-
Notifications
You must be signed in to change notification settings - Fork 1
/
LeetCode-983-Minimum-Cost-For-Tickets.java
246 lines (192 loc) · 8.55 KB
/
LeetCode-983-Minimum-Cost-For-Tickets.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
class Solution {
// 1. DP. doesn't work
// public int mincostTickets(int[] days, int[] costs) {
// if (days == null || costs == null || days.length == 0 || costs.length == 0) return 0;
// int[][] state = new int[costs.length][days.length];
// state[0][0] = costs[0];
// state[1][0] = costs[1];
// state[2][0] = costs[2];
// int[] s = new int[costs.length];
// for (int j = 1; j < days.length; j++) {
// for (int i = 0; i < costs.length; i++) {
// // update the left boundary if we purchase a costs[i] ticket
// int minVal = 0;
// if (i == 0) {
// minVal = days[j] - 1;
// } else if (i == 1) {
// minVal = days[j] - 7;
// } else if (i == 3) {
// minVal = days[j] - 30;
// }
// int k = s[i];
// while (days[k] < minVal) {
// if (i == 0) {
// System.out.println("i: " + i + " j: " + j + " " + days[k] + " " + minVal);
// }
// k++;
// }
// s[i] = k;
// if (k > 0) {
// state[i][j] = Math.min(state[0][k - 1], Math.min(state[1][k - 1], state[2][k - 1])) + costs[i];
// } else {
// state[i][j] = costs[i];
// }
// }
// System.out.println(Arrays.toString(s));
// }
// System.out.println(Arrays.deepToString(state));
// return Math.min(state[0][days.length - 1], Math.min(state[1][days.length - 1], state[2][days.length - 1]));
// }
/*
1.DP
https://leetcode.com/problems/minimum-cost-for-tickets/discuss/226659/Two-DP-solutions-with-pictures
https://leetcode.com/problems/minimum-cost-for-tickets/discuss/226670/Java-DP-Solution-with-explanation-O(n)
Tracking the costs for all calendar days.
1.Travel day. Minimal cost is min of costs for three tickets.
2.Non travel day. Minimal cost is same as previous day.
*/
// public int mincostTickets(int[] days, int[] costs) {
// if (days == null || costs == null || days.length == 0 || costs.length == 0) return 0;
// int[] state = new int[366];
// state[0] = 0;
// int j = 0;
// for (int i = 1; i <= 365; i++) {
// if (j < days.length && i == days[j]) {
// // find a travel day
// int c1 = state[i - 1] + costs[0];
// int c7 = state[Math.max(0, i - 7)] + costs[1];
// int c30 = state[Math.max(0, i - 30)] + costs[2];
// state[i] = Math.min(c1, Math.min(c7, c30));
// j++;
// } else {
// // a non-travel day
// state[i] = state[i - 1];
// }
// }
// return state[365];
// }
/*
subproblem:
dp[i] - the min cost of tickets til ith day
recurrence relation:
Assume dp[0, i - 1] has been optimized amount, then ith days's ticket cost cound be transit from i-1 days ticket cost, or i-7, i-30 days ticket cost.
if ith day has travel:
dp[i] = min{dp[i - 1] + costs[0], dp[i - 7] + costs[1], dp[i - 30] + costs[2]}
if ith day doesn't have travel:
dp[i] = dp[i - 1]
init:
dp[0] = 0 // if no travel, then ticket cost is 0
ans:
dp[365]
*/
// public int mincostTickets(int[] days, int[] costs) {
// if (days == null || costs == null || days.length == 0 || costs.length == 0) return 0;
// // subproblem
// int[] dp = new int[365 + 1];
// // init
// dp[0] = 0;
// HashSet<Integer> set = new HashSet<>();
// for (int d : days) set.add(d);
// for (int i = 1; i <= 365; i++) {
// if (set.contains(i)) {
// int c1 = dp[Math.max(0, i - 1)] + costs[0];
// int c2 = dp[Math.max(0, i - 7)] + costs[1];
// int c3 = dp[Math.max(0, i - 30)] + costs[2];
// dp[i] = Math.min(c1, Math.min(c2, c3));
// } else {
// dp[i] = dp[i - 1];
// }
// }
// return dp[365];
// }
/*
Optimized version. Since we only look 30 days back, we can just store the past 30 days' ticket cost.
