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why don't you use binarySearch for left side

class Solution { public String removeOccurrences(String s, String part) { StringBuilder str=new StringBuilder(); for(int i=0;i<s.length();i++){ str.append(s.charAt(i)); int j=str.length()-part.length(); if(j>=0&&str.substring(j).equals(part)){ str.setLength(j); } } return str.toString(); } } runs in 5 ms time done by myself had a great boost in confidence ..code is small but it took me 6 hours to come up to this solution

Easy to understand. Thanks man .

Hey, This is the best explanation for this question. That DP table made all of my doubts cleared.

Eclipse debug window is not showing in video

Excellent explanation!!!!

very good explained😊

Nice Solution and code.

code- public static boolean iscycle(boolean[] visited,Map<Integer,Boolean> map,int graph[][],int curr ){ if(map.containsKey(curr)) return map.get(curr); if(visited[curr]){ return false; } visited[curr]=true; for(int i: graph[curr]){ if(iscycle(visited, map, graph, i)) map.put(i, true); else return false; } visited[curr]=false; return true; } public static List<Integer> eventualSafeNodes(int[][] graph) { int size=graph.length; boolean visited[] =new boolean[size]; @SuppressWarnings("rawtypes") Map<Integer,Boolean> map=new HashMap(); List<Integer> L=new ArrayList<>(); for(int i=0;i<size;i++){ if(graph[i].length==0) map.put(i, true); } for(int i=0;i<size;i++){ boolean result=iscycle(visited, map, graph, i); map.put(i, result); } for(int i=0;i<size;i++){ if (map.get(i)) L.add(i); } return L; }

That many DFS method, definitely give wrong impression to interviewer

Oh, so just reverse dfs, makes sense! Sadly would not scale really well on higher sized puzzles I think you need 20 TB of memory to make it work at classical 4x4

excellent

nice solution thanks

Thanks i was bit conf

the code doesnot pass all the test cases!!

what will happen if a B is surrounded by G in all four sides?

Doesn't is consider that plus have only one good cells?

Thank you sir crystal clear explanation

Nice solution

This is such a clean and understandable code, thanks

i still didnt understand why we decremented b count

Thank you

worst video, presenter is very confused about what he is presenting .

if(left[1] >= root.val || right[0] <= root.val){ return new int[]{Integer.MIN_VALUE, Integer.MAX_VALUE, 0}; // return new int[]{0, 0, 0}; } i not undersatnd, if minimum is set to Integer,Min AND MAXIMUM IS SET tO iN.MAX , THAN how it is working , for max and maxsum ??

Rani

I bet Takahashi's teeth won't be happy with all these candies..

Time Complexity?

very bad explanation ,nhi samjha kuch

Very easy to understand. Thank you😊

Thank you

Amazing explanation

Great explanation brother than you

excellent explanation!

i like it your chanal, i want to solve all question of DSA. that's why i'll show your all video. nice work. thak for videos.

Very clear explanation ! Great

Thanks man !

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Good explanation

Thank you good work

nice explanation sir, keep posting content like this