121. The Best Time to Buy and Sell Stock, the Best opportunity to Buy and Sell stocks

You are given an array prices where prices[i] is the price of a given stock on the ith day.

Given an array prices, its i element prices[i] represents a given stock i days price.

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.

You can only choose one day to buy the stock and choose to sell the stock on a different day in the future.Design an algorithm to calculate the maximum profit you can make.

Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.

Returns the maximum profit you can make on this trade.If you cannot make any profit, return 0 .

Example 1: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.

Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (stock price = 1), buy on day 5 (stock price = 1)Sell ​​when price = 6), maximum profit = 6-1 = 5.
Note that the profit cannot be 7-1 = 6, because the selling price needs to be greater than the buying price; at the same time, you cannot sell the stock before buying.

Example 2:Example 2:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.

Enter:prices = [7,6,4,3,1]Output:0Explanation: In this case, no transaction is completed, so the maximum profit is 0.

Constraints:Hint:

1 <= prices.length <=
0 <= prices[i] <=

C language:

int maxProfit(int* prices, int pricesSize){int diff=0;int min=prices[0];for(int i=1;iprices[i])min=prices[i];elsediff=(prices[i]-min)>diff?prices[i]-min:diff;}return diff;}

Execution result: pass

Execution time: 120 ms, beat 43.28% of users in all C commits

Memory consumption: 12.6 MB, beats 62.32% of users in all C commits

Passed test case: 211 / 211

C language:

int maxProfit(int* prices, int pricesSize){int diff=0;int min=prices[0];for(int i=1;iprices[i])min=prices[i];if(prices[i]-min>diff)diff=prices[i]-min;}return diff;}

Execution result: pass

Execution time: 124 ms, beat 32.14% of users in all C commits

Memory consumption: 12.5 MB, beat 89.69% of users in all C commits

Passed test case: 211 / 211

C language:

int maxProfit(int* prices, int pricesSize){int diff=0;int min=prices[0];for(int i=1;iprices[i])min=prices[i];elsediff=(prices[i]-min)>diff?prices[i]-min:diff;}return diff;}

Execution result: pass

Execution time: 120 ms, beat 43.28% of users in all C commits

Memory consumption: 12.6 MB, beats 62.32% of users in all C commits

Passed test case: 211 / 211

C language:

int maxProfit(int* prices, int pricesSize){int max=0;int sum=0;for(int i=1;imax)max=sum;if(sum<0)sum=0;}return max;}

Execution result: pass

Execution time: 116 ms, beat 61.56% of users in all C commits

Memory consumption: 12.8 MB, beats 15.39% of users in all C commits

Passed test case: 211 / 211

7 1 1 5 5 3 3 6 4

-6 4 4 -2 3 3 -2

4+(-2)+3=5