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Developer Coder
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เข้าร่วมเมื่อ 3 ม.ค. 2021
🚀 Welcome to DEVELOPER CODER! 🖥️
Dive into the world of coding mastery with our channel dedicated to bringing you top-notch coding tutorials, development insights, and daily LeetCode challenges! 💻🌟
Explore the vast realm of Java and stay updated on the latest in the tech world with our engaging content. Whether you're a seasoned developer or just starting your coding journey, DEVELOPER CODER is your go-to source for learning, growing, and staying ahead in the dynamic field of programming. 🌐🔍
Join our community of passionate coders, sharpen your skills, and embark on a journey of continuous learning. Subscribe now and unlock the door to endless possibilities in the coding universe! 🚀🔐 #CodeWithDeveloperCoder #TechTutorials #JavaDevelopment
Dive into the world of coding mastery with our channel dedicated to bringing you top-notch coding tutorials, development insights, and daily LeetCode challenges! 💻🌟
Explore the vast realm of Java and stay updated on the latest in the tech world with our engaging content. Whether you're a seasoned developer or just starting your coding journey, DEVELOPER CODER is your go-to source for learning, growing, and staying ahead in the dynamic field of programming. 🌐🔍
Join our community of passionate coders, sharpen your skills, and embark on a journey of continuous learning. Subscribe now and unlock the door to endless possibilities in the coding universe! 🚀🔐 #CodeWithDeveloperCoder #TechTutorials #JavaDevelopment
Step-By-Step Directions From a Binary Tree Node to Another | Leetcode 2096 | Java | Developer Coder
In this video, we tackle the LeetCode problem "Step-By-Step Directions From a Binary Tree Node to Another" (Leetcode 2096). We'll walk through the Java solution step-by-step, explaining the logic and code in detail. If you're preparing for coding interviews or simply want to improve your problem-solving skills, this video is perfect for you! 🚀
👨💻 Java Solution:
We'll start by understanding the problem statement and constraints. Then, we'll dive into the Java implementation, breaking down each part of the code to ensure you grasp the solution fully. By the end of this video, you'll have a clear understanding of how to solve this problem using Java. 💻
📝 What's Covered:
Problem analysis and approach
Step-by-step coding in Java
Explanation of key concepts
Tips and tricks for similar problems
👍 Why Watch?
Improve your coding skills
Learn effective problem-solving techniques
Get ready for coding interviews with top companies
📢 Don't forget to like, share, and subscribe to DEVELOPER CODER for more LeetCode solutions and coding tips!
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👨💻 Java Solution:
We'll start by understanding the problem statement and constraints. Then, we'll dive into the Java implementation, breaking down each part of the code to ensure you grasp the solution fully. By the end of this video, you'll have a clear understanding of how to solve this problem using Java. 💻
📝 What's Covered:
Problem analysis and approach
Step-by-step coding in Java
Explanation of key concepts
Tips and tricks for similar problems
👍 Why Watch?
Improve your coding skills
Learn effective problem-solving techniques
Get ready for coding interviews with top companies
📢 Don't forget to like, share, and subscribe to DEVELOPER CODER for more LeetCode solutions and coding tips!
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มุมมอง: 0
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Great video , super helpful explanations , thanks!
Glad it was helpful!
helpful video
Thanks a lot! I’m glad you found it helpful!” 😊
Bhai mere code run nahi ho raha error aa raha hai
mera submission ke time error aa rha h . don't know what's wrong with my code though
can you please share your code? I will review from my side, if it's still not working.
@@DeveloperCoder okay
Keep it up bro 💪🏻💪🏻 you can do it ⭐⭐
Thank you 🙌
4th ?
Time Complexity: Sorting the Positions Array: Sorting takes O(nlogn) time, where n is the number of baskets. Binary Search on the Answer: The range of possible minimum forces is from 1 to position[position.length−1]−position[0]. Let's denote this maximum possible distance as D. Binary search on this range takes O(logD) iterations. Feasibility Check:- For each mid value in the binary search, we perform a linear scan of the positions array to check if we can place all m balls with at least mid distance. This takes O(n) time. Combining these complexities, the overall time complexity of the algorithm is: O(nlogn+logD⋅n) Sorting the positions array: O(nlogn). Binary search iterations: O(logD). Feasibility check for each iteration: O(n). logD is effectively a constant factor (around 30 for base-2 logarithm). Therefore, the dominant term is O(nlogn) from the sorting step. Final Time Complexity O(nlogn+nlogD) Where: n is the length of the positions array. D is the difference between the maximum and minimum positions. Given that logD is constant in practical scenarios, the time complexity can be considered O(nlogn).
