Flip

[@avinashvishwakarma6488]

The best part about your explanation is that you always develop the intuition first. That helps me a lot in solving other similar problems. Keep making more videos of interviewbit problems. Thanks a lot.

[@probabilitycodingisfunis1]

Thanks Avinash!

[@manikmalhotra3903]

Just came across this channel. One of the best and easy-to-understand content. Thank you

[@probabilitycodingisfunis1]

Thank you Manik!

[@attyuttamsaha8]

Really loved the way you have explained the concept and related the two problems.

[@mrinmoyhalder7293]

thanks for the approach, I understand the concept clearly, but one thing I want to say is that changing input may not be feasible always, so it's better to keep a variable which ++ and -- accordingly.

[@annanyamathur8869]

I really like your way of explaining

[@akashnigam5321]

Just got to know that you are an alumni of my college IIT Bombay, your videos are really helpful.

[@ankit_irl]

Top notch explanation. Thank you for existing

[@probabilitycodingisfunis1]

Glad you liked it!

[@mrinalmadhav8119]

Tysm mam, it was great, loved it, big fan of u mam

[@raju_bugude]

nice explanation(intution part)

[@probabilitycodingisfunis1]

Thank you so much

[@pharmacybyanuj2459]

your way of teaching is really superb when ever i understand any problem by you it clear my concepts. thanks Alisha didi di can you tell me how should i make dsa notes ? or make a video how to make dsa notes

[@rithikrk3695]

Beauty with brain....very nice explanation mam!

[@namanagrawal1908]

Great Explanation

[@truptisatsangi6043]

very nice explaination along with related topic so you to go nowhere for 1 question. Great job

[@vikaskumar-kc4jv]

Make more videos you are really awesome

[@jatinbhatoya8420]

for java users - public static int[] flip(String A) { int left = -1, right = -1; int sum = 0, maxSum = 0, currleft = 0; for (int i = 0; i < A.length(); i++) { sum += A.charAt(i) == '0' ? 1 : -1; if (sum < 0) { sum = 0; currleft = i + 1; } if (sum > maxSum) { maxSum = sum; left = currleft; right = i; } } if (left == -1 && right == -1) return new int[0]; return new int[]{left + 1, right + 1}; }

[@rajbhardwaj2180]

Hey Alisha Plz tell the way how to remember ,truth table of all flip flops quite confusing

[@raj_kundalia]

Thanks for the video!

[@rohankumarshah5679]

fantastic !!

[@aniketkumarsinha2537]

I guess you skip this part "return the lexicographically smallest pair of L and R."

[@avinashvishwakarma6488]

but code still runs. Why?

[@dishantmeena1621]

Actually in this approach, we are considering only the best flipping which will give us the maximum no. of ones. So, we are constantly updating our L and R values and only return the best pos. Hope this will help.

[@deepakkapasia1]

More than half test cases failed with this solution. Explaination good for the question and kadane algo but figuring out index not explained.

[@siddharthjain4361]

hi , answer could be one or more right ...how did u check lexicographically for the same

[@Hadle695]

can u plzz share your code

[@rizwanpolice2741]

At the timestamp 7:22 in the kadane algorighm. the solution will not work for all the negative integer array. For example if the array is {-2,-4,-9,-3,-1} The maximum subarray sum will be -1. But with the above pseudo code, it will be 0.

[@sambhavsharma2780]

just init maxSum with arr[0]

[@neerajjoshi1249]

for that take the value of max_sum as INT_MAX

[@chandraprakashsahu8557]

can you please drop your linkedIn id

[@govinth2462]

hi mam I wanna Connect With you can i get your LinkedIn ID