SPOJ - RACETIME

Problem description

You are given an array (size <= 100000). There are 2 kind of operations you need to perform

1. Change the value at some position.
2. Print the number of elements is <= to some number in a subarray.

I thought of using the idea of MKTHNUM (merge sort). The problem is, with every operation 1, you'll have to update the sorted array all the way. The second option is to split the whole array into sqrt(n) segments. Operations can be performed in the following manner:

1. i. Find the segment which contains the position (a binary search can be employed, linear search is fine too)
   ii. Binary search for the position of the number in that segment (It can be done with a linear search too)
  iii. Change the initial array, and make the segment sorted by right to left and left to right passes on the sub arrays.

2. i. Find out the segments which partially or wholly intersect the query segment.
   ii. Brute force for partially intersected segments, use binary search for wholly intersected segments.

I got many wrong answers for mistake number 3. I traced it only after testing the solution with small random test cases with brute force results.

Code

Mistakes:

1. Again X(: Mixing subarray indices with input array indices.
2. Binary search errors: Made a logical error on binary search check.
3. When checking how many numbers less/greater than some number, we need to check whether the binary search (lower/upper bound) result yields a valid answer.

SPOJ - MKTHNUM

Problem statement

You are given an array. It's size can be as big as 10^5. You'll also be given a some (<=5000) queries. Each query will ask you about the k-th element of the sorted array between indices a and b. How to solve that?

Naive idea is to sort the query subarray and find the answer. As pointed out in the problem statement, naive algorithm will not work. How about keeping some presorted intervals? Hmm...

Remember the way we did merge sort, we divide each interval, sort each of them and merge them. We can pre-process the sorted intervals that way.

How to use the presorted arrays to answer queries? At each query, it's easy to find the intervals of the merge sort tree which are completely contained in the query. We'll have to find the least x for which the number of elements with value <= x is less than or equal to (actually equal to) x. for fixing the x (log n), for each interval (log n intervals), binary search for each interval (log n). So (log n)^3 for each query. So complexity for answering queries is m (log n)^3. With n log n merge sort, the overall complexity is O(n log n + m (log n)^3).

Code

Mistakes:

1. Mistaking it for k-th smallest element finding at first.
2. Annoying mistake: If an element can't be found in an interval, lo of binary search should be decreased by one.
3. Binary search on negative numbers: (a+b)/2 rounds towards zero (ceiling) instead of floor. A workaround is to binary search on min-value + |min-value| and max-value + |min-value| which converts to to a non-negative binary search.