Best Time to Buy and Sell Stock
LeetCode #121 · Easy · Greedy · C++ solution with worked-out approach and complexity analysis.
Approach
One Pass with Min Tracking
Strategy
- 1Track the minimum price seen so far.
- 2For each price, calculate the profit if we sell at this price (price -
- 3Update the maximum profit if the current profit is higher.
Time
O(n)
Space
O(1)
Problem description(from LeetCode)
You are given an array prices where prices[i] is the price of a given stock on the ith day. You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock. Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.
Examples
Example 1
- Input:
- prices = [7,1,5,3,6,4]
- Output:
- 5
- Note:
- Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5. Input: prices = [7,6,4,3,1] Output: 0 Explanation: In this case, no transactions are done and the max profit = 0.
Constraints
- •1 <= prices.length <= 10^5
- •0 <= prices[i] <= 10^4
C++ Solution
solution.cpp
class Solution {
public:
int maxProfit(vector<int> &prices) {
int mn = INT_MAX, res = 0;
for (int price : prices) {
mn = min(mn, price);
res = max(res, price - mn);
}
return res;
}
};