122, Best Time to Buy and Sell Stock II

I Problem

You are given an integer array prices where prices[i] is the price of a given stock on the iᵗʰ day.

On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day.

Find and return the maximum profit you can achieve.

Example 1
Input: prices = [7, 1, 5, 3, 6, 4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5 - 1 = 4. Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6 - 3 = 3. Total profit is 4 + 3 = 7.

Example 2
Input: prices = [1, 2, 3, 4, 5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5 - 1 = 4. Total profit is 4.

Example 3
Input: prices = [7, 6, 4, 3, 1]
Output: 0
Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.

Constraints

• 1 <= prices.length <= 3 * 10⁴
• 0 <= prices[i] <= 10⁴

Related Topics

• Greedy
• Array
• Dynamic Programming

II Solution

Approach 1: Greedy

pub fn max_profit(prices: Vec<i32>) -> i32 {
    let mut res = 0;

    for i in 1..prices.len() {
        res += std::cmp::max(0, prices[i] - prices[i - 1]);
    }

    res
}

public int maxProfit(int[] prices) {
    int res = 0;
    
    for (int i = 1; i < prices.length; i++) {
        res += Math.max(0, prices[i] - prices[i - 1]);
    }

    return res;
}

func maxProfit(prices []int) int {
    res := 0

    for i, size := 1, len(prices); i < size; i++ {
        res += max(0, prices[i]-prices[i-1])
    }

    return res
}

Approach 2: Dynamic Programming

pub fn max_profit(prices: Vec<i32>) -> i32 {
    let len = prices.len();
    let mut dp = vec![[0; 2]; len];
    (dp[0][0], dp[0][1]) = (0, -prices[0]);

    for i in 1..len {
        dp[i][0] = std::cmp::max(dp[i-1][0], dp[i-1][1] + prices[i]);
        dp[i][1] = std::cmp::max(dp[i-1][1], dp[i-1][0] - prices[i]);
    }

    dp[len -1][0]
}

public int maxProfit(int[] prices) {
    int len = prices.length;
    int[][] dp = new int[len][2];
    dp[0][0] = 0;
    dp[0][1] = -prices[0];

    for (int i = 1; i < len; i++) {
        dp[i][0] = Math.max(dp[i-1][0], dp[i-1][1] + prices[i]);
        dp[i][1] = Math.max(dp[i-1][1], dp[i-1][0] - prices[i]);
    }

    return dp[len-1][0];
}

func maxProfit(prices []int) int {
    size := len(prices)
    dp := make([][2]int, size)
    dp[0][0], dp[0][1] = 0, -prices[0]

    for i := 1; i < size; i++ {
        dp[i][0] = max(dp[i-1][0], dp[i-1][1]+prices[i])
        dp[i][1] = max(dp[i-1][1], dp[i-1][0]-prices[i])
    }

    return dp[size-1][0]
}

Approach 3: Optimized Dynamic Programming

pub fn max_profit(prices: Vec<i32>) -> i32 {
    let (mut dp0, mut dp1) = (0, -prices[0]);

    for i in 1..prices.len() {
        (dp0, dp1) = (
            std::cmp::max(dp0, dp1 + prices[i]),
            std::cmp::max(dp1, dp0 - prices[i]),
        );
    }

    dp0
}

public int maxProfit(int[] prices) {
    int dp0 = 0, dp1 = -prices[0];

    for (int i = 1; i < prices.length; i++) {
        int newDp0 = Math.max(dp0, dp1 + prices[i]);
        int newDp1 = Math.max(dp1, dp0 - prices[i]);
        dp0 = newDp0;
        dp1 = newDp1;
    }

    return dp0;
}

func maxProfit(prices []int) int {
    dp0, dp1 := 0, -prices[0]

    for i, size := 1, len(prices); i < size; i++ {
        dp0, dp1 = max(dp0, dp1+prices[i]), max(dp1, dp0-prices[i])
    }

    return dp0
}