pb7 + readme

Author: EISAWESOME

README.md

@@ -86,3 +86,17 @@ Implement car and cdr.
 
 [JS](pb5/answer.js) (Not timed)
 
+## 6
+> This problem was asked by Google.
+
+An XOR linked list is a more memory efficient doubly linked list. Instead of each node holding `next` and `prev` fields, it holds a field named both, which is an XOR of the next node and the previous node. Implement an XOR linked list; it has an `add(element)` which adds the element to the end, and a `get(index)` which returns the node at index.
+If using a language that has no pointers (such as Python), you can assume you have access to get_pointer and dereference_pointer functions that converts between nodes and memory addresses.
+
+
+### 7 
+> This problem was asked by Facebook.
+
+Given the mapping `a = 1, b = 2, ... z = 26`, and an encoded message, count the number of ways it can be decoded.
+For example, the message '111' would give 3, since it could be decoded as 'aaa', 'ka', and 'ak'.
+You can assume that the messages are decodable. For example, '001' is not allowed.
+

pb7/Problem.md

@@ -0,0 +1,5 @@
+> This problem was asked by Facebook.
+
+Given the mapping `a = 1, b = 2, ... z = 26`, and an encoded message, count the number of ways it can be decoded.
+For example, the message '111' would give 3, since it could be decoded as 'aaa', 'ka', and 'ak'.
+You can assume that the messages are decodable. For example, '001' is not allowed.

pb7/answer.js

@@ -0,0 +1,72 @@
+const m1 = 111;
+const m2 = 12345;
+const m3 = 4;
+
+// Less optimized, clearer version
+const decode = (message) => {
+  message = message.toString();
+  
+  const split1 = [];
+  const split2 = [];
+  //Split1, start at index 0
+  for (var i = 0; i < message.length; i+=2) {
+    // If we have enough letters left to make a par
+    if(message.charAt(i+1)){
+      split1.push(message.charAt(i) + message.charAt(i+1));
+    } else {
+      split1.push(message.charAt(i));
+    }
+  }
+
+    //Split1, start at index 1
+    split2.push(message.charAt(0));
+    for (var i = 1; i < message.length; i+=2) {
+      // If we have enough letters left to make a par
+      if(message.charAt(i+1)){
+        split2.push(message.charAt(i) + message.charAt(i+1))
+      } else {
+        split2.push(message.charAt(i))
+      }
+      
+    }
+    console.log(split1, split2)
+    // base : one way to solve
+    // Every time we encounter an index that is <= 26, that's another way 
+    const nbWays = 1 + split1.concat(split2).filter( x => x <= 26 && x >= 10).length ;
+    return nbWays
+}
+
+// Optimized, more cryptic version
+const decode2 = (message) => {
+  message = message.toString();
+  
+  const split1 = [];
+  const split2 = [];
+  
+  split2.push(message.charAt(0));
+  for (var i = 0; i < message.length; i++) {
+    if(i % 2 === 0) {
+      if(message.charAt(i+1)){
+        split1.push(message.charAt(i) + message.charAt(i+1));
+      } else {
+        split1.push(message.charAt(i));
+      }
+    } else {
+      if(message.charAt(i+1)){
+        split2.push(message.charAt(i) + message.charAt(i+1))
+      } else {
+        split2.push(message.charAt(i))
+      }
+    }
+  }
+  console.log(split1, split2)
+  const nbWays = 1 + split1.concat(split2).filter( x => x <= 26 && x >= 10).length ;
+  return nbWays
+}
+
+/*
+console.log(decode(m2));
+console.log('---------')
+*/
+console.log(decode2(m2));
+