LeetCode #188 — HARD

Best Time to Buy and Sell Stock IV

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Find the maximum profit you can achieve. You may complete at most k transactions: i.e. you may buy at most k times and sell at most k times.

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

Example 1:

Input: k = 2, prices = [2,4,1]
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.

Example 2:

Input: k = 2, prices = [3,2,6,5,0,3]
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4. Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.

Constraints:

1 <= k <= 100
1 <= prices.length <= 1000
0 <= prices[i] <= 1000

Step 01

Brute Force Baseline

Problem summary: You are given an integer array prices where prices[i] is the price of a given stock on the ith day, and an integer k. Find the maximum profit you can achieve. You may complete at most k transactions: i.e. you may buy at most k times and sell at most k times. Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

2
[2,4,1]

Example 2

2
[3,2,6,5,0,3]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #188: Best Time to Buy and Sell Stock IV
class Solution {
    private Integer[][][] f;
    private int[] prices;
    private int n;

    public int maxProfit(int k, int[] prices) {
        n = prices.length;
        this.prices = prices;
        f = new Integer[n][k + 1][2];
        return dfs(0, k, 0);
    }

    private int dfs(int i, int j, int k) {
        if (i >= n) {
            return 0;
        }
        if (f[i][j][k] != null) {
            return f[i][j][k];
        }
        int ans = dfs(i + 1, j, k);
        if (k > 0) {
            ans = Math.max(ans, prices[i] + dfs(i + 1, j, 0));
        } else if (j > 0) {
            ans = Math.max(ans, -prices[i] + dfs(i + 1, j - 1, 1));
        }
        return f[i][j][k] = ans;
    }
}

// Accepted solution for LeetCode #188: Best Time to Buy and Sell Stock IV
func maxProfit(k int, prices []int) int {
	n := len(prices)
	f := make([][][2]int, n)
	for i := range f {
		f[i] = make([][2]int, k+1)
		for j := range f[i] {
			f[i][j] = [2]int{-1, -1}
		}
	}
	var dfs func(i, j, k int) int
	dfs = func(i, j, k int) int {
		if i >= n {
			return 0
		}
		if f[i][j][k] != -1 {
			return f[i][j][k]
		}
		ans := dfs(i+1, j, k)
		if k > 0 {
			ans = max(ans, prices[i]+dfs(i+1, j, 0))
		} else if j > 0 {
			ans = max(ans, -prices[i]+dfs(i+1, j-1, 1))
		}
		f[i][j][k] = ans
		return ans
	}
	return dfs(0, k, 0)
}

# Accepted solution for LeetCode #188: Best Time to Buy and Sell Stock IV
class Solution:
    def maxProfit(self, k: int, prices: List[int]) -> int:
        @cache
        def dfs(i: int, j: int, k: int) -> int:
            if i >= len(prices):
                return 0
            ans = dfs(i + 1, j, k)
            if k:
                ans = max(ans, prices[i] + dfs(i + 1, j, 0))
            elif j:
                ans = max(ans, -prices[i] + dfs(i + 1, j - 1, 1))
            return ans

        return dfs(0, k, 0)

// Accepted solution for LeetCode #188: Best Time to Buy and Sell Stock IV
struct Solution;

impl Solution {
    fn max_profit(k: i32, prices: Vec<i32>) -> i32 {
        let n = prices.len();
        let mut k = k as usize;
        k = k.min(n / 2);
        if k == 0 {
            return 0;
        }
        let mut min_costs = vec![std::i32::MAX; k];
        let mut max_profits = vec![0; k];
        for price in prices {
            min_costs[0] = min_costs[0].min(price);
            max_profits[0] = max_profits[0].max(price - min_costs[0]);
            for i in 1..k {
                min_costs[i] = min_costs[i].min(price - max_profits[i - 1]);
                max_profits[i] = max_profits[i].max(price - min_costs[i]);
            }
        }
        max_profits[k - 1] as i32
    }
}

#[test]
fn test() {
    let prices = vec![2, 4, 1];
    let k = 2;
    let res = 2;
    assert_eq!(Solution::max_profit(k, prices), res);
    let prices = vec![3, 2, 6, 5, 0, 3];
    let k = 2;
    let res = 7;
    assert_eq!(Solution::max_profit(k, prices), res);
}

// Accepted solution for LeetCode #188: Best Time to Buy and Sell Stock IV
function maxProfit(k: number, prices: number[]): number {
    const n = prices.length;
    const f = Array.from({ length: n }, () =>
        Array.from({ length: k + 1 }, () => Array.from({ length: 2 }, () => -1)),
    );
    const dfs = (i: number, j: number, k: number): number => {
        if (i >= n) {
            return 0;
        }
        if (f[i][j][k] !== -1) {
            return f[i][j][k];
        }
        let ans = dfs(i + 1, j, k);
        if (k) {
            ans = Math.max(ans, prices[i] + dfs(i + 1, j, 0));
        } else if (j) {
            ans = Math.max(ans, -prices[i] + dfs(i + 1, j - 1, 1));
        }
        return (f[i][j][k] = ans);
    };
    return dfs(0, k, 0);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × k)

Space

O(n × k)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.