2192. All Ancestors of a Node in a Directed Acyclic Graph | Topological Sort | Merge Two Sorted List

In this video, I'll talk about how to solve Leetcode 2192. All Ancestors of a Node in a Directed Acyclic Graph | Topological Sort | Merge Two Sorted List
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I am Aryan Mittal - A Software Engineer in Goldman Sachs, Speaker, Creator & Educator. During my free time, I create programming education content on this channel & also how to use that to grow :)
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[@ARYANMITTAL]

Master Leetcode in 2024 - kzbin.info/www/bejne/kJPOeGtjnsuhf80

[@rajrajesh1669]

Bro, I forgot that problem you had a editorial which can be solved using multi bfs traversal. Can someone give me that problem name/link.

[@nooob_coder]

thanks for all the efforts

[@mohammedansil2302]

GREAT

[@Vishalgupta-cs9cg]

100% for your efforts bhai but fir bhi smz nhi aya .....hindi me daloge to jada understanding bethegi and grow bhi jada kr paoge bhai..e.g., love babbar and striver have same content but lovebabbar grow more becuase of his hindi language .

[@The-coding-cracker28]

Hi Aryan, I've been working on the same problem you discussed in your video about finding ancestors in a directed graph. Here’s the code I wrote to solve it using a topological sort approach. Could you please check it and let me know if there are any optimizations or improvements I can make? Thanks! class Solution { public: vector getAncestors(int n, vector& edges) { unordered_map mp; vector indegree(n, 0); for (const auto& edge : edges) { int u = edge[0]; int v = edge[1]; mp[u].push_back(v); indegree[v]++; } queue q; for (int i = 0; i < n; ++i) { if (indegree[i] == 0) { q.push(i); } } vector v(n); // Perform a topological sort while (!q.empty()) { int top = q.front(); q.pop(); for (int x : mp[top]) { indegree[x]--; if (indegree[x] == 0) { q.push(x); } v[x].insert(v[top].begin(), v[top].end()); v[x].insert(top); } } vector ans; for (const auto& x : v) { ans.push_back(vector(x.begin(), x.end())); } return ans; } };

[@codencode5546]

i didn't understand the 2 consq. sortedAncestors line

[@mohit7717]

Hello Hello Hello Aryan Good morning

[@Munchen888]

Dfs here is bad because it will be timeout (

[@KADOfficial23]

honestly, didnot understand anything.

[@Vishalgupta-cs9cg]

same hindi me smzane ki kosis kerta to ata bhi yrr

[@limitless6189]

explain in java. i am sure u r reach and subscribers will be doubled

[@aayushyavajpayee4989]

do not put such strong hints in video title, It spoils the question pretty quickly