121. Best Time to Buy and Sell Stock π
Date: Jun 03, 2024
Tags: two-pointers
Description
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
Optimal
function maxProfit(prices: number[]): number {
let l = 0;
let r = 1;
let diff = -Infinity;
while (r < prices.length) {
const x = prices[l];
const y = prices[r];
if (y < x) {
l = r;
} else {
diff = Math.max(y - x, diff);
}
r++;
}
return diff < 0 ? 0 : diff;
}
export { maxProfit };
Brute-Force
function maxProfit(prices: number[]): number {
let diff = -Infinity;
for (let i = 0; i < prices.length; i++) {
for (let j = i + 1; j < prices.length; j++) {
diff = Math.max(prices[j] - prices[i], diff);
}
}
return diff < 0 ? 0 : diff;
}
export { maxProfit };