Problem Statement

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

Problem Link

Best Time to Buy and Sell Stock - LeetCode

Code

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        
        int n = prices.size();
        
        vector<int> dp(n);
        // dp[i] denotes profit obtained at by selling at ith day and buying at minbuy price
        dp[0] = 0;
        int minbuy = prices[0];
        int res = dp[0];
        for(int i=1;i<n;i++)
        {
            minbuy = min(minbuy,prices[i]);
            dp[i] = prices[i] - minbuy;
            res = max(res,dp[i]);
            
        }
        
        return res;
        
    }
};
// Time Complexity : O(n)
// Space Complexity : O(n)

Space Optimized Version

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

// Time Complexity : O(n)
// Space Complexity : O(1)