# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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; ` `} `

## 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 `

Output:

```2
```

Time Complexity : O(n)

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Improved By : Vikash Kumar 37