LeetCode 0121. Best Time to Buy and Sell Stock Solution in Java, Python, C++, JavaScript, Go & Rust | Explanation + Code

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

Solutions

Solution 1: Enumerate + Maintain the Minimum Value of the Prefix

We can enumerate each element of the array nums as the selling price. Then we need to find a minimum value in front of it as the purchase price to maximize the profit.

Therefore, we use a variable mi to maintain the prefix minimum value of the array nums. Then we traverse the array nums and for each element v, calculate the difference between it and the minimum value mi in front of it, and update the answer to the maximum of the difference. Then update mi = min(mi, v). Continue to traverse the array nums until the traversal ends.

Finally, return the answer.

The time complexity is O(n), where n is the length of the array nums. The space complexity is O(1).

PythonJavaC++GoTypeScriptRustJavaScriptC#PHP

class Solution: def maxProfit(self, prices: List[int]) -> int: ans, mi = 0, inf for v in prices: ans = max(ans, v - mi) mi = min(mi, v) return ans(code-box)

class Solution { public int maxProfit(int[] prices) { int ans = 0, mi = prices[0]; for (int v : prices) { ans = Math.max(ans, v - mi); mi = Math.min(mi, v); } return ans; } }(code-box)

class Solution { public: int maxProfit(vector<int>& prices) { int ans = 0, mi = prices[0]; for (int& v : prices) { ans = max(ans, v - mi); mi = min(mi, v); } return ans; } };(code-box)

func maxProfit(prices []int) (ans int) { mi := prices[0] for _, v := range prices { ans = max(ans, v-mi) mi = min(mi, v) } return }(code-box)

function maxProfit(prices: number[]): number { let ans = 0; let mi = prices[0]; for (const v of prices) { ans = Math.max(ans, v - mi); mi = Math.min(mi, v); } return ans; }(code-box)

impl Solution { pub fn max_profit(prices: Vec<i32>) -> i32 { let mut ans = 0; let mut mi = prices[0]; for &v in &prices { ans = ans.max(v - mi); mi = mi.min(v); } ans } }(code-box)

/** * @param {number[]} prices * @return {number} */ var maxProfit = function (prices) { let ans = 0; let mi = prices[0]; for (const v of prices) { ans = Math.max(ans, v - mi); mi = Math.min(mi, v); } return ans; };(code-box)

public class Solution { public int MaxProfit(int[] prices) { int ans = 0, mi = prices[0]; foreach (int v in prices) { ans = Math.Max(ans, v - mi); mi = Math.Min(mi, v); } return ans; } }(code-box)

class Solution { /** * @param Integer[] $prices * @return Integer */ function maxProfit($prices) { $ans = 0; $mi = $prices[0]; foreach ($prices as $v) { $ans = max($ans, $v - $mi); $mi = min($mi, $v); } return $ans; } }(code-box)