Attempt 1 at assignment 1

Author: smodi23

02_activities/assignments/assignment_1.ipynb

@@ -25,17 +25,25 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 12,
    "metadata": {},
-   "outputs": [],
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "2\n"
+     ]
+    }
+   ],
    "source": [
     "import hashlib\n",
     "\n",
     "def hash_to_range(input_string: str) -> int:\n",
     "     hash_object = hashlib.sha256(input_string.encode())\n",
     "     hash_int = int(hash_object.hexdigest(), 16)\n",
     "     return (hash_int % 3) + 1\n",
-    "input_string = \"your_first_name_here\"\n",
+    "input_string = \"sahil\"\n",
     "result = hash_to_range(input_string)\n",
     "print(result)\n"
    ]
@@ -112,7 +120,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 1,
+   "execution_count": 5,
    "metadata": {},
    "outputs": [],
    "source": [
@@ -169,7 +177,9 @@
    "metadata": {},
    "outputs": [],
    "source": [
-    "# Your answer here"
+    "# Your answer here\n",
+    "# The problem asks us to determine the string consisting of brackets sequence is correct. A valid sequence is that every open bracket\n",
+    "# has a corresponding close bracket of the same type. "
    ]
   },
   {
@@ -185,7 +195,13 @@
    "metadata": {},
    "outputs": [],
    "source": [
-    "# Your answer here"
+    "# Your answer here\n",
+    "\n",
+    "Input: s = \"([{}])\"\n",
+    "Output: True \n",
+    "\n",
+    "Input: s = \"([{)}]\"\n",
+    "Output: False "
    ]
   },
   {
@@ -202,7 +218,23 @@
    "metadata": {},
    "outputs": [],
    "source": [
-    "# Your answer here"
+    "# Your answer here\n",
+    "\n",
+    "def is_valid_brackets(s: str) -> bool:\n",
+    "    stack = []\n",
+    "    matching = {')': '(', ']': '[', '}': '{'}\n",
+    "\n",
+    "    for char in s:\n",
+    "        if char in matching.values():\n",
+    "            stack.append(char)\n",
+    "        elif char in matching:\n",
+    "            if not stack or stack[-1] != matching[char]:\n",
+    "                return False\n",
+    "            stack.pop()\n",
+    "        else:\n",
+    "            return False\n",
+    "\n",
+    "    return not stack"
    ]
   },
   {
@@ -219,7 +251,10 @@
    "metadata": {},
    "outputs": [],
    "source": [
-    "# Your answer here"
+    "# Your answer here\n",
+    "# The solution correctly uses stack to keep track of unmatched opening brackets. The closing bracket is mapped by the dictionary to \n",
+    "# see if it's the correct opening pair. The opening bracket is stored in a list in the stack. The closing bracket is compard to the\n",
+    "# stack, and if it matches, the stack is pop. If it doesn't match, code returns False. "
    ]
   },
   {
@@ -236,7 +271,10 @@
    "metadata": {},
    "outputs": [],
    "source": [
-    "# Your answer here"
+    "# Your answer here\n",
+    "# The time complexity is O(n), where n is the length of the string. The string is only gone through once, performing constant-time \n",
+    "# operations. \n",
+    "# The space complexity is O(n) in the worst case as well, where all characters stored in the stack are opening brackets. "
    ]
   },
   {
@@ -253,7 +291,10 @@
    "metadata": {},
    "outputs": [],
    "source": [
-    "# Your answer here"
+    "# Your answer here\n",
+    "# An alternative solution can be repeatedly removing matching pairs of brackets from string until all matches are removed and no more \n",
+    "# can be removed anymore. This can be done using a loop iteration. All matched pairs will be removed using .replace(). The loop will \n",
+    "# continue until there are no more matched pairs to be removed. In the end, if the string is empty, all brackets matched."
    ]
   },
   {
@@ -301,7 +342,7 @@
  ],
  "metadata": {
   "kernelspec": {
-   "display_name": "Python 3 (ipykernel)",
+   "display_name": "dsi_participant",
    "language": "python",
    "name": "python3"
   },
@@ -315,7 +356,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.11.5"
+   "version": "3.9.15"
   }
  },
  "nbformat": 4,