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Add path crossing
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‎README.md

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| 9 | LRU Cache | [Source](https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/hashtable/9.lruCache/lruCache.js) | [JavaScript](https://livecodes.io/?console&x=https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/hashtable/9.lruCache/lruCache.js) | [Documentation](https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/hashtable/9.lruCache/lruCache.md) | Medium | Combination of Hash Table and doubly linked list |
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| 10 | Maximum number of balloons | [Source](https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/hashtable/10.maximumBalloons/maximumBalloons.js) | [JavaScript](https://livecodes.io/?console&x=https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/hashtable/10.maximumBalloons/maximumBalloons.js) | [Documentation](https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/hashtable/10.maximumBalloons/maximumBalloons.md) | Easy | Hash map on characters |
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| 11 | Isomorphic Strings | [Source](https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/hashtable/11.isomorphicStrings/isomorphicStrings.js) | [JavaScript](https://livecodes.io/?console&x=https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/hashtable/11.isomorphicStrings/isomorphicStrings.js) | [Documentation](https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/hashtable/11.isomorphicStrings/isomorphicStrings.md) | Easy | mapping characters using Hash maps |
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| 12 | Path Crossing | [Source](https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/hashtable/12.pathCrossing/pathCrossing.js) | [JavaScript](https://livecodes.io/?console&x=https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/hashtable/12.pathCrossing/pathCrossing.js) | [Documentation](https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/hashtable/12.pathCrossing/pathCrossing.md) | Easy | Visited set based on path characters |
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## Sorting
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‎src/java1/algorithms/hashmap/pathCrossing/PathCrossing.java

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package java1.algorithms.hashmap.pathCrossing;
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import java.util.HashSet;
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import java.util.Set;
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public class PathCrossing {
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private static boolean isPathCrossing(String path) {
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Set<String> visited = new HashSet<>();
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int x=0, y = 0;
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visited.add(x+","+y);
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for (char move : path.toCharArray()) {
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if(move == 'N') y++;
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if(move == 'S') y--;
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if(move == 'E') x++;
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if(move == 'W') x--;
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if(visited.contains(x+","+y)) {
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return true;
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}
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visited.add(x+","+y);
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}
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return false;
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}
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public static void main(String[] args) {
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String path1="NESE";
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String path2="NEWS";
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System.out.println(isPathCrossing(path1));
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System.out.println(isPathCrossing(path2));
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}
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}

‎src/java1/algorithms/hashmap/pathCrossing/PathCrossing.md

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**Problem statement:**
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Given a string `path`, where `path[i] = 'N', 'S', 'E' or 'W'`, each letter representing moving one unit towards north, south, east, or west, respectively. Assume that you started at the origin (0, 0) and walk on the specified path(`path`).
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Return `true` if the path crosses same point at any point. otherwise, return `false`.
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## Examples:
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Example 1:
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Input: path = "NESE"
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Output: false
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Example 2:
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Input: path = "NESW"
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Output: true
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**Algorithmic Steps**
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This problem is solved with the help of a set, which is used to store the visited co-ordinates and determine if the current point is crossed or not.
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1. Define a set variable(`visited`) to keep track of visited co-oridnates.
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2. Initialize current co-ordinates to (0,0).
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3. Iterate over each character of given path,
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1. Update the current co-ordinates based on each character(`N, E, W, S`).
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2. If the updated location is already visited, return true indicating that path is already crossed.
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3. Add the newly visited co-ordinate/point to the visited set.
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4. Return `false` indicating that path didn't cross the same point.
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**Time and Space complexity:**
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This algorithm has a time complexity of `O(n)`, Where `n` is the length of the path. This is because we traverse the string once and perform constant time operations for lookup in a set.
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Here, it takes time complexity of `O(n)`. This is because of the space required by `visited` hash set.

‎src/javascript/algorithms/hashtable/12.pathCrossing/pathCrossing.js

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function isPathCrossing(path){
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const visited = new Set();
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let x = 0, y = 0;
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visited.add(`${x},${y}`);
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const directions = {
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'N': [0, 1],
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'S': [0, -1],
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'E': [1, 0],
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'W': [-1, 0]
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}
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for (const move of path) {
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x+= directions[move][0],
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y+= directions[move][1]
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if(visited.has(`${x},${y}`)) {
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return true;
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}
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visited.add(`${x},${y}`);
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}
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return false;
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}
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const path1="NESE";
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const path2="NEWS";
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console.log(isPathCrossing(path1));
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console.log(isPathCrossing(path2));

‎src/javascript/algorithms/hashtable/12.pathCrossing/pathCrossing.md

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**Problem statement:**
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Given a string `path`, where `path[i] = 'N', 'S', 'E' or 'W'`, each letter representing moving one unit towards north, south, east, or west, respectively. Assume that you started at the origin (0, 0) and walk on the specified path(`path`).
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Return `true` if the path crosses same point at any point. otherwise, return `false`.
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## Examples:
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Example 1:
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Input: path = "NESE"
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Output: false
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Example 2:
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Input: path = "NESW"
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Output: true
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**Algorithmic Steps**
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This problem is solved with the help of a set, which is used to store the visited co-ordinates and determine if the current point is crossed or not.
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1. Define a set variable(`visited`) to keep track of visited co-oridnates.
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2. Initialize current co-ordinates to (0,0).
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3. Declare a `directions` object to map the direction with the respective unit points
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4. Iterate over each character of given path,
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1. Update the current co-ordinates based on each character.
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2. If the updated location is already visited, return true indicating that path is already crossed.
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3. Add the newly visited co-ordinate/point to the visited set.
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5. Return `false` indicating that path didn't cross the same point.
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**Time and Space complexity:**
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This algorithm has a time complexity of `O(n)`, Where `n` is the length of the path. This is because we traverse the string once and perform constant time operations for lookup in a set.
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Here, it takes time complexity of `O(n)`. This is because of the space required by `visited` hash set.

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