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 by buying and selling stocks at most k times. 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 multiple transactions can be done simultaneously, which is not allowed.
To solve this problem, we need to consider dynamic programming due to the constraints and the need to optimize the solution. A naive approach would involve checking all possible transactions, which is computationally expensive and not feasible for large inputs.
The naive solution involves generating all possible pairs of buy and sell days and calculating the profit for each pair. This approach is not optimal due to its high time complexity of O(n^2).
We can use dynamic programming to optimize the solution. The idea is to maintain two arrays: one for the maximum profit up to the i-th day with at most j transactions, and another for the maximum profit up to the i-th day with exactly j 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 over the days and transactions, updating the `dp` array based on the maximum profit achievable by either not making a transaction or making a transaction on that day.
3. Return the value in `dp[n-1][k]` where `n` is the number of days and `k` is the maximum number of transactions.
def maxProfit(k, prices):
# If there are no prices or k is 0, return 0 profit
if not prices or k == 0:
return 0
n = len(prices)
# If k is greater than n//2, it's equivalent to unlimited transactions
if k >= n // 2:
return sum(max(prices[i] - prices[i - 1], 0) for i in range(1, n))
# Initialize the dp array
dp = [[0] * (k + 1) for _ in range(n)]
for j in range(1, k + 1):
max_diff = -prices[0]
for i in range(1, n):
dp[i][j] = max(dp[i - 1][j], prices[i] + max_diff)
max_diff = max(max_diff, dp[i][j - 1] - prices[i])
return dp[-1][-1]
# Example usage
print(maxProfit(2, [2, 4, 1])) # Output: 2
print(maxProfit(2, [3, 2, 6, 5, 0, 3])) # Output: 7
The time complexity of this solution is O(n*k) where `n` is the number of days and `k` is the number of transactions. The space complexity is also O(n*k) due to the 2D `dp` array.
1. If the prices array is empty, the profit should be 0.
2. If `k` is 0, the profit should be 0.
3. If `k` is greater than or equal to `n//2`, it is equivalent to unlimited transactions.
To test the solution comprehensively, consider the following test cases:
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 develop a deeper understanding of dynamic programming techniques.
Understanding and solving the "Best Time to Buy and Sell Stock" problem is essential for mastering dynamic programming and financial market algorithms. Practice and explore further to enhance your problem-solving skills.