Problem 67 (#18)

* add solution and resources

* update readme

Author: curtisbarnard

README.md

@@ -25,7 +25,7 @@ As I work through this list I figure I would make a GitHub repo with my solution
 17. [Longest Palindrome #409](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/easy/longest-palindrome-409.md)
 18. [Reverse Linked List #206](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/easy/reverse-linked-list-206.md)
 19. [Majority Element #169](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/easy/majority-element-169.md)
-20. Add Binary #67
+20. [Add Binary #67](https://github.com/curtisbarnard/leetcode-grind75-javascript/blob/main/easy/add-binary-67.md)
 21. Diameter of Binary Tree #543
 22. Middle of the Linked List #876
 23. Maximum Depth of Binary Tree #104

easy/add-binary-67.md

@@ -28,16 +28,59 @@ Output: "10101"
 
 ### Pseudocode
 
-### Initial Solution
+1. Convert to decimal
+2. Add numbers
+3. Use toString method base 2 to convert to binary and return
 
-This solution has a time complexity
+### Initial Attempt
 
-```javascript
+This solution has a time complexity O(a+b) however the solution doesn't work due to overflow.
 
+```javascript
+const addBinary = function (a, b) {
+  const aDec = parseInt(a, 2);
+  const bDec = parseInt(b, 2);
+  const sum = aDec + bDec;
+  return sum.toString(2);
+};
 ```
 
 ### Optimized Solution
 
-```javascript
+This solution has a time complexity of O(n) where n is the larger string and a space complexity of O(n). You can see an explanation of this approach here: https://www.youtube.com/watch?v=keuWJ47xG8g
 
+```javascript
+const addBinary = function (a, b) {
+  let result = '',
+    carry = 0,
+    aPoint = a.length - 1,
+    bPoint = b.length - 1;
+
+  for (let i = 0; i < Math.max(a.length, b.length); i++) {
+    let sum = carry;
+
+    // Make sure pointer is inbounds
+    const aVal = a[aPoint] ? +a[aPoint] : 0;
+    const bVal = b[bPoint] ? +b[bPoint] : 0;
+
+    // Update sum and carry based on division and modulo
+    sum += aVal + bVal;
+    carry = Math.floor(sum / 2);
+    sum = sum % 2;
+
+    // Add to sum to front of result string
+    result = String(sum) + result;
+
+    // Move to the next character left in each string
+    aPoint--;
+    bPoint--;
+  }
+
+  // If any leftover carry add a 1 to beginning of result
+  if (carry) {
+    result = '1' + result;
+  }
+
+  return result;
+};
 ```