First element occurring k times in an array

Difficulty Level : Easy
Last Updated : 16 Aug, 2021

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
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

Simple Approach: By using two loops, count the number of times a number appears in the array.
Time complexity: O(n²).
Efficient Approach: Use unordered_map for hashing as range is not known. Steps:

Traverse the array elements from left to right.
While traversing increment their count in the hash table.
Again traverse the array from left to right and check which element has a count equal to k. Print that element and stop.
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:

Time Complexity: O(n)
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Method 3:Using Built-in Python functions:

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

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 usinf 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

Output:

Time Complexity: O(N)

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