SDE - 1

Solution Approach:
- iterate string and check index value whether it's 0 or 1
- check if maxi allowed bit is allowed and set the same index in the result as the opposite
- as if x ith index is 0 set ith index in y to 1 ( so as to get maximum value during xor)
- repeat the above steps till the length of string and if max set bits is over ( set remaining bits in y to 0)
 

By uisng Max Heap for finding the k’th smallest element.

1) Build a Max-Heap MH of the first k elements (arr[0] to arr[k-1]) of the given array. O(k)

2) For each element, after the k’th element (arr[k] to arr[n-1]), compare it with root of MH.
a) If the element is less than the root then make it root and call heapify for MH
b) Else ignore it.

3) Finally, root of the MH is the kth smallest element.

Tip 1: Think, listen and understand the problem well
Solution:
Start both timers together.
When the 4 minute timer is done, flip it. 7 minute timer will have 3 minutes left.
When the 7 minute timer is done, the 4 minute timer will have 1 minute left.
Now you can count to 9 minutes by simply leaving the 4 minute to expire (1 min), flip it and let it expire (4 min), flip it again and let it expire (4 min). 1 + 4 + 4 = 9

Be honest and explain well about your project.
Study in-depth whatever language you have learned.

Tip 1: Be confident and speak clearly 
Tip 2: Be honest and show interest in the company 
Tip 3: Always ask questions at the end and Be friendly