Video

4.22 where did 7 came from. i didnt understood

Thanks

Thanks

you are the best

why are we initialising 0 with count 1 , I couldn't understand that point

thanks

You explained it very nicely. Looking forward to more videos

but what if the [0,0] is not 0?

how to think of this logic sir ?

Keep it going ! :D

you explained the steps but you didn't explain how and why it works

Can you please share the time and space complexities as well

what's the app you using to note?

Fresher's Community Link: chat.whatsapp.com/CsvgJRFNZTQAIcujub1gDe Experienced Community Link: chat.whatsapp.com/IeeHQGlUPK25OFm7mIdv5G

Keep it up man !

Hey bro, will you be uploading answer of daily challenges everyday? and good approach!

Telegram Links: t.me/+FieKB7Ds6j02Y2Y1 t.me/+JU-ZF-oDjX4xYjBl

Similar Problems: leetcode.com/problems/minimum-time-to-collect-all-apples-in-a-tree/ leetcode.com/problems/binary-tree-cameras/

Please improve your pronunciation. I can’t listen you.

why didnt you finish the seaies.. this is exactly what I was looking for

Good explanation

Like the explaination but solution is far from ideal and will timeout if run with higher inputs

Nice explanation

best explaination for the java submission

00:02 Problem requires step-by-step approach 02:20 Finding the shortest path to collect all keys 04:43 Path to obtaining keys to unlock cells. 07:03 Using binary representation to store key configurations efficiently 09:15 Using numbers to represent key configurations simplifies identification 11:37 Algorithm for finding shortest path to get all keys. 13:49 Finding the shortest path to get all keys with specific conditions. 16:03 Traversal and key management in pathfinding algorithm. 18:00 Iterating over elements in the queue and updating steps

code for pair ?

brute force solution is incorrect. It will fail on the test case [1,100,5]. Explanation: One possible case is when player 1 picks 5. Player 2 picks 1. Player 1 picks 100. This situation will arise because you are considering all possible cases and this case will also fall under it. Your solution will return true but expected answer is false

Thank you for your explanation! It's better than on many other channels!

in 10:19 why to take a2 for a7 and for a10 to a3 I did'nt understand this

This is my java solution: Logically It makes sense to me but I am having trouble with the testcases, any pointers will help: class Solution { public int getLengthOfOptimalCompression(String s, int k) { int[][] x = new int[s.length() + 1][k + 1]; for(int[] j : x){ Arrays.fill(j, -1); } return find(s, k, 0, x, Character.MIN_VALUE, 0); } int find(String s, int k, int index, int[][] x, char cur, int charcount){ if(k == 0 || index >= s.length()){ return Calculate_len(s, index); } if(x[index][k] != -1){ return x[index][k]; } int count = 0; if(cur != s.charAt(index)){ String num = String.valueOf(charcount); int val = 0; if(num.equals("1")){ val = 1; } else if(num.equals("0")){ val = 0; } else{ val = num.length() + 1; } // System.out.println(num + " " + val); count = Math.min(find(s, k, index + 1, x, s.charAt(index),1) + val, find(s, k - 1, index + 1, x, cur, charcount)); } else{ count = Math.min(find(s, k, index + 1, x, s.charAt(index),charcount + 1), find(s, k - 1, index + 1, x, cur, charcount)); } x[index][k] = count; // System.out.println(x[index][k] + " " + index + " " + k); return x[index][k]; } int Calculate_len(String s, int index){ if(index >= s.length()){ return 0; } String ns = s.substring(index); int i = 0; int fin_count = 0; while(i < ns.length()){ int count = 1; while(i < ns.length() - 1 && ns.charAt(i) == ns.charAt(i + 1)){ count++; i++; } if(count == 1){ fin_count++; } else{ String num = String.valueOf(count); fin_count += (num.length() + 1); } i++; } return fin_count; } }

Check out our community of 400+ people: t.me/+FieKB7Ds6j02Y2Y1

class Solution { public: int dp[102][102]; int find(int n){ if(n==1) return 1; if(n>1 and n<10) return 2; if(n>=10 and n<100) return 3; return 4; } int solve(string &s,int k,int idx){ if(k<0) return s.size(); if(idx>=s.size() || (s.size()-idx)<=k) return 0; if(dp[idx][k]!=-1) return dp[idx][k]; int ans=s.size(); ans=solve(s,k-1,idx+1); int cons=0,cnt=0,j; for( j=idx;j<s.size() && cnt<=k;j++){ if(s[j]==s[idx]) cons++; else{ ans=min(ans,find(cons)+solve(s,k-cnt,j)); cnt++; } } if(cnt<=k) ans=min(ans,find(cons)+solve(s,k-cnt,j)); return dp[idx][k]=ans; } int getLengthOfOptimalCompression(string s, int k) { memset(dp,-1,size(dp)); return solve(s,k,0); } }; what is the error in my code

Thanks

Wow very simply and beautifully explained!

Thanks bro

Thank you, well explained.

Nice thumbnail ❤

Very energetic and enthusiastic way of explaining the solution. Thanks much.

Simplified and clear explanation.

nice explanatory video

Nice

Bro, this was dope. Thanks a lot.

Amazing Explanation, brooo!!! 💯💯

Nice presentation

Telegram Links: t.me/+FieKB7Ds6j02Y2Y1 t.me/+JU-ZF-oDjX4xYjBl

The first approach is O(mlogm) m = no of batteries and the binary search approach is O(mlogk). It's given that k >> m. It seems that the frist one is better. Why use binary search?

you are helping a lot but sometime your voice buffer so much and sound like not a ear friendly so plz improve your voice quality just a suggestion anyways you are doing great thanks for sharing :)

I wasnt able to think of 1D dp solution so I wrote 2D dp code and converted the dp array to an unordered map and combined the 2 states to a single string in the map , and got AC. will this be a valid solution in the interview ??

can we use recursive DP with Memoization (using a HashMap for storing values) for this problem?

Sir, I thought about this problem like this: If I add up the totalTime and start with that: f(n,totalTime, T). If I went for not_take, totalTime will decrease by one and if I went for take, totalTime will increase by time[i]. If totalTime is greater than the start(T) then return 0 else return 1e9. Hence the dp will be [n+1][2*totalTime+1]. But it gives Memory Limit Exceeded. If I use your way which is [n][n+1], it will work. Mr. how should I know this is the correct way to solve starting from recursion? Thank you for your amazing videos bhaiya!!!