Find the Point Hackerrank Solution Python
In this HackerRank Find Merge Point of Two Lists Interview preparation kit problem, You have Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge.
Problem solution in Python programming.
#!/bin/python3 import math import os import random import re import sys class SinglyLinkedListNode: def __init__(self, node_data): self.data = node_data self.next = None class SinglyLinkedList: def __init__(self): self.head = None self.tail = None def insert_node(self, node_data): node = SinglyLinkedListNode(node_data) if not self.head: self.head = node else: self.tail.next = node self.tail = node def print_singly_linked_list(node, sep, fptr): while node: fptr.write(str(node.data)) node = node.next if node: fptr.write(sep) # Complete the findMergeNode function below. # # For your reference: # # SinglyLinkedListNode: # int data # SinglyLinkedListNode next # # def findMergeNode(headA, headB): while headA: node = headB while node: if headA == node: return headA.data node = node.next headA = headA.next if __name__ == '__main__': fptr = open(os.environ['OUTPUT_PATH'], 'w') tests = int(input()) for tests_itr in range(tests): index = int(input()) llist1_count = int(input()) llist1 = SinglyLinkedList() for _ in range(llist1_count): llist1_item = int(input()) llist1.insert_node(llist1_item) llist2_count = int(input()) llist2 = SinglyLinkedList() for _ in range(llist2_count): llist2_item = int(input()) llist2.insert_node(llist2_item) ptr1 = llist1.head; ptr2 = llist2.head; for i in range(llist1_count): if i < index: ptr1 = ptr1.next for i in range(llist2_count): if i != llist2_count-1: ptr2 = ptr2.next ptr2.next = ptr1 result = findMergeNode(llist1.head, llist2.head) fptr.write(str(result) + '\n') fptr.close()
Problem solution in Java Programming.
import java.io.*; import java.math.*; import java.security.*; import java.text.*; import java.util.*; import java.util.concurrent.*; import java.util.regex.*; public class Solution { static class SinglyLinkedListNode { public int data; public SinglyLinkedListNode next; public SinglyLinkedListNode(int nodeData) { this.data = nodeData; this.next = null; } } static class SinglyLinkedList { public SinglyLinkedListNode head; public SinglyLinkedListNode tail; public SinglyLinkedList() { this.head = null; this.tail = null; } public void insertNode(int nodeData) { SinglyLinkedListNode node = new SinglyLinkedListNode(nodeData); if (this.head == null) { this.head = node; } else { this.tail.next = node; } this.tail = node; } } public static void printSinglyLinkedList(SinglyLinkedListNode node, String sep, BufferedWriter bufferedWriter) throws IOException { while (node != null) { bufferedWriter.write(String.valueOf(node.data)); node = node.next; if (node != null) { bufferedWriter.write(sep); } } } // Complete the findMergeNode function below. /* * For your reference: * * SinglyLinkedListNode { * int data; * SinglyLinkedListNode next; * } * */ static int findMergeNode(SinglyLinkedListNode head1, SinglyLinkedListNode head2) { SinglyLinkedListNode temp1=head1; SinglyLinkedListNode temp2=head2; List<SinglyLinkedListNode>list=new ArrayList<SinglyLinkedListNode>(); while(temp1!=null){ list.add(temp1); temp1=temp1.next; } while(temp2!=null){ if(list.contains(temp2)){ break; } temp2=temp2.next; } return temp2.data; } private static final Scanner scanner = new Scanner(System.in); public static void main(String[] args) throws IOException { BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH"))); int tests = scanner.nextInt(); scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); for (int testsItr = 0; testsItr < tests; testsItr++) { int index = scanner.nextInt(); scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); SinglyLinkedList llist1 = new SinglyLinkedList(); int llist1Count = scanner.nextInt(); scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); for (int i = 0; i < llist1Count; i++) { int llist1Item = scanner.nextInt(); scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); llist1.insertNode(llist1Item); } SinglyLinkedList llist2 = new SinglyLinkedList(); int llist2Count = scanner.nextInt(); scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); for (int i = 0; i < llist2Count; i++) { int llist2Item = scanner.nextInt(); scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); llist2.insertNode(llist2Item); } SinglyLinkedListNode ptr1 = llist1.head; SinglyLinkedListNode ptr2 = llist2.head; for (int i = 0; i < llist1Count; i++) { if (i < index) { ptr1 = ptr1.next; } } for (int i = 0; i < llist2Count; i++) { if (i != llist2Count-1) { ptr2 = ptr2.next; } } ptr2.next = ptr1; int result = findMergeNode(llist1.head, llist2.head); bufferedWriter.write(String.valueOf(result)); bufferedWriter.newLine(); } bufferedWriter.close(); scanner.close(); } } Problem solution in C++ programming.
