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 |
Output:
7
Time Complexity: O(n)
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