Given a linked list, the task is to find the sum of distances between the two nearest perfect squares for all the nodes of the given linked list.
Input: 3 -> 15 -> 7 -> NULL
For 3: closest left perfect square is 1 and closest right 4 i.e. 4-1 = 3
For 15: 16 – 9 = 7
For 7: 9 – 4 = 5
3 + 7 + 5 = 15
Input: 1 -> 5 -> 10 -> 78 -> 23 -> NULL
Approach: Initialise sum = 0 and for every node, if the current node’s value is a perfect square itself then the left and right closest perfect square will be the value itself and distance will be 0. Else, find the left and right closest perfect squares say leftPS and rightPS and update sum = sum + (rightPS – leftPS).
Below is the implementation of the above approach:
Time Complexity: O(n)
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- Linked List Product of Nodes Between 0s
- Segregate even and odd nodes in a Linked List
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- Sum of the nodes of a Singly Linked List
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- Sum of the nodes of a Circular Linked List
- Print nodes of linked list at given indexes
- Append the last M nodes to the beginning of the given linked list
- Replace nodes with duplicates in linked list
- Alternate Odd and Even Nodes in a Singly Linked List
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