# Best Time To Buy And Sell Stock IV Problem

## Description

LeetCode Problem 188.

You are given an integer array prices where prices[i] is the price of a given stock on the i^th 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:

``````1
2
3
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:

``````1
2
3
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

## Sample C++ Code

``````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
class Solution {
public:
int maxProfit(int k, vector<int>& prices) {
int maxProfit=0;

if (prices.size()<2)
return 0;
if (k>prices.size()/2) {
for (int i=1; i<prices.size(); i++)
maxProfit += max(prices[i]-prices[i-1], 0);
return maxProfit;
}

int hold[k+1];
int rele[k+1];
for (int i=0;i<=k;++i) {
hold[i] = INT_MAX;
rele[i] = 0;
}

for (int i=0; i<prices.size(); i++) {
for (int j=k; j>=1; j--) {
rele[j] = max(rele[j], prices[i]-hold[j]);
hold[j] = min(hold[j], prices[i]-rele[j-1]);
maxProfit = max(maxProfit, rele[j]);
}
}
return maxProfit;
}
};
``````