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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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

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

For each day as a potential buy day, check all future days as potential sell days and track the maximum profit found.

One Pass Tracking

⏱️ Time: O(n) Space: O(1)

Keep track of the minimum price seen so far and calculate the profit if we sell on the current day. Update the maximum profit as we go.

Brute Force — Algorithm Steps

For each day i (buy day)
Check all days j > i (sell days)
Calculate profit = prices[j] - prices[i]
Keep track of maximum profit

Visualization

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Step-by-Step Walkthrough

Check All Pairs

For each buy day, check all future sell days

Calculate Profit

Profit = sell_price - buy_price

Track Maximum

Keep track of the highest profit found

Code -

solution.c — C

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

int solution(int* prices, int n) {
    int maxProfit = 0;
    
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            int profit = prices[j] - prices[i];
            if (profit > maxProfit) {
                maxProfit = profit;
            }
        }
    }
    
    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(n²)

Nested loops: outer loop n times, inner loop up to n times

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to track maximum profit

✓ Linear Space

892.4K Views

Very High Frequency

~15 min Avg. Time

19.8K Likes

Ln 1, Col 1

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Best Time to Buy and Sell Stock - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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