### Problem:-

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

__Solution:-__

__Solution:-__

class Solution {

public:

int maxProfit(vector<int>& prices) {

int min_num = prices[0];

int max_num = prices[0];

int max_ans = max_num -min_num;

for(int i=0;i<prices.size() ; i++){

if(prices[i] < min_num){

min_num = prices[i];

max_num = prices[i];

}

if(prices[i] > max_num){

max_num = prices[i];

}

if(max_ans < max_num - min_num)

max_ans = max_num - min_num;

}

return max_ans;

}

};

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