Given a singly linked list. The task is to find the sum and product of all of the nodes of the given linked list which are divisible by a given number k.
Input : List = 7->60->8->40->1 k = 10 Output : Product = 2400, Sum = 100 Product of nodes: 60 * 40 = 2400 Input : List = 15->7->3->9->11->5 k = 5 Output : Product = 75, Sum = 20
- Initialize a pointer ptr with the head of the linked list, a product variable with 1 and a sum variable with 0.
- Start traversing the linked list using a loop until all the nodes get traversed.
- For every node:
- Multiply the value of the current node to the product if current node is divisible by k.
- Add the value of the current node to the sum if current node is divisible by k.
- Increment the pointer to the next node of linked list i.e. ptr = ptr ->next.
- Repeat the above steps until end of linked list is reached.
- Finally, print the product and sum.
Below is the implementation of the above approach:
Product = 700 Sum = 80
Time Complexity: O(N), where N is the number of nodes in the linked list.
- Sum and Product of the nodes of a Circular Singly Linked List which are divisible by K
- Product of the nodes of a Singly Linked List
- Sum and Product of all Prime Nodes of a Singly Linked List
- Product of all nodes in a doubly linked list divisible by a given number K
- Sum of the nodes of a Singly Linked List
- Alternate Odd and Even Nodes in a Singly Linked List
- Reverse alternate K nodes in a Singly Linked List
- Delete all Non-Prime Nodes from a Singly Linked List
- Delete all Prime Nodes from a Singly Linked List
- Count of Prime Nodes of a Singly Linked List
- Find the common nodes in two singly linked list
- Delete all Prime Nodes from a Circular Singly Linked List
- Update adjacent nodes if the current node is zero in a Singly Linked List
- Linked List Product of Nodes Between 0s
- Sum and Product of all the nodes which are less than K in the linked list
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