Best Time to Buy and Sell Stock III

LeetCode 123 | Difficulty: Hard

Hard

Problem Description

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

Find the maximum profit you can achieve. You may complete at most two transactions.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

Example 1:

Input: prices = [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

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.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Constraints:

- `1 <= prices.length <= 10^5`

- `0 <= prices[i] <= 10^5`

Topics: Array, Dynamic Programming

---

Approach

Dynamic Programming

Break the problem into overlapping subproblems. Define a state (what information do you need?), a recurrence (how does state[i] depend on smaller states?), and a base case. Consider both top-down (memoization) and bottom-up (tabulation) approaches.

When to use

Optimal substructure + overlapping subproblems (counting ways, min/max cost, feasibility).

---

Solutions

Solution 1: C# (Best: 96 ms)

Metric	Value
Runtime	96 ms
Memory	23.4 MB
Date	2019-02-24

Solution

public class Solution {
    public int MaxProfit(int[] prices) {
        if (prices == null || prices.Length == 0)
    {
        return 0;
    }

        int k=2, n=prices.Length;
    int[,] profit = new int[k + 1, n+1];   
  
    // fill the table in bottom-up fashion 
    for (int i = 1; i <= k; i++) { 
        int prevDiff = Int32.MinValue; 
        for (int j = 1; j < n; j++) { 
            prevDiff = Math.Max(prevDiff, profit[i - 1,j - 1] - prices[j - 1]); 
            profit[i,j] = Math.Max(profit[i,j - 1], prices[j] + prevDiff); 
        } 
    }

return profit[k,n - 1];
    }
}

📜 2 more C# submission(s)

Submission (2019-02-24) — 100 ms, 23.6 MB

public class Solution {
    public int MaxProfit(int[] prices) {
        if (prices == null || prices.Length == 0)
    {
        return 0;
    }

    int n = prices.Length;
    int profit = 0;

    // scan from left
    // left[i] keeps the max profit from 0 to i
    int[] left = new int[n];
    int min = prices[0];

    for (int i = 1; i < n; i++)
    {
        left[i] = Math.Max(left[i - 1], prices[i] - min);
        min = Math.Min(min, prices[i]);
    }

    // scan from right
    // right[i] keeps the max profit from i to n - 1
    int[] right = new int[n];
    int max = prices[n - 1];

    for (int i = n - 2; i >= 0; i--)
    {
        right[i] = Math.Max(right[i + 1], max - prices[i]);
        max = Math.Max(max, prices[i]);
        profit = Math.Max(profit, left[i] + right[i]);
    }

    return profit;
    }
}

Submission (2019-02-24) — 104 ms, 23.6 MB

public class Solution {
    public int MaxProfit(int[] prices) {
        if (prices == null || prices.Length == 0)
    {
        return 0;
    }

        int k=2, n=prices.Length;
    int[,] profit = new int[k + 1, n+1]; 
  
    // For day 0, you can't earn money 
    // irrespective of how many times you trade 
    for (int i = 0; i <= k; i++) 
        profit[i,0] = 0; 
  
    // profit is 0 if we don't do any transation 
    // (i.e. k =0) 
    for (int j = 0; j <= n; j++) 
        profit[0,j] = 0; 
  
    // fill the table in bottom-up fashion 
    for (int i = 1; i <= k; i++) { 
        int prevDiff = Int32.MinValue; 
        for (int j = 1; j < n; j++) { 
            prevDiff = Math.Max(prevDiff, profit[i - 1,j - 1] - prices[j - 1]); 
            profit[i,j] = Math.Max(profit[i,j - 1], prices[j] + prevDiff); 
        } 
    }

return profit[k,n - 1];
    }
}

---

Complexity Analysis

Approach	Time	Space
DP (2D)	$O(n × m)$	$O(n × m)$

---

Interview Tips

Key Points

• Break the problem into smaller subproblems. Communicate your approach before coding.
• Define the DP state clearly. Ask: "What is the minimum information I need to make a decision at each step?"
• Consider if you can reduce space by only keeping the last row/few values.