# O(nk)-time O(k)-space easy to understand 6ms Java solution

• ``````  public int maxProfit(int k, int[] prices) {
int len = prices.length;
if (len <= 1 || k == 0) {
return 0;
}
if (k >= len / 2) {
int max = 0;
for (int i = 1; i < len; i++) {
if (prices[i] > prices[i - 1]) max += prices[i] - prices[i - 1];
}
return max;
}
// Suppose state is defined as f(i,k) where i is the time and k is the transactions remained.
// The problem is how do we know if we are allowed to perform a sell operation or not at a specific state.
// Because with f(i,k), we can't jump into f(i+1 , k-1) because k, representing transactions, involves two operations happened at different time(different i), it doesn't clearly shows the holding status.

// One solution is to expand the state to f(i,k,h) where h could only be 0 or 1, shows whether we are holding a stock or not,
// but this will complicate the calculations and may using extra time and space.

// What we do is expanding k to 2*k to substitute h.
// So totally, including both buying and selling, we are allowed to do 2*k operations.
// To prevent selling before buying, we define even number of k as the time we hold nothing and can only buy a stock.
// and therefore odd number of k as the time we are holding something in hand and can only sell.
// f(i , k) = if k%2==0 we can buy the stock or not
//              buy        f(i+1 , k-1) - prices[i]
//              do nothing f(i+1 , k) (k doesn't change so we still have the right to buy)
//
//            if k%2==1 we are holding a stock right now so only sell it or do nothing
//              sell       f(i+1 , k-1) + prices[i]
//              do nothing f(i+1 , k) (k doesn't change so we are still holding a stock)

// O(nk) Space without optimization
//
// int[][] dp= new int[len+1][k*2+2];
// for(int i = len-1; i>=0; i--){
//     for(int j = k*2; j>0; j--){
//         dp[i][j] = dp[i+1][j];
//         if(j%2==1){
//             dp[i][j] = Math.max(dp[i][j], dp[i+1][j-1]+prices[i]);
//         }else{
//             dp[i][j] = Math.max(dp[i][j], dp[i+1][j-1]-prices[i]);
//         }
//     }
// }

// optimized in space, using O(k) instead of O(nk)
int[] dp = new int[k * 2 + 2];

for (int i = len - 1; i >= 0; i--) {
for (int j = k * 2; j > 0; j--) {
if (j % 2 == 1) {
dp[j] = Math.max(dp[j], dp[j - 1] + prices[i]);
} else {
dp[j] = Math.max(dp[j], dp[j - 1] - prices[i]);
}
}
}
return dp[k * 2];
}``````

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