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suriyakrishnasuriyakrishna
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‎src/main/scala/com/kishan/scala/leetcode/juneChallenges/ReconstructItinerary.scala‎

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package com.kishan.scala.leetcode.juneChallenges
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import java.util
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import scala.collection.JavaConverters._
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import scala.collection.mutable
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/*
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* Given a list of airline tickets represented by pairs of departure and arrival airports [from, to], reconstruct the itinerary in order.
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* All of the tickets belong to a man who departs from JFK. Thus, the itinerary must begin with JFK.
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*
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* Note:
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1. If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string.
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For example, the itinerary ["JFK", "LGA"] has a smaller lexical order than ["JFK", "LGB"].
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2. All airports are represented by three capital letters (IATA code).
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3. You may assume all tickets form at least one valid itinerary.
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4. One must use all the tickets once and only once.
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*
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* */
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/*
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* Input: [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
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Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]
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*
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*
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* Input: [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
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Output: ["JFK","ATL","JFK","SFO","ATL","SFO"]
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Explanation: Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"].
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But it is larger in lexical order.
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*
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* */
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/*
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* Approach: DFS - Depth First Search
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*
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* 1. We have to build HashMap with Source as Key and Queue of Destination - By Implementing Priority Queue as value.
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* 2. We have to Start from JFK and find the city which doesn't have any more destination and add to linked list at head position.
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* 3. We have to recursively go through each Priority Queue and have to add each destination at the head of Linked List
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* 4. Finally convert LinkedList asScala List and return the result.
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*
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* */
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object ReconstructItinerary {
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private var map: mutable.HashMap[String, mutable.PriorityQueue[String]] = mutable.HashMap()
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private var result: util.LinkedList[String] = new util.LinkedList()
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def findItinerary(tickets: List[List[String]]): List[String] = {
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tickets.foreach(a => {
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if (!map.contains(a.head)) {
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var pq: mutable.PriorityQueue[String] = mutable.PriorityQueue.empty[String](Ordering.String.reverse)
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map += (a.head -> pq)
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}
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map(a.head).enqueue(a.last)
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})
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dfs("JFK")
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return result.asScala.toList
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}
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def dfs(startCity: String): Unit = {
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var arrivals = map.getOrElse(startCity, null)
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while (arrivals != null && arrivals.nonEmpty) {
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dfs(arrivals.dequeue())
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}
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result.add(0, startCity)
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}
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def main(args: Array[String]): Unit = {
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val input1: List[List[String]] = List(List("MUC", "LHR"), List("JFK", "MUC"), List("SFO", "SJC"), List("LHR", "SFO"))
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val input2: List[List[String]] = List(List("JFK", "SFO"), List("JFK", "ATL"), List("SFO", "ATL"), List("ATL", "JFK"), List("ATL", "SFO"))
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val input3: List[List[String]] = List(List("JFK", "KUL"), List("JFK", "NRT"), List("NRT", "JFK"))
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val input4: List[List[String]] = List(List("EZE", "AXA"), List("TIA", "ANU"), List("ANU", "JFK"), List("JFK", "ANU"), List("ANU", "EZE"), List("TIA", "ANU"), List("AXA", "TIA"), List("TIA", "JFK"), List("ANU", "TIA"), List("JFK", "TIA"))
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val input5: List[List[String]] = List(List("EZE", "TIA"), List("EZE", "HBA"), List("AXA", "TIA"), List("JFK", "AXA"), List("ANU", "JFK"), List("ADL", "ANU"), List("TIA", "AUA"), List("ANU", "AUA"), List("ADL", "EZE"), List("ADL", "EZE"), List("EZE", "ADL"), List("AXA", "EZE"), List("AUA", "AXA"), List("JFK", "AXA"), List("AXA", "AUA"), List("AUA", "ADL"), List("ANU", "EZE"), List("TIA", "ADL"), List("EZE", "ANU"), List("AUA", "ANU"))
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val result = findItinerary(input5)
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println(s"Result: ${result.mkString(",")}, ${result.mkString(",") == "JFK,AXA,AUA,ADL,ANU,AUA,ANU,EZE,ADL,EZE,ANU,JFK,AXA,EZE,TIA,AUA,AXA,TIA,ADL,EZE,HBA"}")
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}
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// ["JFK","ANU","EZE","AXA","TIA","ANU","JFK","TIA","ANU","TIA","JFK"]
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// ["JFK","AXA","AUA","ADL","ANU","AUA","ANU","EZE","ADL","EZE","ANU","JFK","AXA","EZE","TIA","AUA","AXA","TIA","ADL","EZE","HBA"]
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}

‎src/main/scala/com/kishan/scala/leetcode/juneChallenges/UniquePaths.scala‎

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package com.kishan.scala.leetcode.juneChallenges
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/*
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* A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
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The robot can only move either down or right at any point in time.
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The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
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How many possible unique paths are there?
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*
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* */
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/*
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* Input: m = 3, n = 2
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Output: 3
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Explanation:
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From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
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1. Right -> Right -> Down
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2. Right -> Down -> Right
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3. Down -> Right -> Right
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*
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* Input: m = 7, n = 3
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Output: 28
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*
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* Constraints:
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1. 1 <= m, n <= 100
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2. It's guaranteed that the answer will be less than or equal to 2 * 10 ^ 9.
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*
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* */
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/*
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* Approach: Dynamic Programming
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*
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* 1. For i = 0 or j = 0. Maximum possible steps will be 1.
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* 2. For i >= 1 and j >= 1, we can reach from top and we can also reach from left. So it will be sum of top + left cell value.
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*
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* m = 3
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* n = 7
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*
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* Iteration:
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*
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* 0 -> 1 1 1 1 1 1 1
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* 1 -> 1 2 3 4 5 6 7
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* 2 -> 1 3 6 10 15 21 28
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*
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* return the last element of array which will be result
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* */
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object UniquePaths {
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//Using Two Dimensional Array
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def uniquePathsTwoDimensional(m: Int, n: Int): Int = {
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var dp = Array.ofDim[Int](m, n)
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for (i <- dp.indices) {
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for (j <- dp(i).indices) {
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if (i == 0 || j == 0) {
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dp(i)(j) = 1
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}
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if (i >= 1 && j >= 1) {
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dp(i)(j) = dp(i - 1)(j) + dp(i)(j - 1)
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}
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}
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}
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return dp(m - 1)(n - 1)
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}
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def main(args: Array[String]): Unit = {
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val m = 3
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val n = 7
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val result = uniquePaths(m, n)
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println(s"Result: ${result}, ${result == 28}")
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val result1 = uniquePaths(3, 2)
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println(s"Result: ${result1}, ${result1 == 3}")
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}
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//Using One Dimensional Array
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def uniquePaths(m: Int, n: Int): Int = {
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var dp = Array.ofDim[Int](n)
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for (i <- 0 until m) {
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var j = 0
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while (j < n) {
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if (i == 0 || j == 0) dp(j) = 1 else if (i >= 1) dp(j) += dp(j - 1)
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j += 1
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}
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}
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return dp(n - 1)
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}
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}

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