This page explains Java solution to problem `Best Time to Buy and Sell Stock IV`

using `Dynamic Programming`

algorithm.

Say you have an array for which the `i-th`

element is the price of a given stock on day `i`

.

Design an algorithm to find the maximum profit. You may complete at most `k`

transactions.

**Note**: You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

Input: [2,4,1], k = 2Output: 2Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.

Input: [3,2,6,5,0,3], k = 2Output: 7Explanation:

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.

```
package com.vc.hard;
import java.util.*;
class BestTimeToBuyAndSellStockIv {
public int maxProfit(int k, int[] prices) {
int n = prices.length;
if(n == 0) return 0;
if(2 * k >= n) {// i.e. we can perform unlimited number of transactions
int dp_i_k_0 = 0;
int dp_i_k_1 = Integer.MIN_VALUE;
for(int i = 0; i < prices.length; i++) {
int dp_i_k_0_old = dp_i_k_0;
//Rest or Sell on given Day
dp_i_k_0 = Math.max(dp_i_k_0, dp_i_k_1 + prices[i]);
//Rest or Buy on given Day
dp_i_k_1 = Math.max(dp_i_k_1, dp_i_k_0_old - prices[i]);
}
return dp_i_k_0;
}
/**
Ideally logic below should be enough for this problem but since very large value of K gives us Memory Limit Exceeded exception, so we need logic above
If value of K larger than prices.length / 2 then we know for sure that we are allowed to do unlimited number of transaction, because on a given day we can either Buy, Rest or Sell.
dp_i_0[k]
on ith day, we can perform k at most transaction and we should have zero stock left in our hand
dp_i_1[k]
on ith day, we can perform k at most transaction and we should have one stock left in our hand
*/
int[] dp_i_0 = new int[k + 1];
int[] dp_i_1 = new int[k + 1];
Arrays.fill(dp_i_1, Integer.MIN_VALUE);
for(int i = 0; i < prices.length; i++) {
for(int j = 1; j <= k; j++) {
//Rest or Sell so that we should have zero stock in our hand
dp_i_0[j] = Math.max(dp_i_0[j], dp_i_1[j] + prices[i]);
//Rest or Buy so that we should have one stock in our hand
dp_i_1[j] = Math.max(dp_i_1[j], dp_i_0[j - 1] - prices[i]);
}
}
return dp_i_0[k];
}
}
```

O(N * K) Where

N is total number of days in an input array.

K is total number of allowed transactions.

O(K) Where

K is total number of allowed transactions.