# 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++

`// 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

`# 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)

## Recommended Posts:

- Find minimum and maximum elements in singly Circular Linked List
- Sum of all odd frequency nodes of the Linked List
- Sum of all minimum frequency elements in Matrix
- Partitioning a linked list around a given value and If we don't care about making the elements of the list "stable"
- Count duplicates in a given linked list
- Reverse first K elements of given linked list
- Reverse even elements in a Linked List
- Count nodes in Circular linked list
- Sum of smaller elements of nodes in a linked list
- Find unique elements in linked list
- Pairwise swap elements of a given linked list
- Maximum and Minimum element of a linked list which is divisible by a given number k
- Minimum and Maximum Prime Numbers of a Singly Linked List
- Count of Prime Nodes of a Singly Linked List
- Replace each node with its Surpasser Count in Linked List

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