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: 5 max. 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: 0 In this case, no transaction is done, i.e. max profit = 0.
Solution 1: tranverse the array, track the current min value, update the max difference with the max of current price-current min and previous max difference. The time complexity is O(n)
1 class Solution { 2 public: 3 int maxProfit(vector<int>& prices) { 4 int maxdiff=0; 5 int minprice=INT_MAX; 6 for (int price:prices){ 7 minprice=min(price,minprice); 8 maxdiff=max(price-minprice,maxdiff); 9 } 10 return maxdiff; 11 } 12 };