#include <bits/stdc++.h> using namespace std; class SinglyLinkedListNode { public: int data; SinglyLinkedListNode *next; SinglyLinkedListNode(int node_data) { this->data = node_data; this->next = nullptr; } }; class SinglyLinkedList { public: SinglyLinkedListNode *head; SinglyLinkedListNode *tail; SinglyLinkedList() { this->head = nullptr; this->tail = nullptr; } void insert_node(int node_data) { SinglyLinkedListNode* node = new SinglyLinkedListNode(node_data); if (!this->head) { this->head = node; } else { this->tail->next = node; } this->tail = node; } }; void print_singly_linked_list(SinglyLinkedListNode* node, string sep, ofstream& fout) { while (node) { fout << node->data; node = node->next; if (node) { fout << sep; } } } void free_singly_linked_list(SinglyLinkedListNode* node) { while (node) { SinglyLinkedListNode* temp = node; node = node->next; free(temp); } } // Complete the findMergeNode function below. /* * For your reference: * * SinglyLinkedListNode { * int data; * SinglyLinkedListNode* next; * }; * */ int findMergeNode(SinglyLinkedListNode* headA, SinglyLinkedListNode* headB) { while(headA){ SinglyLinkedListNode *tmp = headA->next; headA->next = NULL; headA = tmp; } while(headB){ if(headB->next == NULL){ return headB->data; } headB = headB->next; } return 0; } int main() { ofstream fout(getenv("OUTPUT_PATH")); int tests; cin >> tests; cin.ignore(numeric_limits<streamsize>::max(), '\n'); for (int tests_itr = 0; tests_itr < tests; tests_itr++) { int index; cin >> index; cin.ignore(numeric_limits<streamsize>::max(), '\n'); SinglyLinkedList* llist1 = new SinglyLinkedList(); int llist1_count; cin >> llist1_count; cin.ignore(numeric_limits<streamsize>::max(), '\n'); for (int i = 0; i < llist1_count; i++) { int llist1_item; cin >> llist1_item; cin.ignore(numeric_limits<streamsize>::max(), '\n'); llist1->insert_node(llist1_item); } SinglyLinkedList* llist2 = new SinglyLinkedList(); int llist2_count; cin >> llist2_count; cin.ignore(numeric_limits<streamsize>::max(), '\n'); for (int i = 0; i < llist2_count; i++) { int llist2_item; cin >> llist2_item; cin.ignore(numeric_limits<streamsize>::max(), '\n'); llist2->insert_node(llist2_item); } SinglyLinkedListNode* ptr1 = llist1->head; SinglyLinkedListNode* ptr2 = llist2->head; for (int i = 0; i < llist1_count; i++) { if (i < index) { ptr1 = ptr1->next; } } for (int i = 0; i < llist2_count; i++) { if (i != llist2_count-1) { ptr2 = ptr2->next; } } ptr2->next = ptr1; int result = findMergeNode(llist1->head, llist2->head); fout << result << "\n"; } fout.close(); return 0; } Problem solution in C programming.
