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.
Note : If multiple element occurs K number of time. then the element which occurs first will be our answer.
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.
Algorithm:
- Define a function firstElement that takes an integer array arr, an integer n representing the size of the array, and an integer k representing the number of times an element must occur in the array.
- Iterate through each element in the array arr using a for loop with index i.
- For each element arr[i], count how many times it occurs in the array by iterating through the array again using another for loop with index j.
- If the count of the current element arr[i] is equal to k, return the element arr[i].
- If no element occurs k times in the array, return -1.
- In the main function, initialize an integer array arr, its size n, and the value of k.
- Call the firstElement function with arr, n, and k as arguments, and print the returned value.
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. |
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:
- Traverse the array of 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> |
7
Time Complexity: O(n)
Auxiliary Space: O(n) because we are using an auxiliary array of size n to store the count
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