188. Best Time to Buy and Sell Stock IV.cpp

[kaif969]

PR Title Format: 188. Best Time to Buy and Sell Stock IV.cpp

Intuition

We want to maximize profit by making at most k stock transactions. Each day, we can choose to buy, sell, or skip. Using recursion with memoization (3D DP), we track the day, buy/sell state, and remaining transactions.

Approach

Use a recursive function profit(prices, i, isBuy, k):

Base case: if we reach the end of prices or no transactions left, return 0.

• If isBuy == 1: choose to buy today or skip.

• If isBuy == 0: choose to sell today or skip.

Store results in a 3D DP array dp[i][isBuy][k] to avoid recomputation.

Finally, call profit(prices, 0, 1, k) starting from day 0 with buy allowed.

Code Solution (C++)

class Solution {
public:
    vector<vector<vector<int>>> dp;

    int profit(vector<int>& prices, int i, int isBuy, int k) {
        if (i == prices.size() || k == 0) return 0;

        if (dp[i][isBuy][k] != -1) return dp[i][isBuy][k];

        int a, b;
        if (isBuy) {
            a = profit(prices, i + 1, 1, k);
            b = profit(prices, i + 1, 0, k) - prices[i];
        } else {
            a = profit(prices, i + 1, 0, k);
            b = profit(prices, i + 1, 1, k - 1) + prices[i];
        }

        return dp[i][isBuy][k] = max(a, b);
    }

    int maxProfit(int k, vector<int>& prices) {
        int n = prices.size();
        dp = vector<vector<vector<int>>>(n, vector<vector<int>>(2, vector<int>(k + 1, -1)));
        return profit(prices, 0, 1, k);
    }
};
    

Related Issues

#261

By submitting this PR, I confirm that:

• This is my original work not totally AI generated
• I have tested the solution thoroughly on leetcode
• I have maintained proper PR description format
• This is a meaningful contribution, not spam

Summary by Sourcery

• Adds a C++ solution for LeetCode 188 (Best Time to Buy and Sell Stock IV).
• Uses recursive memoization with a 3D DP array dp[i][isBuy][k] and a helper profit(prices, i, isBuy, k).
• Time complexity O(n * k), space O(n * k). Starts from profit(prices, 0, 1, k).
• Suggestion: if k >= n/2, reduce to the unlimited-transactions case for O(n) solution.

Summary by Sourcery

New Features:

• Introduces a C++ solution for LeetCode 188 (Best Time to Buy and Sell Stock IV) using recursion with memoization.

[sourcery-ai]

Reviewer's Guide

This PR adds a new C++ solution for LeetCode 188 using a 3D memoized recursion that tracks the current day, buy/sell state, and remaining transactions to compute maximum profit.

Class diagram for the new Solution class (Best Time to Buy and Sell Stock IV)

classDiagram
class Solution {
  +vector<vector<vector<int>>> dp
  +int profit(vector<int>& prices, int i, int isBuy, int k)
  +int maxProfit(int k, vector<int>& prices)
}
Solution : profit(prices, i, isBuy, k) uses dp
Solution : maxProfit(k, prices) uses profit(prices, i, isBuy, k)

Loading

File-Level Changes

Change	Details	Files
Introduce recursive memoization with 3D DP to handle at most k stock transactions	Declare dp as a 3D vector initialized to -1 for memo storage Implement profit(prices, i, isBuy, k) with base cases and buy/sell decision branches Store and reuse computed results in dp[i][isBuy][k] to avoid recomputation Implement maxProfit() to set up dp dimensions and invoke the recursion	188. Best Time to Buy and Sell Stock IV.cpp

Possibly linked issues

• 242 : Valid Anagram #188: PR adds recursive DP for LeetCode 188, Best Time to Buy and Sell Stock IV, matching issue's problem.
• 242 : Valid Anagram #188: The PR provides a C++ dynamic programming solution to find the maximum profit for at most k transactions, directly addressing issue 188.

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[sourcery-ai]

Hey there - I've reviewed your changes and they look great!

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[kaif969]

@SjxSubham Please 🙏 check this PR also.

[kaif969]

@SjxSubham, please merge this PR.

[SjxSubham]

@kaif969 U r only allowed to contribute upto two solutions only...