-
Notifications
You must be signed in to change notification settings - Fork 0
/
Copy path123. Best Time to Buy and Sell Stock III.cpp
193 lines (152 loc) · 6.21 KB
/
123. Best Time to Buy and Sell Stock III.cpp
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
// -----Approach 1: Memoization ------------------------------------------------------------
/*
Problem Link: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii
Time: 844 ms (Beats 12.74%), Space: 218 MB (Beats 18.18%)
*/
class Solution {
private:
int sol(int idx, int buy, int limit, vector<vector<vector<int>>>& dp, vector<int>& prices){
if(limit == 0 || idx == prices.size()) return 0;
if(dp[idx][buy][limit] != -1) return dp[idx][buy][limit];
if(buy){
return dp[idx][buy][limit]= max(-prices[idx] + sol(idx+1, 0, limit, dp, prices), sol(idx+1, 1, limit, dp, prices));
} else {
return dp[idx][buy][limit]= max(prices[idx] + sol(idx+1, 1, limit-1, dp, prices), sol(idx+1, 0, limit, dp, prices));
}
}
public:
int maxProfit(vector<int>& prices) {
int n= prices.size();
vector<vector<vector<int>>> dp(n, vector<vector<int>>(2, vector<int>(3, -1)));
// buy/not buy->2 size & transaction till two->3 size
return sol(0, 1, 2, dp, prices);
}
};
// -----Approach 2: Memoization ------------------------------------------------------------
/*
Problem Link: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii
Time: 399 ms (Beats 53.97%), Space: 132.7 MB (Beats 53.66%)
*/
class Solution {
private:
int sol(int idx, int limit, vector<vector<int>>& dp, vector<int>& prices){
if(limit == 4 || idx == prices.size()) return 0;
if(dp[idx][limit] != -1) return dp[idx][limit];
if(limit%2 == 0){
return dp[idx][limit]= max(-prices[idx] + sol(idx+1, limit+1, dp, prices), sol(idx+1, limit, dp, prices));
} else {
return dp[idx][limit]= max(prices[idx] + sol(idx+1, limit+1, dp, prices), sol(idx+1, limit, dp, prices));
}
}
public:
int maxProfit(vector<int>& prices) {
int n= prices.size();
vector<vector<int>> dp(n, vector<int>(4, -1));
// we have a limit of two transacations: buy (0:even) , sell (1:odd), buy(0:even), sell(1:odd) -> so size 4
return sol(0, 0, dp, prices);
}
};
// -----Approach 3: Tabulation ------------------------------------------------------------
/*
Problem Link: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii
Time: 711 ms (Beats 46.94%), Space: 207.2 MB (Beats 36.3%)
*/
class Solution {
public:
int maxProfit(vector<int>& prices) {
int n= prices.size();
vector<vector<vector<int>>> dp(n+1, vector<vector<int>>(2, vector<int>(3, 0)));
// these base cases are not necessary since dp is initialised to 0, just for practice
// --> the case of limit == 0 in memoization
// for(int idx=0; idx<n; idx++){
// for(int buy=0; buy<=1; buy++){
// dp[idx][buy][0]= 0;
// }
// }
// --> the case of idx == n in memoization
// for(int buy=0; buy<=1; buy++){
// for(int limit=0; limit<=2; limit++){
// dp[n][buy][limit]= 0;
// }
// }
for(int idx=n-1; idx>=0; idx--){
for(int buy=0; buy<=1; buy++){
for(int limit=1; limit<=2; limit++){ // limit=1 case already covered by base case
if(buy){
dp[idx][buy][limit]= max(-prices[idx] + dp[idx+1][0][limit], dp[idx+1][1][limit]);
} else {
dp[idx][buy][limit]= max(prices[idx] + dp[idx+1][1][limit-1], dp[idx+1][0][limit]);
}
}
}
}
return dp[0][1][2];
}
};
// -----Approach 4: Tabulation ------------------------------------------------------------
/*
Problem Link: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii
Time: 354 ms (Beats 57.46%), Space: 135.2 MB (Beats 52.3%)
*/
class Solution {
public:
int maxProfit(vector<int>& prices) {
int n= prices.size();
vector<vector<int>> dp(n+1, vector<int>(5, 0));
// for(int idx=1; idx<=n; idx++) dp[idx][4]= 0; // base case
for(int idx=n-1; idx>=0; idx--){
for(int limit=3; limit>=0; limit--){
if(limit%2 == 0){
dp[idx][limit]= max(-prices[idx] + dp[idx+1][limit+1], dp[idx+1][limit]);
} else {
dp[idx][limit]= max(prices[idx] + dp[idx+1][limit+1], dp[idx+1][limit]);
}
}
}
return dp[0][0];
}
};
// -----Approach 5: Space Optimization ------------------------------------------------------------
/*
Problem Link: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii
Time: 239 ms (Beats 66.15%), Space: 75.3 MB (Beats 89.39%)
*/
class Solution {
public:
int maxProfit(vector<int>& prices) {
int n= prices.size();
vector<vector<int>> front(2, vector<int>(3, 0));
vector<vector<int>> curr(2, vector<int>(3, 0));
for(int idx=n-1; idx>=0; idx--){
for(int limit=1; limit<=2; limit++){ // limit=0 case already covered by base case
curr[1][limit]= max(-prices[idx] + front[0][limit], front[1][limit]);
curr[0][limit]= max(prices[idx] + front[1][limit-1], front[0][limit]);
}
front= curr;
}
return front[1][2];
}
};
// -----Approach 6: Space Optimization ------------------------------------------------------------
/*
Problem Link: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii
Time: 187 ms (Beats 73.55%), Space: 75.2 MB (Beats 97.22%)
*/
class Solution {
public:
int maxProfit(vector<int>& prices) {
int n= prices.size();
vector<int> front(5, 0), curr(5, 0);
for(int idx=n-1; idx>=0; idx--){
for(int limit=3; limit>=0; limit--){
if(limit%2 == 0){
curr[limit]= max(-prices[idx] + front[limit+1], front[limit]);
} else {
curr[limit]= max(prices[idx] + front[limit+1], front[limit]);
}
}
front= curr;
}
return front[0];
}
};