# Easy understanding and can be easily modified to different situations Java Solution

• The basic idea is to create two tables. hold and unhold.

hold[i][j] means the maximum profit with at most j transaction for 0 to i-th day. hold means you have a stock in your hand.

unhold[i][j] means the maximum profit with at most j transaction for 0 to i-th day. unhold means you don't have a stock in your hand.

The equation is

hold[i][j] = Math.max(unhold[i-1][j]-prices[i],hold[i-1][j]);

unhold[i][j] = Math.max(hold[i-1][j-1]+prices[i],unhold[i-1][j]);

when you sell your stock this is a transaction but when you buy a stock, it is not considered as a full transaction. so this is why the two equation look a little different.

And we have to initiate hold table when k = 0.

When the situation is you can not buy a new stock at the same day when you sell it. For example you can only buy a new stock after one day you sell it. The same idea. Another situation is when you have to pay a transaction fee for each transaction, just make a modification when you sell it, So just change the unhold equation a little.

``````public class Solution {
//hold[i][k]  ith day k transaction have stock and maximum profit
//unhold[i][k] ith day k transaction do not have stock at hand and maximum profit
public int maxProfit(int k, int[] prices) {
if(k>prices.length/2) return maxP(prices);
int[][] hold = new int[prices.length][k+1];
int[][] unhold = new int[prices.length][k+1];
hold[0][0] = -prices[0];
for(int i=1;i<prices.length;i++) hold[i][0] = Math.max(hold[i-1][0],-prices[i]);
for(int j=1;j<=k;j++) hold[0][j] = -prices[0];
for(int i=1;i<prices.length;i++){
for(int j=1;j<=k;j++){
hold[i][j] = Math.max(unhold[i-1][j]-prices[i],hold[i-1][j]);
unhold[i][j] = Math.max(hold[i-1][j-1]+prices[i],unhold[i-1][j]);
}
}
return Math.max(hold[prices.length-1][k],unhold[prices.length-1][k]);
}
public int maxP(int[] prices){
int res =0;
for(int i=0;i<prices.length;i++){
if(i>0 && prices[i] > prices[i-1]){
res += prices[i]-prices[i-1];
}
}
return res;
}
}``````

• nice dynamic programming function, easy understanding, THX :)

• Very nice and generalize solution. This can be extended to more complicated questions.

Based on your idea, this is the shorter version in Java:

``````public int maxProfit(int k, int[] prices) {
if(k == 0 || prices.length < 2)
return 0;
if(k > prices.length / 2)
return noLimit(prices);

// hold[i][j]: For at most i transactions, the max profit on jth day with a stock in hand.
// unhold[i][j]: For at most i transactions, the max profit on jth day without a stock in hand
int[][] hold = new int[k+1][prices.length];
int[][] unhold = new int[k+1][prices.length];
for(int i = 1; i <= k; i++) {
hold[i][0] = -prices[0];
unhold[i][0] = 0;
for(int j = 1; j < prices.length; j++) {
unhold[i][j] = Math.max(prices[j] + hold[i][j-1], unhold[i][j-1]); // Sell or not sell
}
}
return unhold[k][prices.length-1];
}
private int noLimit(int[] prices) { // Solution from Best Time to Buy and Sell Stock II
int max = 0;
for(int i = 0; i < prices.length-1; i++) {
if(prices[i+1] > prices[i])
max += prices[i+1] - prices[i];
}
return max;
}
``````

This solution can also be applied to Best Time to Buy and Sell Stock II & III

• Nice solution! My solution is similar, but only takes o(k) extra space.

``````public class Solution {
public int maxProfit(int k, int[] prices) {
if (prices == null || prices.length <= 1 || k == 0) { return 0;}
if (k >= prices.length / 2) {return unlimited(prices);}
int[] soldN = new int[k + 1], buyN = new int[k + 1];
for (int i = 0; i <= k; i++) {
}
for (int i = 0; i < prices.length; i++) {
for (int j = 1; j <= k; j++) {
soldN[j] = Math.max(soldN[j], buyN[j] + prices[i]);
}
}
return soldN[k];
}

private int unlimited(int[] prices) {
int sum = 0;
for (int i = 0; i < prices.length - 1; i++) {
sum += prices[i + 1] - prices[i] > 0 ? prices[i + 1] - prices[i] : 0;
}
return sum;
}
}
``````

• You can just simply return `unhold[prices.length-1][k]` instead of `Math.max(hold[prices.length-1][k],unhold[prices.length-1][k])`.

`unhold[prices.length-1][k]` must be greater than or equal to `hold[prices.length-1][k]`

• This will cost O(k * n) space, did you get Memory-Outof-limit errror?

• @guangstick could you explain what do soldN and buyN mean ?

• I have been looking through solutions in the forum for two days, yours is the only solution that I an understand quickly, and most importantly, what I could probably think of by myself during a real interview.

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