Count minimum frequency elements in a linked list

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++

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// C++ program to find count of minimum
// frequqncy 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 list
int 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;
}

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Python3

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# Python3 program to find count of minimum 
# frequqncy elements in Linked list 
import sys
import math
  
# Link list node
class 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 Contribute by Vikash Kumar 37

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Output:

2

Time Complexity : O(n)



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Improved By : Vikash Kumar 37