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 <= 1050 <= prices[i] <= 104class Solution {
public:
int maxProfit(vector<int>& prices) {
int maxm = INT_MIN, minm = INT_MAX;
for (int p: prices){
minm = min(minm, p);
maxm = max(maxm, p-minm);
}
return maxm;
}
};class Solution {
public:
int maxProfit(vector<int>& prices) {
int maxm = INT_MIN, minm = INT_MAX;
for (int p: prices){
minm = min(minm, p);
maxm = max(maxm, p-minm);
}
return maxm;
}
};class Solution {
public int maxProfit(int[] prices) {
int minYet = Integer.MAX_VALUE, maxProfit = 0;
for (int p : prices) {
minYet = Math.min(minYet, p);
maxProfit = Math.max(maxProfit, p - minYet);
}
return maxProfit;
}
}class Solution {
public int maxProfit(int[] prices) {
int n = prices.length, maxDiff = 0, runningMin = Integer.MAX_VALUE;
for (int p : prices) {
runningMin = Math.min(runningMin, p);
maxDiff = Math.max(maxDiff, p - runningMin);
}
return maxDiff;
}
}class Solution {
public int maxProfit(int[] prices) {
int minYet = Integer.MAX_VALUE, maxProfit = 0;
for (int p : prices) {
minYet = Math.min(minYet, p);
maxProfit = Math.max(maxProfit, p - minYet);
}
return maxProfit;
}
}