# Best Time to Buy and Sell Stock IV

This page explains Java solution to problem Best Time to Buy and Sell Stock IV using Dynamic Programming algorithm.

## Problem Statement

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).

Example 1:

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

Example 2:

Input: [3,2,6,5,0,3], k = 2
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.

## Solution

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package com.vc.hard;

import java.util.*;

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];
}
}

## Time Complexity

O(N * K) Where
N is total number of days in an input array.
K is total number of allowed transactions.

## Space Complexity

O(K) Where
K is total number of allowed transactions.