在o（1）的整数数组中找到i和j之间的元素数

Question

Given an unsorted integer array and 2 numbers i and j such that 0 <= i <= j <= C(a constant say MAX_INTEGER) what kind of pre-processing can be performed on it so that you will be able to find the number of numbers between i and j(both inclusive) in o(1) time. The array can also have duplicates.

I had thought of building a frequency array f[] for the elements in the array(space o(C)) and also another array cf[] for cumulative frequency(space o(C)).

So given i and j, i can look up the cumulative frequency array and do cf[j] - cf[i] - This will give the number of elements between i and j. To include i and j, look up the frequency array and add the values. ie cf[j] - cf[i] + f[i]+f[j] Time complexity will be o(1) * 4 = constant time.

The look up in the frequency array can be avoided by finding the previous non zero cf array element for both i and j in the respective direction. This will increase the time complexity but will reduce the space complexity.

Wanted to know if there is a better solution for this problem.

Note - Values of i and j will be available to you only after the pre-processing is completed.

-Vijay

Answer 1

I can't imagine how you'd do this in O(1) without using O(C) additional space.

You can do the lookup in O(log n) very easily if you just create a sorted copy of the array on startup. (O(n log n)).

The lookup then becomes:

Binary search to find the first occurrence of i
Binary search to find the last occurrence of j
result = position_of_j - position_of_i + 1

Now, if the range of items in the array is relatively small, you could do it in O(max - min + 1) extra space and get O(1) lookup. But worst case, (max - min + 1) == C .

Answer 2

How about this,

First off, sort the integer array. and create a hash table with a key for every unique integer in the array and the value as indices in the array at which that integer occurs first and last in the sorted array,(since dup is possible). Space complexity of hash table would be O(n) and access complexity is constant, you have to allocate the space for hash table accordingly.

Given these extra data structures, if you want to find out the range of numbers between i and j, get first index of i and get last index of j, from the hash table and subtract the first from the latter to get the result.

Answer 3

Dim result() as integer
Dim C(1000) as integer
Dim Buff() as integer
Dim i as integer=50 
dim j as integer=450 
for count =0 to c.count-1
    if c(count)>i and c(count)<j then 
       dim length as integer=buff.count-1
       redim result(lenght+1)
       for buffcount=0 to buff.count
           result(buffcount)=buff(buffcount)
       next  
       result(result.count-1)=c(count)
       redim buff(result.lenght-1)
       buff=result
    end if 
next 

Answer 4

Idea:

 1. Binary search to find the first occurrence of i               
 2. Binary search to find the last occurrence of j       
 3. result = position_of_j - position_of_i + 1

or

1. first enter size of array    
2. then enter elements of array    
3. then query size    
4. then enter left, right

code link : HackerEarth