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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Note that you cannot sell a stock before you buy one.
Example 1:
Input: [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. Not 7-1 = 6, as selling price needs to be larger than buying price.Example 2:
Input: [7,6,4,3,1] Output: 0 Explanation: In this case, no transaction is done, i.e. max profit = 0.Solution
C++
class Solution { public: //Kadane's Algorithm for {1, 7, 4, 11}, if he gives {0, 6, -3, 7} int maxProfit(vector<int>& prices) { int cur= 0, res = 0; for(int i = 1; i < prices.size(); ++i) { cur= max(cur+prices[i]-prices[i-1], 0); res= max(res, cur); } return res; } };Explanation
Similar Questions
Best Time to Buy and Sell Stock II
class Solution { public: int maxProfit(vector<int> &prices) { int res = 0; for (int i = 1; i < prices.size(); ++i) res += max(prices[i] - prices[i - 1], 0); return res; } };Best Time to Buy and Sell Stock III
class Solution { public: int maxProfit(vector<int>& prices) { int onebuy = INT_MAX, onesell = 0, twobuy = INT_MAX, twosell = 0; for(auto& p: prices) { onebuy = min(onebuy, p); onesell = max(onesell, p - onebuy); twobuy = min(twobuy, p - onesell); twosell = max(twosell, p - twobuy); } return twosell; } };Best Time to Buy and Sell Stock IV
class Solution { public: int maxProfit(int k, vector<int>& prices) { int maxProfit=0; if(prices.size()<2) return 0; if(k>prices.size()/2){ for(int i=1; i<prices.size(); i++) maxProfit += max(prices[i]-prices[i-1], 0); return maxProfit; } int buy[k+1]; int sell[k+1]; for (int i=0;i<=k;++i){ buy[i] = INT_MAX; sell[i] = 0; } for(int i=0; i<prices.size(); i++){ for(int j=1; j<=k; j++){ buy[j] = min(buy[j], prices[i]-sell[j-1]); sell[j] = max(sell[j], prices[i]-buy[j]); } } return sell[k]; } };