Question

Given a list of Numbers along with an index i and integer k , I wish the find the index of the number that is farthest (towards the left) from the number at index i and is less than k.

eg
if the array is
Index :0 1 2 3 4 5 6 7 .....
Array :3 4 1 5 5 4 3 7 .....
Assuming i = 7 and k = 4 , the answer would be 0

I have been trying to implement this using Red Black Trees, but I couldnt go any lower than O(n) . Is there any way I can reduce the complexity to O(logn) by using a different Data Structure ?

Answer 1

Actually, if the array is static and you want to make multiple queries for O(log n) each, you don’t need that complicated data structures.

What you’re actually asking for – “the number that is farthest (towards the left) from the number at index i and is less than k.” – can be transformed to “the leftmost number before i that is less than k“. Then you can see this as the following:

• find the leftmost number that is less than K – say it’s at position j;
• if j < i, j is the answer to the question
• otherwise, there is no such number – all the entries before position i are larger than or equal to k.

To answer the first of these questions, all you need to know is: for position i, what is the smallest number on positions 0..i – let’s call this min(i). Notice that min is a monotonically decreasing function of i – if the min(5) = 10, there is no way that min(6) = 15, since min(6) is the smallest number on positions 0 to 6, and that necessarily includes the smallest number on positions 0 to 5, which we know to be 10. (min is fairly trivial to construct – if we call the array a, then: min(0) = a[0], and min(i) = minimum(min(i - 1), a[i]) for i > 0.)

With this information, you can perform a binary search for the leftmost index i such that min(i) < k. Then, by the construction of min, we know that all numbers on positions from 0 to i - 1 are greater than or equal to k. So i must be the answer of the question.