Integer> count_map = new HashMap<>();
    for (int i = 0; i < n; i++)
    {
      if(count_map.containsKey(arr[i]))
      {

        count_map.put(arr[i], count_map.get(arr[i])+1);
      }
      else
        count_map.put(arr[i], 1);
    }
    //count_map[arr[i]]++;

    for (int i = 0; i < n; i++) // if count of element == k ,then
      // it is the required first element
    {
      if (count_map.containsKey(arr[i]) )
      {
        if(count_map.get(arr[i]) == k)
          return arr[i];
      }
    }

    // no element occurs k times
    return -1;
  }

  // Driver code
  public static void main(String[] args)
  {
    int []arr = {1, 7, 4, 3, 4, 8, 7};
    int n = arr.length;
    int k = 2;
    System.out.print(firstElement(arr, n, k));
  }
}

// This code contributed by Rajput-Ji

Python3

# importing counter from collections
from collections import Counter

# Python3 implementation to find
# first element occurring k times
# function to find the  first element
# occurring  k number of times
def firstElement(arr, n, k):

    # calculating frequencies using Counter
    count_map = Counter(arr)

    for i in range(0, n):

        # If count of element == k ,
        # then it is the required
        # first element
        if (count_map[arr[i]] == k):
            return arr[i]
        i += 1

    # No element occurs k times
    return -1


# Driver Code
if __name__ == "__main__":

    arr = [1, 7, 4, 3, 4, 8, 7]
    n = len(arr)
    k = 2
    print(firstElement(arr, n, k))

# This code is contributed by vikkycirus

C#

// C# implementation to find first
// element occurring k times
using System;
using System.Collections.Generic;

class GFG
{

    // function to find the first element
    // occurring k number of times
    static int firstElement(int []arr, int n, int k)
    {
        // unordered_map to count
        // occurrences of each element

        Dictionary<int, int> count_map = new Dictionary<int,int>();
        for (int i = 0; i < n; i++)
        {
            int a = 0;
            if(count_map.ContainsKey(arr[i]))
            {
                a = count_map[arr[i]];
                count_map.Remove(arr[i]);
                count_map.Add(arr[i], a+1);
            }
            else
                count_map.Add(arr[i], 1);
        }
        //count_map[arr[i]]++;

        for (int i = 0; i < n; i++) // if count of element == k ,then
        // it is the required first element
        {
            if (count_map[arr[i]] == k)
            {
                return arr[i];
            }
        }

        // no element occurs k times
        return -1;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {1, 7, 4, 3, 4, 8, 7};
        int n = arr.Length;
        int k = 2;
        Console.WriteLine(firstElement(arr, n, k));
    }
}

// This code is contributed by avijitmondal1998.

Javascript

<script>
// javascript implementation to find first
// element occurring k times

    // function to find the first element
    // occurring k number of times
    function firstElement(arr , n , k) {

        // unordered_map to count
        // occurrences of each element
        var count_map = new Map();
        for (i = 0; i < n; i++) {
            if (count_map.has(arr[i])) {

                count_map.set(arr[i], count_map.get(arr[i]) + 1);
            } else
                count_map.set(arr[i], 1);
        }
        // count_map[arr[i]]++;

        for (i = 0; i < n; i++) // if count of element == k ,then
        // it is the required first element
        {
            if (count_map.has(arr[i])) {
                if (count_map.get(arr[i]) == k)
                    return arr[i];
            }
        }

        // no element occurs k times
        return -1;
    }

    // Driver code
        var arr = [ 1, 7, 4, 3, 4, 8, 7 ];
        var n = arr.length;
        var k = 2;
        document.write(firstElement(arr, n, k));

// This code is contributed by Rajput-Ji
</script>

Output

7

Time Complexity: O(n*log(n))
Auxiliary Space: O(n)

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