Testbook

Question

# plz explain??? i am not understand question??

Comments

In each row 1's and 0's are there such that all 1's ahead of all 0's ...

So we need to check only (moddle+1) bit of Each row ... One loop is sufficient... from row1 to rown.

I think most means, the number of 1's are more than 0's...

---

What is given answer O(n log n)???

---

@Gabbar  what is complexity??

---

@Rahul its given O(n)??

---

@gabbar what if all rows have last 1 before the middle index????

@hardesh what is explanation given??

---

solution which is given??

---

I think that was really a bad solution. They are asking to go n times(possible in worst case) in each row which show make it n² and not 2n.

My approach is to find leftmost 1 or rightmrightmist 0 in each row that is going to take logn time. For each row do the same O(n log n)