Jump Game 3 (LeetCode 1306) | Simplified solution with diagrams | Level Order Traversal | BFS

Actual problem on LeetCode: leetcode.com/problems/jump-ga...
Chapters:
00:00 - Intro
00:48 - Problem Statement
04:50 - How to attack the problem
09:50 - Conversion to level order traversal
13:28 - Dry-run of Code
19:33 - Final Thoughts
📚 Links to topics I talk about in the video:
Level Order Traversal: • Level order traversal ...
Binary Trees: • Binary Trees Introduct...
Jump Game 1: • Jump Game (LeetCode 55...
Jump Game 2: • Jump Game 2 (LeetCode ...
Other problems on Trees: • Trees
📘 A text based explanation is available at: studyalgorithms.com
Code on Github: github.com/nikoo28/java-solut...
Test-cases on Github: github.com/nikoo28/java-solut...
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Starting Learn to Code: amzn.to/3sJm8Wl
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Get started for interview preparation: amzn.to/44Nn5du
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#leetcode #programming #interview

[@MrRahulmalik]

please keep uploading more videos, you are the best explainer I have found on youtube! kudos man

[@Yaduvanshi6532-Rohit-Yadav]

Best content on youtube ❤

[@ridj41]

The best I've ever seen

[@Anonymous____________A721]

Great sir

[@midkay]

Absolutely excellent explanation, thank you so much

[@user-od5nj6pd7q]

Instead of taking one random problem if u take one by one concept postting problems on that topic it would be nice so that we can say ourselves we know DSA ...

[@nikoo28]

there are a lof of playlists, that I am creating on the way. I am still adding a lot videos, and slowly will structure them nicely too...such that they can become a complete course.

[@bidishadas832]

Easy JS solution:
/**
* @param {number[]} arr
* @param {number} start
* @return {boolean}
*/
var canReach = function(arr, start) {
let queue = [];
let visited = new Set();
queue.push(start);
visited.add(start);
while (queue.length > 0) {
let position = queue.shift();
// Check if we have reached a position with value 0
if (arr[position] === 0) return true;
// Calculate new positions to move to
let newLeftPosition = position - arr[position];
let newRightPosition = position + arr[position];
// Check and add new left position if it's within bounds and not visited
if (newLeftPosition >= 0 && !visited.has(newLeftPosition)) {
queue.push(newLeftPosition);
visited.add(newLeftPosition);
}
// Check and add new right position if it's within bounds and not visited
if (newRightPosition < arr.length && !visited.has(newRightPosition)) {
queue.push(newRightPosition);
visited.add(newRightPosition);
}
}
return false;
};