188. Best Time to Buy and Sell Stock IV
You are given an integer array prices
where prices[i]
is the price of a given stock on the ith
day, and an integer k
.
Find the maximum profit you can achieve. You may complete at most k
transactions.
Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
Example 1:
Input: k = 2, prices = [2,4,1] Output: 2 Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.
Example 2:
Input: k = 2, prices = [3,2,6,5,0,3] Output: 7 Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4. Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
Constraints:
0 <= k <= 100
0 <= prices.length <= 1000
0 <= prices[i] <= 1000
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class Solution { | |
public: | |
/* | |
Framework DP (with Doing Nothing) | |
1. function | |
- states | |
i: ith day | |
t: remaining transaction | |
h: if we are holding stock or not | |
- dp[i][t][h]: ith day, t transaction remains, with hold or not = max profit | |
- answer | |
- dp[0][k][0] | |
2. relation | |
- (holding) dp[i][t][1] = max(dp[i+1][t][1], dp[i+1][t-1][0] - prices[i]) | |
- (not holding) dp[i][t][0] = max(dp[i+1][t][0], dp[i+1][t-1][1] + prices[i]) | |
3. base | |
- (t <= 0) 0 | |
- (i > prices.leng() - 1) 0 | |
*/ | |
vector<vector<vector<int>>> memo; | |
vector<int> p; | |
int dp(int i, int t, int h){ | |
// base | |
if(i == p.size() or t == 0) | |
return 0; | |
// relation | |
if(memo[i][t][h] == INT_MIN){ | |
int donothing = dp(i + 1, t, h); | |
// sell | |
if(h == 1) | |
memo[i][t][h] = max(donothing, p[i] + dp(i + 1, t - 1, 0)); | |
// buy | |
else | |
memo[i][t][h] = max(donothing, -p[i] + dp(i + 1, t, 1)); | |
} | |
return memo[i][t][h]; | |
} | |
int maxProfit(int k, vector<int>& prices) { | |
int n = prices.size(); | |
p = prices; | |
memo.resize(n); | |
for(auto &m: memo){ | |
m.resize(k + 1); | |
for(auto &i: m) | |
i.resize(2, INT_MIN); | |
}; | |
return dp(0, k, 0); | |
} | |
}; |
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