Best Time to Buy and Sell Stock II
LeetCode #122 · Medium · Greedy · C++ solution with worked-out approach and complexity analysis.
Approach
Greedy - Collect All Upward Slopes
Key insight
Since we can make unlimited transactions, we should capture every profit opportunity. Any time the price goes up from one day to the next, we should "buy" the previous day and "sell" the current day.
Strategy
- 1Iterate through prices starting from day 1.
- 2If price[i] > price[i-1], add the difference to our profit.
- 3This effectively captures all upward movements.
Time
O(n)
Space
O(1)
Problem description(from LeetCode)
You are given an integer array prices where prices[i] is the price of a given stock on the ith day. On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day. Find and return the maximum profit you can achieve.
Examples
Example 1
- Input:
- prices = [7,1,5,3,6,4]
- Output:
- 7
- Note:
- Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4. Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3. Total profit is 4 + 3 = 7. Input: prices = [1,2,3,4,5] Output: 4 Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4. Total profit is 4. Input: prices = [7,6,4,3,1] Output: 0 Explanation: There is no way to make a positive profit, so we never buy the stock.
Constraints
- •1 <= prices.length <= 3 * 10^4
- •0 <= prices[i] <= 10^4
C++ Solution
solution.cpp
class Solution {
public:
int maxProfit(vector<int> &prices) {
int res = 0;
for (int i = 1, n = prices.size(); i < n; i++) {
if (prices[i] > prices[i - 1]) res += (prices[i] - prices[i - 1]);
}
return res;
}
};