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.
The core challenge of this problem is to maximize the profit from stock transactions given a constraint on the number of transactions. The significance of this problem lies in its application to financial markets where traders aim to maximize their returns. A common pitfall is to assume that buying and selling on every profitable opportunity is optimal, which may not be the case when transaction limits are imposed.
To solve this problem, we need to consider both the number of transactions and the prices of the stocks. A naive approach would be to try all possible combinations of buy and sell days, but this is computationally expensive. Instead, we can use dynamic programming to optimize our solution.
The naive solution involves checking all possible pairs of buy and sell days, which results in a time complexity of O(n^2). This is not feasible for large inputs.
We can use dynamic programming to keep track of the maximum profit at each day for up to k transactions. The idea is to maintain two arrays: one for the maximum profit up to the current day with at most j transactions, and another for the maximum profit up to the previous day with at most j-1 transactions.
1. Initialize a 2D array `dp` where `dp[i][j]` represents the maximum profit up to day `i` with at most `j` transactions.
2. Iterate through each day and each transaction, updating the `dp` array based on the maximum profit achievable by either not making a transaction or making a transaction on that day.
3. The final answer will be in `dp[n-1][k]` where `n` is the number of days.
public class Solution {
public int maxProfit(int k, int[] prices) {
if (prices == null || prices.length == 0 || k == 0) {
return 0;
}
int n = prices.length;
if (k >= n / 2) {
return quickSolve(prices);
}
int[][] dp = new int[k + 1][n];
for (int i = 1; i <= k; i++) {
int maxDiff = -prices[0];
for (int j = 1; j < n; j++) {
dp[i][j] = Math.max(dp[i][j - 1], prices[j] + maxDiff);
maxDiff = Math.max(maxDiff, dp[i - 1][j] - prices[j]);
}
}
return dp[k][n - 1];
}
private int quickSolve(int[] prices) {
int profit = 0;
for (int i = 1; i < prices.length; i++) {
if (prices[i] > prices[i - 1]) {
profit += prices[i] - prices[i - 1];
}
}
return profit;
}
}
The time complexity of the optimized solution is O(kn), where `k` is the number of transactions and `n` is the number of days. The space complexity is also O(kn) due to the 2D array used for dynamic programming.
1. If the number of transactions `k` is greater than or equal to half the number of days, we can use a simplified solution that allows unlimited transactions.
2. If the prices array is empty or contains only one element, the maximum profit is 0.
To test the solution, we should consider various scenarios including:
When approaching such problems, it is crucial to break down the problem into smaller subproblems and use dynamic programming to optimize the solution. Practice similar problems to improve your problem-solving skills and understand the underlying concepts.
Understanding and solving the "Best Time to Buy and Sell Stock II" problem helps in grasping dynamic programming concepts and their applications in financial problems. Practice and explore further to enhance your skills.