My LeetCode Diary - Day50 Dynamic Programming

123. Best Time to Buy and Sell Stock III

Link

class Solution {
    public int maxProfit(int[] prices) {
        int n = prices.length;
        int max_k = 2;
        int[][][] dp = new int [n][max_k+1][2];

        for (int i = 0; i<n;i++) {
            for(int k = max_k; k >= 1; k--) {
                if (i-1 == -1) {
                    dp[i][k][0] = 0;
                    dp[i][k][1] = -prices[i];
                    continue;
                }
                if (i-2 == -1) {
                    dp[i][k][0] = Math.max(dp[i-1][k][0],dp[i-1][k][1]+prices[i]);
                    dp[i][k][1] = Math.max(dp[i-1][k][1], -prices[i]);
                    continue;
                }
                dp[i][k][0] = Math.max(dp[i-1][k][0],dp[i-1][k][1]+prices[i]);
                dp[i][k][1] = Math.max(dp[i-1][k][1], dp[i-1][k-1][0]-prices[i]);
            }
        }
        return dp[n-1][max_k][0];
    }
}

188. Best Time to Buy and Sell Stock IV

Link

class Solution {
    public int maxProfit(int k, int[] prices) {
        if (prices.length == 0) return 0;

        int len = prices.length;
        int[][][] dp = new int[len][k + 1][2];
        
        for (int i = 0; i <= k; i++) {
            dp[0][i][1] = -prices[0];
        }

        for (int i = 1; i < len; i++) {
            for (int j = 1; j <= k; j++) {
                dp[i][j][0] = Math.max(dp[i - 1][j][0], dp[i - 1][j][1] + prices[i]);
                dp[i][j][1] = Math.max(dp[i - 1][j][1], dp[i - 1][j - 1][0] - prices[i]);
            }
        }
        return dp[len - 1][k][0];
    }
}