# Question

Formatted question description: https://leetcode.ca/all/121.html

121. Best Time to Buy and Sell Stock

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock),
design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
Not 7-1 = 6, as selling price needs to be larger than buying price.

Example 2:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

@tag-array



# Algorithm

Traverse the array once, use a variable to record the minimum value of the traversed number, and then calculate the most profit each time the difference between the current value and this minimum value, and then select the larger profit each time to update. When the traversal is completed, the current profit is the desired.

# Code

Java

• public class Best_Time_to_Buy_and_Sell_Stock {

public class Solution_optimize {
public int maxProfit(int[] prices) {
int res = 0, buy = Integer.MAX_VALUE;
for (int price : prices) {
res = Math.max(res, price - buy);

//                // also working
//                res = Math.max(res, price - buy);

}
return res;
}
}

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

int[] profit = new int[prices.length];
for (int i = 1; i < prices.length; i++) {
profit[i] = prices[i] - prices[i - 1];
}

// find maximum subarray
int max = profit; // profit==1
int sum = profit;
for (int i = 1; i < profit.length; i++) {
sum = sum + profit[i];
max = Math.max(max, sum);

if (sum < 0) {
sum = 0;
}
}

return max;

}
}
}

• // OJ: https://leetcode.com/problems/best-time-to-buy-and-sell-stock
// Time: O(N)
// Space: O(1)
class Solution {
public:
int maxProfit(vector<int>& prices) {
int minVal = INT_MAX, ans = 0;
for (int p : prices) {
minVal = min(minVal, p);
ans = max(ans, p - minVal);
}
return ans;
}
};

• class Solution(object):
def maxProfit(self, prices):
"""
:type prices: List[int]
:rtype: int
"""
if not prices:
return 0
ans = 0
pre = prices
for i in range(1, len(prices)):
pre = min(pre, prices[i])
ans = max(prices[i] - pre, ans)
return ans