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mock3
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‎Mock Interviews/Ride Share Start-Up Company/Ride Share Company - Interview Questions - SOLUTIONS/.ipynb_checkpoints/On-Site Question 2 - SOLUTION-checkpoint.ipynb‎

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"source": [
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"## Solution\n",
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"\n",
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"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:"
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"This is a classic interview problem, so classic that you've already seen a very similar problem in the recursion section! Make sure to review that problem first before reading our solution here!\n",
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"\n",
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"In this solution we will use a [bottom-up](https://en.wikipedia.org/wiki/Top-down_and_bottom-up_design) algorithm.\n",
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"\n",
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"* As we iterate through each coin, we are adding the ways of making arr[i - coin] to arr[i]\n",
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"* If we have 2 ways of making 4, and are now iterating on a coin of value 3, there should be 2 ways of making 7.\n",
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"* We are essentially adding the coin we are iterating on to the number of ways of making arr[i]."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"execution_count": 5,
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"metadata": {
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"collapsed": true
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},
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"outputs": [],
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"source": [
35-
"def rec_coin_dynam(target,coins,known_results):\n",
36-
" '''\n",
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" INPUT: This funciton takes in a target amount and a list of possible coins to use.\n",
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" It also takes a third parameter, known_results, indicating previously calculated results.\n",
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" The known_results parameter shoud be started with [0] * (target+1)\n",
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" \n",
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" OUTPUT: Minimum number of coins needed to make the target.\n",
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" '''\n",
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"def solution(n, coins):\n",
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" \n",
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" # Default output to target\n",
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" min_coins = target\n",
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" # Set up our array for trakcing results\n",
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" arr = [1] + [0] * n\n",
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" \n",
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" # Base Case\n",
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" if target in coins:\n",
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" known_results[target] = 1\n",
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" return 1\n",
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" \n",
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" # Return a known result if it happens to be greater than 1\n",
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" elif known_results[target] > 0:\n",
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" return known_results[target]\n",
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" \n",
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" else:\n",
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" # for every coin value that is <= than target\n",
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" for i in [c for c in coins if c <= target]:\n",
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" for coin in coins:\n",
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" for i in range(coin, n + 1):\n",
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" arr[i] += arr[i - coin]\n",
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" \n",
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" # Recursive call, note how we include the known results!\n",
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" num_coins = 1 + rec_coin_dynam(target-i,coins,known_results)\n",
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" \n",
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" # Reset Minimum if we have a new minimum\n",
64-
" if num_coins < min_coins:\n",
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" min_coins = num_coins\n",
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" \n",
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" # Reset the known result\n",
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" known_results[target] = min_coins\n",
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" \n",
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" return min_coins"
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" if n == 0:\n",
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" return 0\n",
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" else:\n",
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" return arr[n]\n",
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" "
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]
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},
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{
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"cell_type": "code",
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"execution_count": 6,
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"collapsed": false
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"outputs": [
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{
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"data": {
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"text/plain": [
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"884"
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"execution_count": 6,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"solution(100, [1, 2, 3])"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"This solution results in O((m)(n)) with m being the number of coins, where we iterate about n operations. This is O(n) space."
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]
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},
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{

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

Lines changed: 48 additions & 35 deletions
Original file line numberDiff line numberDiff line change
@@ -21,53 +21,66 @@
2121
"source": [
2222
"## Solution\n",
2323
"\n",
24-
"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:"
24+
"This is a classic interview problem, so classic that you've already seen a very similar problem in the recursion section! Make sure to review that problem first before reading our solution here!\n",
25+
"\n",
26+
"In this solution we will use a [bottom-up](https://en.wikipedia.org/wiki/Top-down_and_bottom-up_design) algorithm.\n",
27+
"\n",
28+
"* As we iterate through each coin, we are adding the ways of making arr[i - coin] to arr[i]\n",
29+
"* If we have 2 ways of making 4, and are now iterating on a coin of value 3, there should be 2 ways of making 7.\n",
30+
"* We are essentially adding the coin we are iterating on to the number of ways of making arr[i]."
2531
]
2632
},
2733
{
2834
"cell_type": "code",
29-
"execution_count": 1,
35+
"execution_count": 5,
3036
"metadata": {
3137
"collapsed": true
3238
},
3339
"outputs": [],
3440
"source": [
35-
"def rec_coin_dynam(target,coins,known_results):\n",
36-
" '''\n",
37-
" INPUT: This funciton takes in a target amount and a list of possible coins to use.\n",
38-
" It also takes a third parameter, known_results, indicating previously calculated results.\n",
39-
" The known_results parameter shoud be started with [0] * (target+1)\n",
40-
" \n",
41-
" OUTPUT: Minimum number of coins needed to make the target.\n",
42-
" '''\n",
41+
"def solution(n, coins):\n",
4342
" \n",
44-
" # Default output to target\n",
45-
" min_coins = target\n",
43+
" # Set up our array for trakcing results\n",
44+
" arr = [1] + [0] * n\n",
4645
" \n",
47-
" # Base Case\n",
48-
" if target in coins:\n",
49-
" known_results[target] = 1\n",
50-
" return 1\n",
51-
" \n",
52-
" # Return a known result if it happens to be greater than 1\n",
53-
" elif known_results[target] > 0:\n",
54-
" return known_results[target]\n",
55-
" \n",
56-
" else:\n",
57-
" # for every coin value that is <= than target\n",
58-
" for i in [c for c in coins if c <= target]:\n",
46+
" for coin in coins:\n",
47+
" for i in range(coin, n + 1):\n",
48+
" arr[i] += arr[i - coin]\n",
5949
" \n",
60-
" # Recursive call, note how we include the known results!\n",
61-
" num_coins = 1 + rec_coin_dynam(target-i,coins,known_results)\n",
62-
" \n",
63-
" # Reset Minimum if we have a new minimum\n",
64-
" if num_coins < min_coins:\n",
65-
" min_coins = num_coins\n",
66-
" \n",
67-
" # Reset the known result\n",
68-
" known_results[target] = min_coins\n",
69-
" \n",
70-
" return min_coins"
50+
" if n == 0:\n",
51+
" return 0\n",
52+
" else:\n",
53+
" return arr[n]\n",
54+
" "
55+
]
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},
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{
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"cell_type": "code",
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"execution_count": 6,
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"metadata": {
61+
"collapsed": false
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},
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"outputs": [
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{
65+
"data": {
66+
"text/plain": [
67+
"884"
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]
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},
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"execution_count": 6,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"solution(100, [1, 2, 3])"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
82+
"source": [
83+
"This solution results in O((m)(n)) with m being the number of coins, where we iterate about n operations. This is O(n) space."
7184
]
7285
},
7386
{

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