#include <assert.h> #include <limits.h> #include <math.h> #include <stdbool.h> #include <stddef.h> #include <stdint.h> #include <stdio.h> #include <stdlib.h> #include <string.h> char* readline(); typedef struct SinglyLinkedListNode SinglyLinkedListNode; typedef struct SinglyLinkedList SinglyLinkedList; struct SinglyLinkedListNode { int data; SinglyLinkedListNode* next; }; struct SinglyLinkedList { SinglyLinkedListNode* head; SinglyLinkedListNode* tail; }; SinglyLinkedListNode* create_singly_linked_list_node(int node_data) { SinglyLinkedListNode* node = malloc(sizeof(SinglyLinkedListNode)); node->data = node_data; node->next = NULL; return node; } void insert_node_into_singly_linked_list(SinglyLinkedList** singly_linked_list, int node_data) { SinglyLinkedListNode* node = create_singly_linked_list_node(node_data); if (!(*singly_linked_list)->head) { (*singly_linked_list)->head = node; } else { (*singly_linked_list)->tail->next = node; } (*singly_linked_list)->tail = node; } void print_singly_linked_list(SinglyLinkedListNode* node, char* sep, FILE* fptr) { while (node) { fprintf(fptr, "%d", node->data); node = node->next; if (node) { fprintf(fptr, "%s", sep); } } } void free_singly_linked_list(SinglyLinkedListNode* node) { while (node) { SinglyLinkedListNode* temp = node; node = node->next; free(temp); } } // Complete the findMergeNode function below. /* * For your reference: * * SinglyLinkedListNode { * int data; * SinglyLinkedListNode* next; * }; * */ int findMergeNode(SinglyLinkedListNode* headA, SinglyLinkedListNode* headB) { while(headA){ SinglyLinkedListNode *tmp = headA->next; headA->next = NULL; headA = tmp; } while(headB){ if(headB->next == NULL){ return headB->data; } headB = headB->next; } return 0; } int main() { FILE* fptr = fopen(getenv("OUTPUT_PATH"), "w"); char* tests_endptr; char* tests_str = readline(); int tests = strtol(tests_str, &tests_endptr, 10); if (tests_endptr == tests_str || *tests_endptr != '\0') { exit(EXIT_FAILURE); } for (int tests_itr = 0; tests_itr < tests; tests_itr++) { char* index_endptr; char* index_str = readline(); int index = strtol(index_str, &index_endptr, 10); if (index_endptr == index_str || *index_endptr != '\0') { exit(EXIT_FAILURE); } SinglyLinkedList* llist1 = malloc(sizeof(SinglyLinkedList)); llist1->head = NULL; llist1->tail = NULL; char* llist1_count_endptr; char* llist1_count_str = readline(); int llist1_count = strtol(llist1_count_str, &llist1_count_endptr, 10); if (llist1_count_endptr == llist1_count_str || *llist1_count_endptr != '\0') { exit(EXIT_FAILURE); } for (int i = 0; i < llist1_count; i++) { char* llist1_item_endptr; char* llist1_item_str = readline(); int llist1_item = strtol(llist1_item_str, &llist1_item_endptr, 10); if (llist1_item_endptr == llist1_item_str || *llist1_item_endptr != '\0') { exit(EXIT_FAILURE); } insert_node_into_singly_linked_list(&llist1, llist1_item); } SinglyLinkedList* llist2 = malloc(sizeof(SinglyLinkedList)); llist2->head = NULL; llist2->tail = NULL; char* llist2_count_endptr; char* llist2_count_str = readline(); int llist2_count = strtol(llist2_count_str, &llist2_count_endptr, 10); if (llist2_count_endptr == llist2_count_str || *llist2_count_endptr != '\0') { exit(EXIT_FAILURE); } for (int i = 0; i < llist2_count; i++) { char* llist2_item_endptr; char* llist2_item_str = readline(); int llist2_item = strtol(llist2_item_str, &llist2_item_endptr, 10); if (llist2_item_endptr == llist2_item_str || *llist2_item_endptr != '\0') { exit(EXIT_FAILURE); } insert_node_into_singly_linked_list(&llist2, llist2_item); } SinglyLinkedListNode* ptr1 = llist1->head; SinglyLinkedListNode* ptr2 = llist2->head; for (int i = 0; i < llist1_count; i++) { if (i < index) { ptr1 = ptr1->next; } } for (int i = 0; i < llist2_count; i++) { if (i != llist2_count-1) { ptr2 = ptr2->next; } } ptr2->next = ptr1; int result = findMergeNode(llist1->head, llist2->head); fprintf(fptr, "%d\n", result); } fclose(fptr); return 0; } char* readline() { size_t alloc_length = 1024; size_t data_length = 0; char* data = malloc(alloc_length); while (true) { char* cursor = data + data_length; char* line = fgets(cursor, alloc_length - data_length, stdin); if (!line) { break; } data_length += strlen(cursor); if (data_length < alloc_length - 1 || data[data_length - 1] == '\n') { break; } size_t new_length = alloc_length << 1; data = realloc(data, new_length); if (!data) { break; } alloc_length = new_length; } if (data[data_length - 1] == '\n') { data[data_length - 1] = '\0'; } data = realloc(data, data_length); return data; } Problem solution in JavaScript programming.
