Member of Technical Staff

Ninja is in the mood for a walk over the city, but being a ninja he prefers jumping over building roofs instead of walking through the streets.

The height of the buildings in his city can be represented through an array ‘HEIGHTS’ where ‘HEIGHT[i]’ is the height of the ith building. Ninja starts his journey from the 1st building and in one step can only travel to the roof of the next building.

While traveling from the ‘i’th to (i+1)th building:

1. If the ith building has a height greater than or equal to the next i.e (i+1)th building then he simply jumps to the next building.

2. Otherwise he uses either {‘HEIGHTS[i+1] -‘HEIGHTS[i]} bricks or just 1 ladder to climb up to the next building.

Having a limited number of bricks say ‘BRICKS’ and a limited number of ladders say ‘LADDERS’ in his Ninja pocket, he wants to know which is the farthest building he can travel up to if he uses the bricks and ladders optimally.

As Ninja is weak in mathematics so he asks for your help, can you help Ninja to find the maximum index of the building he can reach up to(1 based indexing)?

You are given two strings, ‘s1’ and ‘s2’, each of length ‘N’. The strings ‘s1’ and ‘s2’ are anagrams (i.e., they contain the same characters). You can swap the position of any two characters in ‘s1’ any number of times. The task is to find the minimum number of swaps we need to perform to make ‘s1’ equal to ‘s2’.

Example :

s1 = “bac”, s2 = “acb”, n = 3

We need to perform at least two swaps to make ‘s1 = s2’:
1. Swap ‘b’ and ‘a’, so:
    s1 = “abc”
2. Swap ‘b’ and ‘c’, so:
    s1 = “acb”
So, the answer is ‘2’.

It was more of a open discussion round.
Interviewer asked me about my previous experiences and follow up questions were there.