CareerCup

1) Iterate through array and check if the element is equals to 'e'.
Time Complexity : O(n)
Space Complexity: O(1)
2) Sorting the array (eg. quick sort, merge or heap based on the sorted elements).
Use binary search to find first and last location of 'e'
Sorting Time Complexity: O(n log n).
Searching Time Complexity: log n

1) Just iterate over array and check if the element is equal 'e'. O(n) time O(1) space
2) If you sort an array (eg. quick sort) you can use binary search to find first and last location of 'e'.
Whole operation - O(n logn) time, O(1) space , Bin Search O(log(n)) time
So 2 is better only if we are considering searching phase. For whole algo better is 1.

Sorting the array can be O(n) by using

int counter[255];

for all characters.
Then re-scanning 'e' would be O(1) by summarizing all counters before 'e'.

I think, the question should show, if you know about radix-sort algorithm. In this case, the range is small and known in advance (a-Z),, so radix sort can be used and time complexity of "sorting" is 0(n)...