Row with max 1's | Naive-Better-Best Approach | Matrix | Interview Question

In this video I have explained the solution for row with max 1's starting from Naive to Optimal Approach.
DO LIKE AND SHARE THIS VIDEO AND DON'T FORGET TO SUBSCRIBE !!
C++:- github.com/tejasbirsingh/Prog...

[@ayushidalal5488]

nicely explained

[@ServingTech]

Great Explanation 👏

[@barshapal_1227]

superb

[@dharanyuvi6951]

wonderful presentation

[@madhav3444]

Great explanation. Thanks brother

[@gurmukhsinghnirman4935]

Awesome explanation👌👌

[@AddiNarayan]

Great explanation

[@AmanKumar-kz1rn]

thanks buddy good explanation keep making videos

[@ClashingCoder]

Thanks!

[@ankitkumarmahato4318]

I also thought the same naive approach(except i initialized row=-1), i thought it will show tle but it failed in one of the cases.

[@abhishekgupta3698]

Thanks brother

[@ClashingCoder]

Bro do subscribe to support. Thanks!

[@ankitdubey9310]

hello bro we have to return the first instance of max 1's the code you wrote shows the latest instance of the code, please help me with that

[@rohandevaki4349]

at 13:41 , for column if worst case is that all rows have 0 and only one column and one row has 1 (bottom right and bottom left)
then you traverse each row and column , so it is also o(N*M) right?

[@mridulgupta4185]

What if 1st row has all 0 with no 1...then your code must b wrong

[@momozkichutney]

Exactly..

[@sachintendulkar6951]

But I think binary search on everyrow works it takes less tiem

[@shivangisharma3643]

When I m submitting my code it is showing your code failed for multiple testcases. please help

[@ClashingCoder]

Can you share your code?

[@madhav3444]

Bro optimal 1 solution is giving wrong answer in one case where N=1000 and M=1000 , It's correct ouput is 976 but code is giving 0 answer. Don't know why ?

[@mayankverma2356]

how O(n+m) is best approach ???
best will be O(log(m))

[@ClashingCoder]

Using Binary search will be O(Logn) which will be applied on each row therefore total time will be O(totalRows Log (totalCols)) whereas Pointers approach will take O(n + m ) time which is less than n log m.

[@SHASHANKRUSTAGII]

@@ClashingCoder
int rowWithMax1s(vector a, int n, int m)
{
int res=-1;
for(int i=0;i

[@lokeshnegi5051]

your explanation is good but try to reduce the pace.

[@ClashingCoder]

Yea I have got feedback from many viewers on that. In new video I will improve that. Thanks!

[@rishabsharma5307]

Can you explain why the TC of optimal solution is O(N + M)

[@insane2539]

because you travelling m horizontal steps and n vertical steps to traverse the complete array in worst case therefore O(n+m)

[@mikegaming7160]

bro what will be the answer for test case 3 3 -----> 0 1 1 1 0 0 0 0 0

[@ClashingCoder]

This will not be a valid input this question as in a row there can't be 0s after 1. For the input provided by you this algorithm will output 1 as answer as per 0 based indexing. So this answer will be wrong in sense of maximum 1's per row

[@sanyamgoel1259]

Confusing, sorry but it' is truth.

[@ClashingCoder]

I will try to improve video quality and explain properly. Thanks for pointing out the problem!

[@shubhayanai8981]

Most Simple Solution based on the logic
int row=m;
int pos=-1;
for (int i=0;i