Best Time to Buy and Sell Stock - Practice Coding Problems

Best Time to Buy and Sell Stock - Problem

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

Input & Output

Example 1 — Basic Case

$ Input: prices = [7,1,5,3,6,4]

› Output: 5

💡 Note: 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 — No Profit

$ Input: prices = [7,6,4,3,1]

› Output: 0

💡 Note: In this case, no transactions are done and the max profit = 0. Prices are continuously decreasing, so no profit can be made.

Example 3 — Small Profit

$ Input: prices = [1,2]

› Output: 1

💡 Note: Buy on day 1 (price = 1) and sell on day 2 (price = 2), profit = 2-1 = 1.

Constraints

1 ≤ prices.length ≤ 10⁵
0 ≤ prices[i] ≤ 10⁴

Visualization

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Understanding the Visualization

Input

Array of stock prices over time

Process

Find optimal buy and sell days

Output

Maximum profit achievable

Key Takeaway

🎯 Key Insight: Track the minimum price seen so far and calculate profit at each step to find the maximum profit in O(n) time

Asked in

A Amazon 89 M Microsoft 67 🍎 Apple 45 G Google 43 f Facebook 34

The key insight is to track the minimum price seen so far and calculate profit at each step. The optimal approach uses one pass tracking to find the maximum profit by keeping track of the lowest buy price and highest profit. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ One Pass Tracking	O(n)	O(1)	Track minimum price and maximum profit in a single pass
Brute Force	O(n²)	O(1)	Check all possible buy-sell day combinations

One Pass Tracking — Algorithm Steps

Initialize minPrice to first price and maxProfit to 0
For each day, update minPrice if current price is lower
Calculate profit if selling today: currentPrice - minPrice
Update maxProfit if current profit is higher

Visualization

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Step-by-Step Walkthrough

Track Minimum

Keep track of lowest price seen so far

Calculate Profit

At each day, calculate profit if selling today

Update Maximum

Update maximum profit if current profit is higher

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* prices, int n) {
    if (n < 2) {
        return 0;
    }
    
    int minPrice = prices[0];
    int maxProfit = 0;
    
    for (int i = 1; i < n; i++) {
        // Calculate profit if we sell today
        int profit = prices[i] - minPrice;
        if (profit > maxProfit) {
            maxProfit = profit;
        }
        // Update minimum price for future transactions
        if (prices[i] < minPrice) {
            minPrice = prices[i];
        }
    }
    
    return maxProfit;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int prices[1000];
    int n = 0;
    char* token = strtok(line + 1, ",]");
    
    while (token != NULL) {
        prices[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    printf("%d\n", solution(prices, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop through the array once

✓ Linear Growth

Space Complexity

O(1)

Only using two variables: minPrice and maxProfit

✓ Linear Space

892.3K Views

Very High Frequency

~15 min Avg. Time

19.8K Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

One Pass Tracking — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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