LeetCode 188: Best Time to Buy and Sell Stock IV Solution

Master LeetCode problem 188 (Best Time to Buy and Sell Stock IV), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

188. Best Time to Buy and Sell Stock IV

Hard

Array Dynamic Programming

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. We will create a 2D array dp where dp[i][j] represents the maximum profit we can make up to day i with at most j transactions.

Initialize the dp array with dimensions prices.length x k+1.
Perform dynamic programming to fill in the dp array.
Iterate through each day and each possible number of transactions to update the dp array based on two cases: not making a transaction on that day, or making a transaction on that day.
The final answer will be dp[prices.length - 1][k].

Time Complexity: O(nk) Space Complexity: O(nk)

Solution Code

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

        int n = prices.length;
        if (k >= n / 2) {
            int maxProfit = 0;
            for (int i = 1; i < n; i++) {
                if (prices[i] > prices[i - 1]) {
                    maxProfit += prices[i] - prices[i - 1];
                }
            }
            return maxProfit;
        }

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

        return dp[n - 1][k];
    }
}

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