more mocks

Author: jmportilla

Mock Interviews/Ride Share Start-Up Company/Ride Share Company - Interview Questions - SOLUTIONS/.ipynb_checkpoints/On-Site Question 2 - SOLUTION-checkpoint.ipynb

@@ -14,6 +14,68 @@
     "\n",
     "** Write out your work on paper/pencil, then see if you can code up your solution **"
    ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Solution\n",
+    "\n",
+    "This is a classic interview problem, so classic that you've already seen it when we discussed memoization and recursion! Refer to that lecture to see the full explanation! Here it is duplicated below:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": [
+    "def rec_coin_dynam(target,coins,known_results):\n",
+    "    '''\n",
+    "    INPUT: This funciton takes in a target amount and a list of possible coins to use.\n",
+    "    It also takes a third parameter, known_results, indicating previously calculated results.\n",
+    "    The known_results parameter shoud be started with [0] * (target+1)\n",
+    "    \n",
+    "    OUTPUT: Minimum number of coins needed to make the target.\n",
+    "    '''\n",
+    "    \n",
+    "    # Default output to target\n",
+    "    min_coins = target\n",
+    "    \n",
+    "    # Base Case\n",
+    "    if target in coins:\n",
+    "        known_results[target] = 1\n",
+    "        return 1\n",
+    "    \n",
+    "    # Return a known result if it happens to be greater than 1\n",
+    "    elif known_results[target] > 0:\n",
+    "        return known_results[target]\n",
+    "    \n",
+    "    else:\n",
+    "        # for every coin value that is <= than target\n",
+    "        for i in [c for c in coins if c <= target]:\n",
+    "            \n",
+    "            # Recursive call, note how we include the known results!\n",
+    "            num_coins = 1 + rec_coin_dynam(target-i,coins,known_results)\n",
+    "            \n",
+    "            # Reset Minimum if we have a new minimum\n",
+    "            if num_coins < min_coins:\n",
+    "                min_coins = num_coins\n",
+    "                \n",
+    "                # Reset the known result\n",
+    "                known_results[target] = min_coins\n",
+    "                \n",
+    "    return min_coins"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Good Job!"
+   ]
   }
  ],
  "metadata": {

Mock Interviews/Ride Share Start-Up Company/Ride Share Company - Interview Questions - SOLUTIONS/On-Site Question 2 - SOLUTION.ipynb

@@ -14,6 +14,68 @@
     "\n",
     "** Write out your work on paper/pencil, then see if you can code up your solution **"
    ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Solution\n",
+    "\n",
+    "This is a classic interview problem, so classic that you've already seen it when we discussed memoization and recursion! Refer to that lecture to see the full explanation! Here it is duplicated below:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": [
+    "def rec_coin_dynam(target,coins,known_results):\n",
+    "    '''\n",
+    "    INPUT: This funciton takes in a target amount and a list of possible coins to use.\n",
+    "    It also takes a third parameter, known_results, indicating previously calculated results.\n",
+    "    The known_results parameter shoud be started with [0] * (target+1)\n",
+    "    \n",
+    "    OUTPUT: Minimum number of coins needed to make the target.\n",
+    "    '''\n",
+    "    \n",
+    "    # Default output to target\n",
+    "    min_coins = target\n",
+    "    \n",
+    "    # Base Case\n",
+    "    if target in coins:\n",
+    "        known_results[target] = 1\n",
+    "        return 1\n",
+    "    \n",
+    "    # Return a known result if it happens to be greater than 1\n",
+    "    elif known_results[target] > 0:\n",
+    "        return known_results[target]\n",
+    "    \n",
+    "    else:\n",
+    "        # for every coin value that is <= than target\n",
+    "        for i in [c for c in coins if c <= target]:\n",
+    "            \n",
+    "            # Recursive call, note how we include the known results!\n",
+    "            num_coins = 1 + rec_coin_dynam(target-i,coins,known_results)\n",
+    "            \n",
+    "            # Reset Minimum if we have a new minimum\n",
+    "            if num_coins < min_coins:\n",
+    "                min_coins = num_coins\n",
+    "                \n",
+    "                # Reset the known result\n",
+    "                known_results[target] = min_coins\n",
+    "                \n",
+    "    return min_coins"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Good Job!"
+   ]
   }
  ],
  "metadata": {

Recursion/Recursion Interview Problems/Recursion Problems - SOLUTIONS/.ipynb_checkpoints/Recursion Problem 4 - Coin Change - SOLUTION-checkpoint.ipynb

@@ -300,7 +300,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython2",
-   "version": "2.7.10"
+   "version": "2.7.11"
   }
  },
  "nbformat": 4,

Recursion/Recursion Interview Problems/Recursion Problems - SOLUTIONS/Recursion Problem 4 - Coin Change - SOLUTION.ipynb

@@ -300,7 +300,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython2",
-   "version": "2.7.10"
+   "version": "2.7.11"
   }
  },
  "nbformat": 4,