6_buyAndSellStock.md

Best Time to Buy and Sell Stock

You are given an array/list 'prices' where the elements of the array represent the prices of the stock as they were yesterday and indices of the array represent minutes. Your task is to find and return the maximum profit you can make by buying and selling the stock. You can buy and sell the stock only once.

Note:You can’t sell without buying first.

O(N^2) Time and O(1) Space

• Brute force:
• For each day, find the max profit that can be made by buying at that day and selling at the next j days.

Code

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        int maxProfit=0;

        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                maxProfit=max(maxProfit,prices[j]-prices[i]);
            }
        }
        return maxProfit;
    }
};

---

O(N) Time and O(N) Space

• We try to sell stock each day.
• For each day from last we store maximum stock price that will appear.
• then for each day we calculate by selling the stock.

Code

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        int maxProfit = 0;
        vector<int> futureMax(n);
        futureMax[n-1] = prices[n-1];
        for(int i=n-2; i>=0; i--) {
            futureMax[i] = max(prices[i], futureMax[i + 1]);
        }
        for(int i=0; i<n; i++) {
            maxProfit = max(maxProfit, futureMax[i] - prices[i]);
        }
        return maxProfit;
    }
};

---

O(N) Time and O(1) Space

• We try to buy stock each day.
• For each day we keep track of the minimum price of the stock that appeared before it.
• if todays stock price is minimum we will update it.
• return max profit.

Code

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        int maxProfit=0,minStockPrice=1e9;
        for(int i=0;i<n;i++){
            maxProfit=max(maxProfit,prices[i]-minStockPrice);
            minStockPrice=min(minStockPrice,prices[i]);
        }
        return maxProfit;
    }
};

The inner for loop can be optimized and will require 33% less time.

• If the price of the stock that day less than minimum price so far then there is no chance to get profit so we only update minimum price.
• else we can get profit, update maxProfit.

Code

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        int maxProfit=0,minStockPrice=1e9;
        for(int i=0;i<n;i++){ // 33% faster
            if(minStockPrice>prices[i])
                minStockPrice=prices[i];
            else
                maxProfit=max(maxProfit,prices[i]-minStockPrice);
        }
        return maxProfit;
    }
};