Explain Best Time to Buy and Sell Stock (one transaction).

Short answer: Track the minimum price so far while scanning left to right. At each day, compute profit = price − minSoFar and keep the maximum profit. One pass, O(1) extra space.

Problem statement

prices[i] is the stock price on day i. You may buy once and sell once later. Return the maximum profit (0 if no profit).

Interview approach

1. Clarify: only one buy and one sell; sell after buy.
2. Reject nested O(n²) “try every buy/sell pair” as too slow.
3. Maintain minPrice and maxProfit while iterating.
4. Update maxProfit when prices[i] − minPrice is better.

Sample solution

int MaxProfit(int[] prices) {
    int minPrice = int.MaxValue, best = 0;
    foreach (int p in prices) {
        if (p < minPrice) minPrice = p;
        else best = Math.Max(best, p - minPrice);
    }
    return best;
}

Complexity

Time O(n), Space O(1).

Edge cases to mention

• Strictly decreasing prices → 0
• Single day → 0
• All equal prices → 0

Common follow-ups

• Unlimited transactions
• At most k transactions
• Cooldown / fees

Say “running minimum” out loud — it shows you understand the greedy invariant.