Skip to main content

Stock Buy and Sell

Problem Statement:​

You are given an array of prices where prices[i] is the price of a given stock on an 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.

✅ Solution: Greedy / Single Pass (Best Time to Buy and Sell Stock)​

int maximumProfit(vector<int>& stockPrices) {
    int lowestPriceSoFar = INT_MAX;  // Track the minimum price (buying price)
    int highestProfit = 0;           // Track the max profit so far

    for (int price : stockPrices) {
        lowestPriceSoFar = min(lowestPriceSoFar, price);             // Update min price
        highestProfit = max(highestProfit, price - lowestPriceSoFar); // Update max profit
    }

    return highestProfit;
}

📝 How It Works​

  • You're given a list of stock prices, and you must buy once and sell once to maximize profit.
  • At each step:
    • Update the lowest price seen so far (lowestPriceSoFar).
    • Calculate profit: current price - lowestPriceSoFar.
    • Update highestProfit if the profit is better than before.
  • All in one pass, making it highly efficient.

🧩 Key Formula​

maxProfit = max(maxProfit, price[i] - minPriceSoFar)
minPriceSoFar = min(minPriceSoFar, price[i])

⏱️ Time & Space Complexity​

MetricValue
⏱ TimeO(n)
🗂 SpaceO(1)

⚠️ Edge Cases​

  • All decreasing prices → profit = 0
  • All prices same → profit = 0
  • Only one price → profit = 0 (can't sell without buying before)
  • Empty array → return 0 or handle with a guard

💡 Other Approaches​

ApproachTimeSpaceNotes
Brute ForceO(n²)O(1)Compare all pairs, too slow ❌
Greedy (this)O(n)O(1)✅ Best and optimal

🔁 Related Problems​

💬

Discussion & Doubts