Best Time to Buy and Sell Stock IV - Practice Coding Problems

Best Time to Buy and Sell Stock IV - Problem

You're a stock trader with limited transactions - you can make at most k complete buy-sell cycles to maximize your profit.

Given an array prices where prices[i] represents the stock price on day i, and an integer k representing the maximum number of transactions allowed, find the maximum profit you can achieve.

Important constraints:

You must sell before buying again (no simultaneous transactions)
A transaction consists of buying AND then selling a stock
You can complete at most k transactions

Example: If prices = [2,4,1] and k = 2, you can buy at 2, sell at 4 (profit = 2), but there's no profitable second transaction, so maximum profit is 2.

Input & Output

example_1.py — Basic Case

$ Input: k = 2, prices = [2,4,1]

› Output: 2

💡 Note: Buy at price 2, sell at price 4. Profit = 4 - 2 = 2. No profitable second transaction possible.

example_2.py — Multiple Transactions

$ Input: k = 2, prices = [3,2,6,5,0,3]

› Output: 7

💡 Note: Buy at price 2, sell at price 6 (profit = 4). Buy at price 0, sell at price 3 (profit = 3). Total = 7.

example_3.py — Edge Case

$ Input: k = 2, prices = [1,2,3,4,5]

› Output: 4

💡 Note: One transaction is optimal: buy at 1, sell at 5 for profit of 4. Multiple smaller transactions don't yield more profit.

Visualization

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Understanding the Visualization

Initialize Trading Slots

Set up k transaction slots. Each has a 'buy state' (money after buying) and 'sell state' (profit after selling).

Process Daily Prices

For each day's price, update all trading slots optimally from highest to lowest transaction number.

Update Sell Decisions

For each slot, decide: keep previous sell profit or sell today (buy state + current price).

Update Buy Decisions

For each slot, decide: keep previous buy state or buy today (previous transaction's profit - current price).

Extract Maximum Profit

The sell state of the k-th transaction contains the maximum possible profit.

Key Takeaway

🎯 Key Insight: By maintaining separate buy/sell states for each transaction level and updating them optimally, we avoid exploring all combinations while guaranteeing the maximum profit.

Time & Space Complexity

Time Complexity

⏱️

O(n*k)

We process each of n prices and update k transaction states

✓ Linear Growth

Space Complexity

O(k)

We only need to store buy and sell states for k transactions

✓ Linear Space

Constraints

0 ≤ k ≤ 100
0 ≤ prices.length ≤ 1000
0 ≤ prices[i] ≤ 1000
Time limit: 2 seconds per test case

Asked in

G Google 42 a Amazon 38 ⊞ Microsoft 31 Apple 24 f Meta 19

The optimal solution uses dynamic programming to track buy and sell states for each transaction. Key insight: maintain buy[i] (max profit after buying in transaction i) and sell[i] (max profit after selling in transaction i). Time complexity: O(n*k), Space: O(k).

Common Approaches

Approach	Time	Space	Notes
✓ Dynamic Programming (Optimal)	O(n*k)	O(k)	Track buy and sell states for each transaction using DP
Recursive Brute Force	O(3^n)	O(n)	Try all possible combinations of k transactions recursively

Dynamic Programming (Optimal) — Algorithm Steps

Initialize buy[i] = -infinity (no money to buy) and sell[i] = 0 (no profit yet)
For each price, update all transaction states from k down to 1
buy[i] = max(buy[i], sell[i-1] - price) - best profit after buying for transaction i
sell[i] = max(sell[i], buy[i] + price) - best profit after selling for transaction i
Return sell[k] as the maximum profit after k transactions

Visualization

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Step-by-Step Walkthrough

Initialize States

buy[i] = -∞ (can't buy without money), sell[i] = 0 (no profit initially)

Process Each Price

For each price, update all transaction states from k down to 1

Update Sell State

sell[i] = max(previous sell[i], buy[i] + current price)

Update Buy State

buy[i] = max(previous buy[i], sell[i-1] - current price)

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

int max(int a, int b) {
    return a > b ? a : b;
}

int maxProfit(int k, int* prices, int pricesSize) {
    if (pricesSize == 0 || k == 0) return 0;
    
    // Optimization for large k
    if (k >= pricesSize / 2) {
        int profit = 0;
        for (int i = 1; i < pricesSize; i++) {
            if (prices[i] > prices[i-1]) {
                profit += prices[i] - prices[i-1];
            }
        }
        return profit;
    }
    
    int* buy = (int*)malloc((k + 1) * sizeof(int));
    int* sell = (int*)malloc((k + 1) * sizeof(int));
    
    for (int i = 0; i <= k; i++) {
        buy[i] = INT_MIN;
        sell[i] = 0;
    }
    
    for (int p = 0; p < pricesSize; p++) {
        for (int i = k; i >= 1; i--) {
            sell[i] = max(sell[i], buy[i] + prices[p]);
            buy[i] = max(buy[i], sell[i-1] - prices[p]);
        }
    }
    
    int result = sell[k];
    free(buy);
    free(sell);
    return result;
}

int main() {
    int k;
    scanf("%d", &k);
    
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int prices[1000];
    int count = 0;
    char* token = strtok(line, " \n");
    while (token != NULL) {
        prices[count++] = atoi(token);
        token = strtok(NULL, " \n");
    }
    
    printf("%d\n", maxProfit(k, prices, count));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n*k)

We process each of n prices and update k transaction states

✓ Linear Growth

Space Complexity

O(k)

We only need to store buy and sell states for k transactions

✓ Linear Space

Constraints

0 ≤ k ≤ 100
0 ≤ prices.length ≤ 1000
0 ≤ prices[i] ≤ 1000
Time limit: 2 seconds per test case

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Input & Output

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Related Problems

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Common Approaches

Dynamic Programming (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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