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# Best Time to Buy and Sell Stock IV LeetCode Solution

## Problem – Best Time to Buy and Sell Stock IV LeetCode Solution

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`

## Best Time to Buy and Sell Stock IV LeetCode Solution in Java

``````    public int maxProfit(int k, int[] prices) {
int len = prices.length;
if (k >= len / 2) return quickSolve(prices);

int[][] t = new int[k + 1][len];
for (int i = 1; i <= k; i++) {
int tmpMax =  -prices;
for (int j = 1; j < len; j++) {
t[i][j] = Math.max(t[i][j - 1], prices[j] + tmpMax);
tmpMax =  Math.max(tmpMax, t[i - 1][j - 1] - prices[j]);
}
}
return t[k][len - 1];
}

private int quickSolve(int[] prices) {
int len = prices.length, profit = 0;
for (int i = 1; i < len; i++)
// as long as there is a price gap, we gain a profit.
if (prices[i] > prices[i - 1]) profit += prices[i] - prices[i - 1];
return profit;
}
``````

## Best Time to Buy and Sell Stock IV LeetCode Solution in C++

``````class Solution {
public:
int maxProfit(int k, vector<int> &prices) {
int n = (int)prices.size(), ret = 0, v, p = 0;
priority_queue<int> profits;
stack<pair<int, int> > vp_pairs;
while (p < n) {
// find next valley/peak pair
for (v = p; v < n - 1 && prices[v] >= prices[v+1]; v++);
for (p = v + 1; p < n && prices[p] >= prices[p-1]; p++);
// save profit of 1 transaction at last v/p pair, if current v is lower than last v
while (!vp_pairs.empty() && prices[v] < prices[vp_pairs.top().first]) {
profits.push(prices[vp_pairs.top().second-1] - prices[vp_pairs.top().first]);
vp_pairs.pop();
}
// save profit difference between 1 transaction (last v and current p) and 2 transactions (last v/p + current v/p),
// if current v is higher than last v and current p is higher than last p
while (!vp_pairs.empty() && prices[p-1] >= prices[vp_pairs.top().second-1]) {
profits.push(prices[vp_pairs.top().second-1] - prices[v]);
v = vp_pairs.top().first;
vp_pairs.pop();
}
vp_pairs.push(pair<int, int>(v, p));
}
// save profits of the rest v/p pairs
while (!vp_pairs.empty()) {
profits.push(prices[vp_pairs.top().second-1] - prices[vp_pairs.top().first]);
vp_pairs.pop();
}
// sum up first k highest profits
for (int i = 0; i < k && !profits.empty(); i++) {
ret += profits.top();
profits.pop();
}
return ret;
}
};
``````

## Best Time to Buy and Sell Stock IV LeetCode Solution in Python

``````    def maxProfit(self, k, prices):
n = len(prices)
if n < 2:
return 0
dp = [[0 for _ in range(k+1)] for _ in range(n)]
for k1 in range(1, k+1):
for i in range(1, n):
dp[i][k1] = dp[i-1][k1]
for j in range(i):
tmp = prices[i] - prices[j]
tmp += dp[j][k1-1] if j > 0 and k1 - 1 > 0 else 0
dp[i][k1] = max(dp[i][k1], tmp)
return dp[n-1][k]
``````
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