Minimum Lights To Activate

[@hey.............]

public class Solution { public int solve(int[] A, int B) { int n = A.length; int i = 0; int count = 0; while(i < n) { boolean isBulbLighted = false; int lr = Math.max(i - B + 1, 0); int rr = Math.min(i + B - 1, n - 1); // 0, 0, 1, 0, 1, 0, 0, 1 for(int j = rr; j >= lr; j--) { if(A[j] == 1) { count++; i = j; i += B; isBulbLighted = true; break; } } if(!isBulbLighted) { return -1; } } return count; } }

[@niharikathakur4925]

Bro, at first I was reluctant to watch a 19 minute video but the way you deliver the solution is absolute delight. The flow of the code and the explanation were in sync.

[@JagdeepSingh-cd4jk]

though the video was bit long , but it was worth watching it!! so nicely explained every bit of the line of code

[@bhaveshkumar6842]

This is the best explanation for this question. Great job!

[@mrinalmadhav8119]

Mam plz aap continue rakhiye ye series, views ko dhyan mat dijiye Aapke channel bhut precious h Heere ho har koi tarash nhi sakta

[@ferozahmadqureshi8071]

everytime I am stuck, you come up like magic. Thanks ma'm.

[@shimlamam6066]

R u wall clock alwys stuck after putting battary u charged

[@nehanjalimanchikanti3974]

Most underrated channel

[@rohankumarshah5679]

quality of Explanation justifies your hard work !!

[@IIOT_B1_052_Sachit_Gupta]

int Solution::solve(vector &A, int B) { int count =0; int i=0; int n=A.size(); while(i=left){ if(A[right]==1){ found=true; break; } right--; } if(found==false){ return -1; } count++; i=right+B; } return count; } Correct code . i made a small correction i=right +B not right+B-1 as mentioned in the video.

[@tanzeelahmed2055]

What an awesome explanation!

[@shimlamam6066]

Wow an awesome comment

[@ANANDJULU]

Very nicely explained, thank you!

[@surajsingh-sm7qx]

Nice explaination 😊

[@keya.bitspilani5429]

Intuitive solution❤

[@rudragaming5281]

MOST AWSOME CONTENT MAAM...

[@aditikatiyar5933]

solving this question looks a piece of cake when explained by her!!!

[@anikethdeshpande8336]

nicely explained. !

[@sanyamjain7058]

is Time complexity is O(n^2) ?

[@chiragkarnwal6740]

Great Explanation❤

[@ajaywadhwa3398]

Wow !!!!! Really a good explanations.

[@dheerajkumarmeena629]

Nicely explained. Very easy to understand

[@payalpanjwani5332]

Good explanation! Thanks for helping out 🤗

[@VaibhavSingh-vy6gy]

Very nicely explained...Thanks

[@dhruvpurwar6642]

Fantastic explanation!! KUDOS

[@forlaliga7123]

Great explanation

[@ramaniruth6792]

Nicely explained

[@masternik6126]

int Solution::solve(vector &A, int B) { int n = A.size(); int count = 0; int i = 0; bool bulbfound = false; while(i=l){ if(A[r]==1){ count++; i = B+r; bulbfound = true; break; } r--; } if(!bulbfound) return -1; bulbfound = false; } return count; }

[@swatishambhavi3649]

Great explanation, thanks so much!

[@raj_kundalia]

Thanks for the video!

[@mdshoiebiqbal6155]

Thank you ,that was nice application

[@snake_case07]

Thanks for the explanation 😊

[@devanshsharma2106]

such a good video

[@manup7636]

very good explanation the tabs show how much u research for one question🙂

[@jatinbhatoya8420]

for java folks: public static int solve(int[] arr, int k) { if (k==0) return -1; int i = 0, count = 0; while (i < arr.length) { int idx = lightItUp(arr, k, i); if (idx == -1) return -1; else { i = idx + k; count++; } } return count; } public static int lightItUp(int arr[], int k, int i) { int j = Math.min(i + k - 1, arr.length - 1); while (j >= Math.max(i - k + 1, 0)) { if (arr[j] == 1) { return j; } j--; } return -1; }

[@anujjain9273]

i was just doing the mistake in the second while loop , btw nice explanation , but in the starting your voice was little crumbling , and have you made a git hub repo for interviewbit questions?

[@meme_engineering4521]

best explanation!!

[@probabilitycodingisfunis1]

Glad you liked it

[@Your_Boy_Suraj]

Python Solution: class Solution: # @param A : list of integers # @param B : integer # @return an integer def solve(self, A, B): count = 0 i = 0 n = len(A) while(i < n): right = min(n-1, i + B - 1) left = max(i - B + 1, 0) bulbFound = False while(right >= left): if A[right] == 1: bulbFound = True break right -= 1 if bulbFound == False: return -1 else: count += 1 i = right + B return count

[@dilipparmar3175]

Can you please upload your code also? It will help us with revision purposes.

[@aniketsinghvats767]

Wow thanks for the solution. It was an easy question how can't I do that. ;_;

[@avtardeepsingh6715]

super..

[@shilpapatil1963]

❤

[@dhananjoydey1337]

can you please upload solution of Leetcode problem "127. Word Ladder"?

[@shimlamam6066]

Surèeeeeeeeeeee dhanajoy i m redy for u

[@MayankKumar-nn6us]

can anyone tell me what will be the time complexity? Is it O(n^2)?

[@vedantupadhyay42]

O(N). The worst case : - A = [0, 0, 0, 0], B = 1 We would have to iterate over every element..

[@akshayrathod7679]

How much avg time did you take to solve questions?

[@shimlamam6066]

How much u r taking less than that

[@masseffect0128]

Idiots setting the corridor from starting index 1, while the array of lights starts from 0. here is the one pass for starting index 0; int i, end = -1, cnt = 0; for (i = 0; i < A.size();) { if (A[i] == 1) { if (i - B + 1 > end + 1) return -1; end = i + B - 1; i += i + B; ++cnt; } else ++i; } return cnt;

[@Lifeisgood108-33]

Make a tutorial on how to be focussed to see the main screen while watching your lectures.

[@ayushnath3768]

Keep a mirror to the side where you can see your own reflection.

[@RehanShaikh-sz4mh]

thank you so much ma'am , I was not getting the problem I was afraid when I have read this problem need DP and backtracking concept and I have not yet studied it thank you so much for the confidante by your explanation I have got the problem