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.
class Solution {
public int maxProfit(int price[], int i, int n, int K, int dp[][][], int ongoing) {
if (i == n)
return 0;
if (K == 0) {
dp[i][K][ongoing] = 0;
return 0;
}
if (dp[i][K][ongoing] != -1)
return dp[i][K][ongoing];
int ans = maxProfit(price, i + 1, n, K, dp, ongoing);
if (ongoing == 1) {
int option = maxProfit(price, i + 1, n, K - 1, dp, 0) + price[i];
ans = Math.max(ans, option);
} else {
int option = maxProfit(price, i + 1, n, K, dp, 1) - price[i];
ans = Math.max(ans, option);
}
dp[i][K][ongoing] = ans;
return ans;
}
public int maxProfit(int k, int[] prices) {
int n = prices.length;
if (n <= 1)
return 0;
// if k >= n/2, then you can make maximum number of transactions.
// Then this question is similar to infinite transactions
if (k >= n / 2) {
int T_ik0 = 0, T_ik1 = Integer.MIN_VALUE;
for (int price : prices) {
int T_ik0_old = T_ik0;
// If we want 0 stocks in our hand at end of ith day
// Either take ans from 0 stock i-1th day
// Or take ans from 1 stock i-1th day and sell it
T_ik0 = Math.max(T_ik0, T_ik1 + price);
// If we want 1 stock in our hand at end of ith day
// Either take ans from 1 stock i-1th day
// Or take ans from 0 stock i-1th day and but 1 stock at ith day
T_ik1 = Math.max(T_ik1, T_ik0_old - price);
}
return T_ik0;
}
int dp[][][] = new int[n + 1][k + 1][2];
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= k; j++) {
dp[i][j][0] = -1;
dp[i][j][1] = -1;
}
}
return maxProfit(prices, 0, n, k, dp, 0);
}
}