Example 1: Explanation: Sorting the Positions: The sorted positions array: [1, 2, 3, 4, 7]. Binary Search Setup: left = 1, right = 6 (since 7 - 1 = 6). Binary Search Execution: First Iteration: mid = 3. canPlaceBalls(position, 3, 3): Place first ball at position 1. Next valid position is 4 (since 4 - 1 >= 3). Next valid position is 7 (since 7 - 4 >= 3). Successfully placed all 3 balls: positions [1, 4, 7]. Return true. Update best = 3 and left = 4. Second Iteration: mid = 5. canPlaceBalls(position, 3, 5): Place first ball at position 1. Next valid position is 7 (since 7 - 1 >= 5). Only 2 balls placed successfully. Return false. Update right = 4. Third Iteration: mid = 4. canPlaceBalls(position, 3, 4): Place first ball at position 1. Next valid position is 7 (since 7 - 1 >= 4). Only 2 balls placed successfully. Return false. Update right = 3. End of Binary Search: left > right (since left = 4 and right = 3). The best minimum force found is 3.
In 14 th line your using a and b but in the program there is no a and b variables .I am not understanding that line
Thank you for asking. The part (a, b) -> a[0] - b[0] is a lambda expression that acts as the custom comparator. Here’s what it does: Parameters a and b: These are two elements (sub-arrays) from the jobs array that need to be compared. a and b represent two jobs, each being a sub-array of the form [difficulty, profit]. Comparison Logic: a[0] refers to the difficulty of job a. b[0] refers to the difficulty of job b. a[0] - b[0] calculates the difference in difficulty between job a and job b. When a[0] < b[0]: a[0] - b[0] is negative. The comparator returns a negative value. This means job a should come before job b in the sorted array. When a[0] == b[0]: a[0] - b[0] is zero. The comparator returns zero. This means the order of job a and job b relative to each other does not change. When a[0] > b[0]: a[0] - b[0] is positive. The comparator returns a positive value. This means job a should come after job b in the sorted array. Please let me know if you still have any questions.😊
Ap ny reply toh kafi acha dia ha .... nice
@@DeveloperCoderyou have give a nice and lengthy reply .... nice ....
nums = [1,2,31,33] n = 2147483647 Output generates 29 but expected is 28 What the error ??
Hi, I have executed the test case provided by you using the same code provided by me in this tutorial. I am getting output and expected both 28. Could you please double check the code again? There may be some minor mistakes in loops and conditions.
ThankYou for your explanation ...🔥🔥🔥
Glad it was helpful!
It is Better to Explain the intuition instead showing the solution!
Since most people are waiting for solutions to complete the live contest during 1 hour window time, I shared solutions directly for all 4 problems. Once I have some free time, I'll definitely explain this in detail, like any other videos on my TH-cam channel where I explain the POTD (Problem of the Day) on Leetcode.
why do you use publc static void main string but program is still run??
If I understand it correctly, you are asking where public static void main line, right? We need to implement the solution method for the given problem. Input is already handled by the LeetCode platform, where the public static void main code is written. We don't need to write this on these platforms or during competitive exams. Inputs are taken by itself...we just need to write our main approach to solve that. public static void main(String[] args) { Solution solution = new Solution(); String[] words = {"dog", "cat", "dad", "good"}; char[] letters = {'a', 'a', 'c', 'd', 'd', 'd', 'g', 'o', 'o'}; int[] scores = {1, 0, 9, 5, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}; int result = solution.maxScoreWords(words, letters, scores); System.out.println("Maximum score possible: " + result); } Hope this helps.....
Wow. The Asian girl on the thumb has such a deep voice 😄
Ah, you came after seeing her in thumbnail... caught you😉... sorry for disappointment 😂
cool man 😎 😎
☺️🙏
@@DeveloperCoder yes
The use of the double backslash shocked me haha
Hope, you are feeling good now 😅
Display font size is not clear sir
Sure, I will take care of this in the next videos. For this, please zoom the screen to see the code content. Thanks for letting me know. 😊 Keep coding and Keep learning.
great explanation, really appreciable.
Glad it was helpful!
Time complexity is going to be O(n) since we are running for loops 3 times. O(n) for frequency + O(n) for maxFrequency + O(n) for Maximum frequency count. O(n) + O(n) + O(n) = O(3n) ~ O(n). We are using greedy approach because if we don't pick max frequency first, than in that case we could not get minimum number of result value. We have to be greedy so that we can get best result.( Minimum Value ) Hope it helps, thanks for your time. Keep Coding.
It’s an awesome effort , I have few comments too It will be Good to convey how u identified or came up with greedy strategies What is naive approach and time complexity for the same Which helps in differing whether this video is made to show that u can solve this problem vs ur intention is to educate viewers
Time complexity is going to be O(n) since we are running for loops 3 times. O(n) for frequency + O(n) for maxFrequency + O(n) for Maximum frequency count. O(n) + O(n) + O(n) = O(3n) ~ O(n). We are using greedy approach because if we don't pick max frequency first, than in that case we could not get minimum number of result value. We have to be greedy so that we can get best result.( Minimum Value ) Hope it helps, thanks for your time. Keep Coding.
Superb bro🎉
Thanks 🤗
best explaination sir