Time: O(N)
Space: O(30)
*/
// public int mincostTickets(int[] days, int[] costs) {
// if (days == null || costs == null || days.length == 0 || costs.length == 0) return 0;
// int[] state = new int[30];
// state[0] = 0;
// int j = 0;
// for (int i = 1; i <= days[days.length - 1]; i++) {
// if (j < days.length && i == days[j]) {
// // a travel day
// int c1 = state[(i - 1) % 30] + costs[0];
// int c7 = state[Math.max(0, (i - 7) % 30)] + costs[1];
// int c30 = state[Math.max(0, (i - 30) % 30)] + costs[2];
// state[i % 30] = Math.min(c1, Math.min(c7, c30));
// j++;
// } else {
// state[i % 30] = state[(i - 1) % 30];
// }
// }
// return state[days[days.length - 1] % 30];
// }
// 2.
/*
https://leetcode.com/problems/minimum-cost-for-tickets/solution/
Runtime: 0 ms, faster than 100.00% of Java online submissions for Minimum Cost For Tickets.
*/
// int[] days, costs;
// Integer[] memo;
// int[] durations = new int[]{1, 7, 30};
// public int mincostTickets(int[] days, int[] costs) {
// this.days = days;
// this.costs = costs;
// memo = new Integer[days.length];
// return dp(0);
// }
// public int dp(int i) {
// if (i >= days.length)
// return 0;
// if (memo[i] != null)
// return memo[i];
// int ans = Integer.MAX_VALUE;
// int j = i;
// for (int k = 0; k < 3; ++k) {
// while (j < days.length && days[j] < days[i] + durations[k])
// j++;
// ans = Math.min(ans, dp(j) + costs[k]);
// }
// memo[i] = ans;
// return ans;
// }
// DP. Loop the days array (best solution)
/*
https://leetcode.com/problems/minimum-cost-for-tickets/discuss/326266/Only-loop-'days'-array-with-explanation-O(n)-and-comment
subproblem
dp[i] - the ith days's min cost, assuming the dp[o, i - 1] has the most optimized ticket cost
recurrence relation:
dp[i] = min (dp[i - 1] + costs[0], dp[j] + costs[1], dp[k] + costs[2]), j is the day that traveled 7 days ago, k is the day that
traveled 30 days ago
init:
dp[1] = min(costs[0], costs[1], costs[2])
ans:
dp[n - 1]
Runtime: 1 ms, faster than 93.51% of Java online submissions for Minimum Cost For Tickets.
Memory Usage: 34.4 MB, less than 100.00% of Java online submissions for Minimum Cost For Tickets.
*/
public int mincostTickets(int[] days, int[] costs) {
if (days == null || days.length == 0) return 0;
int n = days.length;
// subproblem: dp[i] denotes the min cost travel at the days [0, i]
int[] dp = new int[n];
// init
dp[0] = Math.min(costs[0], Math.min(costs[1], costs[2])); // travel at only one day will only require one day ticket or the cheapest ticket
// days 0 should be calculate again, because it could happen that
// costs [7,2,15], as 7 days ticket is much cheaper than 1 day ticket.
for (int i = 1; i < n; i++) {
dp[i] = dp[i - 1] + costs[0]; // assume we only travel this day using one day cost
// find the day traveled 7 days ago, so we could purchase a 7-day ticket after it
int j = i;
while (j >= 0 && days[i] - days[j] + 1 <= 7) j--;
dp[i] = Math.min(dp[i], j < 0 ? costs[1] : dp[j] + costs[1]);
// find the day traveled 30 days ago, so we could purchase a 30-day ticket after it
// j = i; -> this one is not necessary
while (j >= 0 && days[i] - days[j] + 1 <= 30) j--;
dp[i] = Math.min(dp[i], j < 0 ? costs[2] : dp[j] + costs[2]);
}
return dp[n - 1];
}
}