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 (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Example 1:

Input:[7, 1, 5, 3, 6, 4] Output: 5max. difference = 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: 0In this case, no transaction is done, i.e. max profit = 0.

class Solution {
public:
    inline int max(int a, int b){
        return a > b ? a : b;
    }
    int maxProfit(vector<int>& prices) {
        vector<int> diff = vector<int>();
        for(int i=1;i<prices.size();i++)
            diff.push_back(prices[i] - prices[i-1]);
        int max_so_far = 0, max_ending_here = 0;
        for(int i=0;i<diff.size();i++){
            max_ending_here = max(0, max_ending_here + diff[i]);
            max_so_far = max(max_so_far, max_ending_here);
        }
        return max_so_far;
    }
};