'use strict'; const fs = require('fs'); process.stdin.resume(); process.stdin.setEncoding('utf-8'); let inputString = ''; let currentLine = 0; process.stdin.on('data', inputStdin => { inputString += inputStdin; }); process.stdin.on('end', _ => { inputString = inputString.replace(/\s*$/, '') .split('\n') .map(str => str.replace(/\s*$/, '')); main(); }); function readLine() { return inputString[currentLine++]; } const SinglyLinkedListNode = class { constructor(nodeData) { this.data = nodeData; this.next = null; } }; const SinglyLinkedList = class { constructor() { this.head = null; this.tail = null; } insertNode(nodeData) { const node = new SinglyLinkedListNode(nodeData); if (this.head == null) { this.head = node; } else { this.tail.next = node; } this.tail = node; } }; function printSinglyLinkedList(node, sep, ws) { while (node != null) { ws.write(String(node.data)); node = node.next; if (node != null) { ws.write(sep); } } } /* Find merge point of two linked lists Note that the head may be 'null' for the empty list. Node is defined as var Node = function(data) { this.data = data; this.next = null; } */ // This is a "method-only" submission. // You only need to complete this method. function findMergeNode(a, b) { let arr = []; while (a || b) { if (a && arr.find(n => n === a)) return a.data; else if(a) { arr.push(a) a = a.next; } if (b && arr.find(n => n === b)) return b.data; else if (b) { arr.push(b); b = b.next; } } return 0; } function main() { const ws = fs.createWriteStream(process.env.OUTPUT_PATH); const tests = parseInt(readLine(), 10); for (let testsItr = 0; testsItr < tests; testsItr++) { const index = parseInt(readLine(), 10); const llist1Count = parseInt(readLine(), 10); let llist1 = new SinglyLinkedList(); for (let i = 0; i < llist1Count; i++) { const llist1Item = parseInt(readLine(), 10); llist1.insertNode(llist1Item); } const llist2Count = parseInt(readLine(), 10); let llist2 = new SinglyLinkedList(); for (let i = 0; i < llist2Count; i++) { const llist2Item = parseInt(readLine(), 10); llist2.insertNode(llist2Item); } let ptr1 = llist1.head; let ptr2 = llist2.head; for (let i = 0; i < llist1Count; i++) { if (i < index) { ptr1 = ptr1.next; } } for (let i = 0; i < llist2Count; i++) { if (i != llist2Count-1) { ptr2 = ptr2.next; } } ptr2.next = ptr1; let result = findMergeNode(llist1.head, llist2.head); ws.write(result + "\n"); } ws.end(); } Find the Point Hackerrank Solution Python
Source: https://programs.programmingoneonone.com/2021/03/hackerrank-find-merge-point-of-two-lists-solution.html
0 Response to "Find the Point Hackerrank Solution Python"
Post a Comment