LeetCode – Refresh – Best Time to Buy and Sell Stock iv

This is combination of II and III.

1. when k >= length, it is totally II.

2. when k < length, it is another III. A localMaximum and globalMaximum subarray need to be maintained. But we can reduce the two dimensional DP to one dimensional DP

localMax[i][j] = max(local[i-1][j] + diff, globalMax[i-1][j-1] + max(0, diff))    ==>    localMax[j] = max(localMax[j] + diff, globalMax[j-1] + max(0, diff))

globalMax[i][j] = max(globalMax[i-1][j], localMax[i][j])      ==>    globalMax[j] = max(globalMax[j], localMax[j])

 

For here ,diff = prices[i+1] - prices[i]. max(0, diff) means only when the profit is positive, we add it to globalMax.

 

class Solution {
public:
    int maxProfit(int k, vector<int> &prices) {
        int len = prices.size(), result = 0;
        if (len < 2) return 0;
        if (len <= k) {
            for (int i = 0; i < len-1; i++) {
                if (prices[i+1] - prices[i] > 0) {
                    result += prices[i+1] - prices[i];
                }
            }
        } else {
            vector<int> lMax(k+1, 0), gMax(k+1, 0);
            for (int i = 0; i < len-1; i++) {
                int diff = prices[i+1] - prices[i];
                for (int j = k; j >= 1; j--) {
                    lMax[j] = max(gMax[j-1] + max(0, diff), lMax[j] + diff);
                    gMax[j] = max(lMax[j], gMax[j]);
                }
            }
            result = gMax[k];
        }
        return result;
    }
};