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
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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Simple Approach: By using two loops. It has a time complexity of 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 the implementation of 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  

Output:

Time Complexity: O(n)

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