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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
.
Example 1:
Input: prices = [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.
Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
Example 2:
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 <= 105
0 <= prices[i] <= 104
class Solution:
def maxProfit(self,prices):
left = 0 #Buy
right = 1 #Sell
max_profit = 0
while right < len(prices):
currentProfit = prices[right] - prices[left] #our current Profit
if prices[left] < prices[right]:
max_profit =max(currentProfit,max_profit)
else:
left = right
right += 1
return max_profit
class Solution {
public int maxProfit(int[] prices) {
int lsf = Integer.MAX_VALUE;
int op = 0;
int pist = 0;
for(int i = 0; i < prices.length; i++){
if(prices[i] < lsf){
lsf = prices[i];
}
pist = prices[i] - lsf;
if(op < pist){
op = pist;
}
}
return op;
}
}
class Solution {
public:
int maxProfit(vector<int>& prices) {
int lsf = INT_MAX;
int op = 0;
int pist = 0;
for(int i = 0; i < prices.size(); i++){
if(prices[i] < lsf){
lsf = prices[i];
}
pist = prices[i] - lsf;
if(op < pist){
op = pist;
}
}
return op;
}
};
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