Given a linked list containing duplicate elements. The task is to find the count of all minimum occurring elements in the given linked list. That is the count of all such elements whose frequency is minimum in the matrix.
Examples:
Input : 1-> 2-> 2-> 3 Output : 2 Explanation: 1 and 3 are elements occurs only one time. So, count is 2. Input : 10-> 20-> 20-> 10-> 30 Output : 1
Approach:
- Traverse the linked list and use a hash table to store the frequency of elements of the linked list such that the key of map is the linked list element and value is its frequency in the linked list.
- Then traverse the hash table to find the minimum frequency.
- Finally, traverse the hash table to find the frequency of elements and check if it matches with the minimum frequency obtained in previous step, if yes, then add this frequency to count.
Below is the implementation of the above approach:
C++
// C++ program to find count of minimum// frequency elements in Linked list#include <bits/stdc++.h>using namespace std;/* Link list node */struct Node { int key; struct Node* next;};/* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */void push(struct Node** head_ref, int new_key){ struct Node* new_node = new Node; new_node->key = new_key; new_node->next = (*head_ref); (*head_ref) = new_node;}// Function to count minimum frequency elements// in the linked listint countMinimum(struct Node* head){ // Store frequencies of all nodes. unordered_map<int, int> mp; struct Node* current = head; while (current != NULL) { int data = current->key; mp[data]++; current = current->next; } // Find min frequency current = head; int min_frequency = INT_MAX, countMin = 0; for (auto it = mp.begin(); it != mp.end(); it++) { if (it->second <= min_frequency) { min_frequency = it->second; } } // Find count of min frequency elements for (auto it = mp.begin(); it != mp.end(); it++) { if (it->second == min_frequency) { countMin += (it->second); } } return countMin;}/* Driver program to test count function*/int main(){ /* Start with the empty list */ struct Node* head = NULL; int x = 21; /* Use push() to construct below list 10->10->11->30->10 */ push(&head, 10); push(&head, 30); push(&head, 11); push(&head, 10); push(&head, 10); cout << countMinimum(head) << endl; return 0;} |
Java
// Java program to find count of minimum// frequency elements in Linked listimport java.util.*;class GFG{/* Link list node */static class Node { int key; Node next;};/* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */static Node push(Node head_ref, int new_key){ Node new_node = new Node(); new_node.key = new_key; new_node.next = (head_ref); (head_ref) = new_node; return head_ref;}// Function to count minimum frequency elements// in the linked liststatic int countMinimum( Node head){ // Store frequencies of all nodes. HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>(); Node current = head; while (current != null) { int data = current.key; mp.put(data, (mp.get(data) == null ? 1:mp.get(data) + 1)); current = current.next; } // Find min frequency current = head; int min_frequency = Integer.MAX_VALUE, countMin = 0; for (Map.Entry<Integer,Integer> it :mp.entrySet()) { if (it.getValue() <= min_frequency) { min_frequency = it.getValue(); } } // Find count of min frequency elements for (Map.Entry<Integer,Integer> it :mp.entrySet()) { if (it.getValue() == min_frequency) { countMin += (it.getValue()); } } return countMin;}/* Driver code*/public static void main(String args[]){ /* Start with the empty list */ Node head = null; int x = 21; /* Use push() to construct below list 10.10.11.30.10 */ head = push(head, 10); head = push(head, 30); head = push(head, 11); head = push(head, 10); head = push(head, 10); System.out.println( countMinimum(head) );}}// This code is contributed by andrew1234 |
Python3
# Python3 program to find count of minimum # frequency elements in Linked list import sysimport math# Link list nodeclass Node: def __init__(self,data): self.data = data self.next = None# Given a reference (pointer to pointer) to the head # of a list and an int, push a new node on the front # of the list. def push(head,data): if not head: return Node(data) temp = Node(data) temp.next = head head = temp return head# Function to count minimum frequency elements # in the linked list def countMinimun(head): # Store frequencies of all nodes. freq = {} temp = head while(temp): d = temp.data if d in freq: freq[d] = freq.get(d) + 1 else: freq[d] = 1 temp = temp.next # Find min frequency minimum_freq = sys.maxsize for i in freq: minimum_freq = min(minimum_freq, freq.get(i)) # Find count of min frequency elements countMin = 0 for i in freq: if freq.get(i) == minimum_freq: countMin += 1 return countMin# Driver program to test count function #if __name__=='__main__': # Start with the empty list head = None # Use push() to construct below list #10->10->11->30->10 head = push(head,10) head = push(head,30) head = push(head,11) head = push(head,10) head = push(head,10) print(countMinimun(head))# This code is Contributed by Vikash Kumar 37 |
C#
// C# program to find count of minimum// frequency elements in Linked listusing System;using System.Collections.Generic;class GFG{/* Link list node */public class Node { public int key; public Node next;};/* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */static Node push(Node head_ref, int new_key){ Node new_node = new Node(); new_node.key = new_key; new_node.next = (head_ref); head_ref = new_node; return head_ref;}// Function to count minimum frequency elements// in the linked liststatic int countMinimum( Node head){ // Store frequencies of all nodes. IDictionary<int,int> mp = new Dictionary<int,int>(); Node current = head; while (current != null) { int data = current.key; int val = 0; mp[data] = (!mp.TryGetValue(data,out val) ? 1:mp[data]+ 1); current = current.next; } // Find min frequency current = head; int min_frequency = int.MaxValue, countMin = 0; foreach (KeyValuePair<int, int> it in mp) { if (it.Value <= min_frequency) { min_frequency = it.Value; } } // Find count of min frequency elements foreach (KeyValuePair<int, int> it in mp) { if (it.Value == min_frequency) { countMin += (it.Value); } } return countMin;}/* Driver code*/public static void Main(){ /* Start with the empty list */ Node head = null; int x = 21; /* Use push() to construct below list 10.10.11.30.10 */ head = push(head, 10); head = push(head, 30); head = push(head, 11); head = push(head, 10); head = push(head, 10); Console.WriteLine( countMinimum(head) );}}// This code is contributed by SoumikMondal |
Javascript
<script> // JavaScript program to find count of minimum // frequency elements in Linked list /* Link list node */ class Node { constructor() { this.key = 0; this.next = null; } } /* Given a reference (pointer to pointer) to the headof a list and an int, push a new node on the frontof the list. */ function push(head_ref, new_key) { var new_node = new Node(); new_node.key = new_key; new_node.next = head_ref; head_ref = new_node; return head_ref; } // Function to count minimum frequency elements // in the linked list function countMinimum(head) { // Store frequencies of all nodes. var mp = {}; var current = head; while (current != null) { var data = current.key; var val = 0; mp[data] = !mp.hasOwnProperty(data) ? 1 : mp[data] + 1; current = current.next; } // Find min frequency current = head; var min_frequency = 2147483647, countMin = 0; for (const [key, value] of Object.entries(mp)) { if (value <= min_frequency) { min_frequency = value; } } // Find count of min frequency elements for (const [key, value] of Object.entries(mp)) { if (value == min_frequency) { countMin += value; } } return countMin; } /* Driver code*/ /* Start with the empty list */ var head = null; var x = 21; /* Use push() to construct below list 10.10.11.30.10 */ head = push(head, 10); head = push(head, 30); head = push(head, 11); head = push(head, 10); head = push(head, 10); document.write(countMinimum(head) + "<br>"); // This code is contributed by rdtank. </script> |
Output
2
Time Complexity : O(n)
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