152. Best Time to Buy and Sell Stock IV

Generalize the two-transaction trick to k: keep a buy and a sell value for each of the k transactions. For every price, update transaction j's buy from transaction j-1's sell, and its sell from its own buy. If k is large enough to allow a trade on every rise, fall back to the unlimited-transaction greedy sum.

Idea

With at most k transactions, keep two arrays: buy[j] = best balance after the jth buy, sell[j] = best after the jth sell. For each price, transaction j buys off the profit of transaction j-1's sell:

buy[j]  = max(buy[j],  sell[j-1] - price)
sell[j] = max(sell[j], buy[j]  + price)

Shortcut: if k >= n/2, there aren't enough days to constrain us — any number of trades is allowed — so just sum every positive day-to-day rise (the Stock II answer).

Algorithm

1. If k >= n/2, return the sum of all positive consecutive differences.
2. Otherwise buy[1..k] = -∞, sell[0..k] = 0. For each price, update buy[j], sell[j] for j from 1 to k. Return sell[k].

Worked example: k = 2, prices = [3,2,6,5,0,3]

Buy at 2 sell at 6 (+4); buy at 0 sell at 3 (+3). Total 7.

Complexity

• Time: O(n·k) (or O(n) via the shortcut).
• Space: O(k).

import java.util.Arrays;

class Solution {
    public int maxProfit(int k, int[] prices) {
        int n = prices.length;
        if (n == 0) {
            return 0;
        }
        if (k >= n / 2) {
            int profit = 0;
            for (int i = 1; i < n; i++) {
                if (prices[i] > prices[i - 1]) {
                    profit += prices[i] - prices[i - 1];
                }
            }
            return profit;
        }
        int[] buy = new int[k + 1];
        int[] sell = new int[k + 1];
        Arrays.fill(buy, Integer.MIN_VALUE);
        for (int p : prices) {
            for (int j = 1; j <= k; j++) {
                buy[j] = Math.max(buy[j], sell[j - 1] - p);
                sell[j] = Math.max(sell[j], buy[j] + p);
            }
        }
        return sell[k];
    }
}