• Home
  • Top Posts
  • Code Solutions
  • How to
  • News
  • Trending
  • Anime
  • Health
  • Education
Wednesday, February 8, 2023
  • Login
Zeroplusfour
No Result
View All Result
  • Home
  • Top Posts
  • Code Solutions
  • How to
  • News
  • Trending
  • Anime
  • Health
  • Education
  • Home
  • Top Posts
  • Code Solutions
  • How to
  • News
  • Trending
  • Anime
  • Health
  • Education
No Result
View All Result
Zeroplusfour
No Result
View All Result
Home Code Solutions

Best Time to Buy and Sell Stock IV – LeetCode Solution

Best Time to Buy and Sell Stock IV - LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

BhautikBhalala by BhautikBhalala
September 5, 2022
Reading Time: 2 mins read
0
Leetcode All Problems Solutions

Leetcode All Problems Solutions

Spread the love

Table of Contents

  • Best Time to Buy and Sell Stock IV – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • C++ Best Time to Buy and Sell Stock IV LeetCode Solution
  • Java Best Time to Buy and Sell Stock IV LeetCode Solution
  • Python 3 Best Time to Buy and Sell Stock IV LeetCode Solution
    • Leave a comment below
      • Related posts:

Best Time to Buy and Sell Stock IV – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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.

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:

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

C++ Best Time to Buy and Sell Stock IV LeetCode Solution


Copy Code Copied Use a different Browser

class Solution {
public:
    int maxProfit(int k, vector<int> &prices) {
        int n = (int)prices.size(), ret = 0, v, p = 0;
        priority_queue<int> profits;
        stack<pair<int, int> > vp_pairs;
        while (p < n) {
            // find next valley/peak pair
            for (v = p; v < n - 1 && prices[v] >= prices[v+1]; v++);
            for (p = v + 1; p < n && prices[p] >= prices[p-1]; p++);
            // save profit of 1 transaction at last v/p pair, if current v is lower than last v
            while (!vp_pairs.empty() && prices[v] < prices[vp_pairs.top().first]) {
                profits.push(prices[vp_pairs.top().second-1] - prices[vp_pairs.top().first]);
                vp_pairs.pop();
            }
            // save profit difference between 1 transaction (last v and current p) and 2 transactions (last v/p + current v/p),
            // if current v is higher than last v and current p is higher than last p
            while (!vp_pairs.empty() && prices[p-1] >= prices[vp_pairs.top().second-1]) {
                profits.push(prices[vp_pairs.top().second-1] - prices[v]);
                v = vp_pairs.top().first;
                vp_pairs.pop();
            }
            vp_pairs.push(pair<int, int>(v, p));
        }
        // save profits of the rest v/p pairs
        while (!vp_pairs.empty()) {
            profits.push(prices[vp_pairs.top().second-1] - prices[vp_pairs.top().first]);
            vp_pairs.pop();
        }
        // sum up first k highest profits
        for (int i = 0; i < k && !profits.empty(); i++) {
            ret += profits.top();
            profits.pop();
        }
        return ret;
    }
};

Java Best Time to Buy and Sell Stock IV LeetCode Solution


Copy Code Copied Use a different Browser

/**
 * dp[i, j] represents the max profit up until prices[j] using at most i transactions. 
 * dp[i, j] = max(dp[i, j-1], prices[j] - prices[jj] + dp[i-1, jj]) { jj in range of [0, j-1] }
 *          = max(dp[i, j-1], prices[j] + max(dp[i-1, jj] - prices[jj]))
 * dp[0, j] = 0; 0 transactions makes 0 profit
 * dp[i, 0] = 0; if there is only one price data point you can't make any transaction.
 */

public int maxProfit(int k, int[] prices) {
	int n = prices.length;
	if (n <= 1)
		return 0;
	
	//if k >= n/2, then you can make maximum number of transactions.
	if (k >=  n/2) {
		int maxPro = 0;
		for (int i = 1; i < n; i++) {
			if (prices[i] > prices[i-1])
				maxPro += prices[i] - prices[i-1];
		}
		return maxPro;
	}
	
    int[][] dp = new int[k+1][n];
    for (int i = 1; i <= k; i++) {
    	int localMax = dp[i-1][0] - prices[0];
    	for (int j = 1; j < n; j++) {
    		dp[i][j] = Math.max(dp[i][j-1],  prices[j] + localMax);
    		localMax = Math.max(localMax, dp[i-1][j] - prices[j]);
    	}
    }
    return dp[k][n-1];
}

