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.
Examples:
Input: 3 -> 15 -> 7 -> NULL
Output: 15
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 = 15Input: 1 -> 5 -> 10 -> 78 -> 23 -> NULL
Output: 38
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).
Algorithm:
- create a function with an int return type that takes the head of the linked list as input.
- Set the base condition that is if the head is equal to null then return 0.
- Now initialize an int variable to store the total distance sum initially assigned 0 to it.
- Now create a pointer named “temp” which points to the head of the linked list.
- start a while loop with the conditioning temp not equal to NULL
- Now for each iteration.
- Find the square root of the node data and store the value in a variable sq_root
- It indicates that “temp->data” is not a perfect square if “sq root” is less than “temp->data”. Get the left and right perfect squares of the given “temp->data” and place them, respectively, in the variables “left ps” and “right ps.”
- To “tsum,” add the discrepancy between “right ps” and “left ps.
- Transfer the “temp” pointer to the following node.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Structure of a node of linked listclass Node { public: int data; Node* next; Node(int data) { this->data = data; this->next = NULL; }}; // Function to find the total distance sumint distanceSum(Node* head){ // If head is null if (head == NULL) return 0; // To store the required sum int tsum = 0; Node* temp = head; // Traversing through all the nodes one by one while (temp != NULL) { double sq_root = sqrt(temp->data); // If current node is not a perfect square // then find left perfect square and // right perfect square if (sq_root < temp->data) { int left_ps = (int)floor(sq_root) * (int)floor(sq_root); int right_ps = (int)ceil(sq_root) * (int)ceil(sq_root); tsum += right_ps - left_ps; } // Get to the next node temp = temp->next; } return tsum;}// Driver codeint main(){ Node* head = new Node(3); head->next = new Node(15); head->next->next = new Node(7); head->next->next->next = new Node(40); head->next->next->next->next = new Node(42); int result = distanceSum(head); cout << result << endl; return 0;}// This code is contributed by divyeshrabadiya07 |
Java
// Java implementation of the approachclass GFG { // Structure of a node of linked list static class Node { int data; Node next; Node(int data) { this.data = data; this.next = null; } } // Function to find the total distance sum static int distanceSum(Node head) { // If head is null if (head == null) return 0; // To store the required sum int tsum = 0; Node temp = head; // Traversing through all the nodes one by one while (temp != null) { double sq_root = Math.sqrt(temp.data); // If current node is not a perfect square // then find left perfect square and // right perfect square if (sq_root < temp.data) { int left_ps = (int)Math.floor(sq_root) * (int)Math.floor(sq_root); int right_ps = (int)Math.ceil(sq_root) * (int)Math.ceil(sq_root); tsum += right_ps - left_ps; } // Get to the next node temp = temp.next; } return tsum; } // Driver code public static void main(String[] args) { Node head = new Node(3); head.next = new Node(15); head.next.next = new Node(7); head.next.next.next = new Node(40); head.next.next.next.next = new Node(42); int result = distanceSum(head); System.out.println(result); }} |
Python3
# Python3 implementation of the approachimport sysimport math# Structure for a nodeclass Node: def __init__(self, data): self.data = data self.next = None# Function to find the total distance sumdef distanceSum(head): # If head is null if not head: return # To store the required sum tsum = 0 temp = head # Traversing through all the nodes one by one while(temp): sq_root = math.sqrt(temp.data) # If current node is not a perfect square # then find left perfect square and # right perfect square if sq_root < temp.data: left_ps = math.floor(sq_root) ** 2 right_ps = math.ceil(sq_root) ** 2 tsum += (right_ps - left_ps) # Get to the next node temp = temp.next return tsum# Driver codeif __name__=='__main__': head = Node(3) head.next = Node(15) head.next.next = Node(7) head.next.next.next = Node(40) head.next.next.next.next = Node(42) result = distanceSum(head) print("{}".format(result)) # This code is contributed by rutvik_56 |
C#
// C# implementation of the approachusing System;using System.Collections;using System.Collections.Generic;class GFG { // Structure of a node of linked list class Node { public int data; public Node next; public Node(int data) { this.data = data; this.next = null; } } // Function to find the total distance sum static int distanceSum(Node head) { // If head is null if (head == null) return 0; // To store the required sum int tsum = 0; Node temp = head; // Traversing through all the nodes one by one while (temp != null) { double sq_root = Math.Sqrt(temp.data); // If current node is not a perfect square // then find left perfect square and // right perfect square if (sq_root < temp.data) { int left_ps = (int)Math.Floor(sq_root) * (int)Math.Floor(sq_root); int right_ps = (int)Math.Ceiling(sq_root) * (int)Math.Ceiling(sq_root); tsum += right_ps - left_ps; } // Get to the next node temp = temp.next; } return tsum; } // Driver code public static void Main(string[] args) { Node head = new Node(3); head.next = new Node(15); head.next.next = new Node(7); head.next.next.next = new Node(40); head.next.next.next.next = new Node(42); int result = distanceSum(head); Console.Write(result); }}// This code is contributed by pratham76 |
Javascript
<script> // JavaScript implementation of the approach // Structure of a node of linked list class Node { constructor(data) { this.data = data; this.next = null; } } // Function to find the total distance sum function distanceSum(head) { // If head is null if (head == null) return 0; // To store the required sum var tsum = 0; var temp = head; // Traversing through all the nodes one by one while (temp != null) { var sq_root = Math.sqrt(temp.data); // If current node is not a perfect square // then find left perfect square and // right perfect square if (sq_root < temp.data) { var left_ps = parseInt(Math.floor(sq_root)) * parseInt(Math.floor(sq_root)); var right_ps = parseInt(Math.ceil(sq_root)) * parseInt(Math.ceil(sq_root)); tsum += right_ps - left_ps; } // Get to the next node temp = temp.next; } return tsum; } // Driver code var head = new Node(3); head.next = new Node(15); head.next.next = new Node(7); head.next.next.next = new Node(40); head.next.next.next.next = new Node(42); var result = distanceSum(head); document.write(result); </script> |
Output:
41
Time Complexity: O(n*sqrt(maximum element in linked list))
Space Complexity: O(1)
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