 



Python 3 Best Time to Buy and Sell Stock IV LeetCode Solution


Copy Code Copied Use a different Browser

def maxProfit4(self, k, prices):
    n = len(prices)
    if n < 2:
        return 0
    # k is big enougth to cover all ramps.
    if k >= n / 2:
        return sum(i - j
                   for i, j in zip(prices[1:], prices[:-1]) if i - j > 0)
    globalMax = [[0] * n for _ in xrange(k + 1)]
    for i in xrange(1, k + 1):
        # The max profit with i transations and selling stock on day j.
        localMax = [0] * n
        for j in xrange(1, n):
            profit = prices[j] - prices[j - 1]
            localMax[j] = max(
                # We have made max profit with (i - 1) transations in
                # (j - 1) days.
                # For the last transation, we buy stock on day (j - 1)
                # and sell it on day j.
                globalMax[i - 1][j - 1] + profit,
                # We have made max profit with (i - 1) transations in
                # (j - 1) days.
                # For the last transation, we buy stock on day j and
                # sell it on the same day, so we have 0 profit, apparently
                # we do not have to add it.
                globalMax[i - 1][j - 1],  # + 0,
                # We have made profit in (j - 1) days.
                # We want to cancel the day (j - 1) sale and sell it on
                # day j.
                localMax[j - 1] + profit)
            globalMax[i][j] = max(globalMax[i][j - 1], localMax[j])
    return globalMax[k][-1]



 

Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264
Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195
Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140
Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66
Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40
Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31
Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21
Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16
Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9
Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7
Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4
Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3
Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1

Leave a comment below

 

Related posts:

Leetcode All Problems SolutionsBest Time to Buy and Sell Stock III – LeetCode Solution Leetcode All Problems SolutionsBest Time to Buy and Sell Stock II – LeetCode Solution Leetcode All Problems SolutionsBest Time to Buy and Sell Stock – LeetCode Solution Leetcode All Problems SolutionsFirst Missing Positive – LeetCode Solution Leetcode All Problems SolutionsPascal’s Triangle – LeetCode Solution Leetcode All Problems SolutionsMaximum Gap – LeetCode Solution
Tags: Best Time to Buy and Sell Stock IVCc++14full solutionGojavajava 15java 7java 8java8javascriptkotlinLeetCodeLeetCodeSolutionpypy 3Python 2python 3rubyrustSolution
ShareTweetPin
BhautikBhalala

BhautikBhalala

Related Posts

Leetcode All Problems Solutions
Code Solutions

Exclusive Time of Functions – LeetCode Solution

by admin
October 5, 2022
0
41

Exclusive Time of Functions - LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions...

Read more
Leetcode All Problems Solutions

Smallest Range Covering Elements from K Lists – LeetCode Solution

October 5, 2022
32
Leetcode All Problems Solutions

Course Schedule III – LeetCode Solution

October 5, 2022
27
Leetcode All Problems Solutions

Maximum Product of Three Numbers – LeetCode Solution

September 11, 2022
53
Leetcode All Problems Solutions

Task Scheduler – LeetCode Solution

September 11, 2022
119
Leetcode All Problems Solutions

Valid Triangle Number – LeetCode Solution

September 11, 2022
28
Next Post
Leetcode All Problems Solutions

Rotate Array - LeetCode Solution

Leetcode All Problems Solutions

House Robber - LeetCode Solution

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

You may also like

Leetcode All Problems SolutionsBest Time to Buy and Sell Stock III – LeetCode Solution Leetcode All Problems SolutionsBest Time to Buy and Sell Stock II – LeetCode Solution Leetcode All Problems SolutionsBest Time to Buy and Sell Stock – LeetCode Solution Leetcode All Problems SolutionsFirst Missing Positive – LeetCode Solution Leetcode All Problems SolutionsPascal’s Triangle – LeetCode Solution Leetcode All Problems SolutionsMaximum Gap – LeetCode Solution

Categories

  • Algorithms
  • Anime
  • Biography
  • Business
  • Code Solutions
  • Cosmos
  • Countdowns
  • Culture
  • Economy
  • Education
  • Entertainment
  • Finance
  • Games
  • Hackerrank
  • Health
  • How to
  • Investment
  • LeetCode
  • Lifestyle
  • LINUX SHELL
  • Manga
  • News
  • Opinion
  • Politics
  • Sports
  • SQL
  • Tech
  • Travel
  • Uncategorized
  • Updates
  • World
  • DMCA
  • Home
  • My account
  • Privacy Policy
  • Top Posts

Recent Blogs

Leetcode All Problems Solutions

Exclusive Time of Functions – LeetCode Solution

October 5, 2022
Leetcode All Problems Solutions

Smallest Range Covering Elements from K Lists – LeetCode Solution

October 5, 2022
Leetcode All Problems Solutions
Code Solutions

Pascal’s Triangle II – LeetCode Solution

September 3, 2022
48
Biography

Amy Huberman Net Worth, Age, Height , Achivements and more

September 22, 2022
3
I love that for you Season 1 episode 3 release date Countdown
Countdowns

I love that for you Season 1 episode 3 release date Countdown

May 10, 2022
1
Leetcode All Problems Solutions
Code Solutions

Merge Intervals – LeetCode Solution

September 3, 2022
91

© 2022 ZeroPlusFour - Latest News & Blog.

No Result
View All Result
  • Home
  • Category
    • Business
    • Culture
    • Economy
    • Lifestyle
    • Health
    • Travel
    • Opinion
    • Politics
    • Tech
  • Landing Page
  • Support Forum
  • Contact Us

© 2022 ZeroPlusFour - Latest News & Blog.

Welcome Back!

Login to your account below

Forgotten Password?

Retrieve your password

Please enter your username or email address to reset your password.

Log In
We use cookies on our website to give you the most relevant experience by remembering your preferences and repeat visits. By clicking “Accept All”, you consent to the use of ALL the cookies. However, you may visit "Cookie Settings" to provide a controlled consent.
Cookie SettingsAccept All
Manage consent

Privacy Overview

This website uses cookies to improve your experience while you navigate through the website. Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. We also use third-party cookies that help us analyze and understand how you use this website. These cookies will be stored in your browser only with your consent. You also have the option to opt-out of these cookies. But opting out of some of these cookies may affect your browsing experience.
Necessary
Always Enabled
Necessary cookies are absolutely essential for the website to function properly. These cookies ensure basic functionalities and security features of the website, anonymously.
CookieDurationDescription
cookielawinfo-checkbox-analytics11 monthsThis cookie is set by GDPR Cookie Consent plugin. The cookie is used to store the user consent for the cookies in the category "Analytics".
cookielawinfo-checkbox-functional11 monthsThe cookie is set by GDPR cookie consent to record the user consent for the cookies in the category "Functional".
cookielawinfo-checkbox-necessary11 monthsThis cookie is set by GDPR Cookie Consent plugin. The cookies is used to store the user consent for the cookies in the category "Necessary".
cookielawinfo-checkbox-others11 monthsThis cookie is set by GDPR Cookie Consent plugin. The cookie is used to store the user consent for the cookies in the category "Other.
cookielawinfo-checkbox-performance11 monthsThis cookie is set by GDPR Cookie Consent plugin. The cookie is used to store the user consent for the cookies in the category "Performance".
viewed_cookie_policy11 monthsThe cookie is set by the GDPR Cookie Consent plugin and is used to store whether or not user has consented to the use of cookies. It does not store any personal data.
Functional
Functional cookies help to perform certain functionalities like sharing the content of the website on social media platforms, collect feedbacks, and other third-party features.
Performance
Performance cookies are used to understand and analyze the key performance indexes of the website which helps in delivering a better user experience for the visitors.
Analytics
Analytical cookies are used to understand how visitors interact with the website. These cookies help provide information on metrics the number of visitors, bounce rate, traffic source, etc.
Advertisement
Advertisement cookies are used to provide visitors with relevant ads and marketing campaigns. These cookies track visitors across websites and collect information to provide customized ads.
Others
Other uncategorized cookies are those that are being analyzed and have not been classified into a category as yet.
SAVE & ACCEPT
Are you sure want to unlock this post?
Unlock left : 0
Are you sure want to cancel subscription?