Array range queries for searching an element

Given an array of N elements and Q queries of the form L R X. For each query, you have to output if the element X exists in the array between the indices L and R(included).

Prerequisite : Mo’s Algorithms 

Examples :  

Input : N = 5
        arr = [1, 1, 5, 4, 5]
        Q = 3
        1 3 2
        2 5 1
        3 5 5         
Output : No
         Yes
         Yes
Explanation :
For the first query, 2 does not exist between the indices 1 and 3.
For the second query, 1 exists between the indices 2 and 5.
For the third query, 5 exists between the indices 3 and 5.

Naive Approach : 
The naive method would be to traverse the elements from L to R for each query, linearly searching for X. In the worst case, there can be N elements from L to R, hence the worst case time complexity for each query would be O(N). Therefore, for all the Q queries, the time complexity would turn out to be O(Q*N).

Using Union-Find Method : 
This method checks only one element among all the consecutive equal values. If X is not equal to these values, then the algorithm skips all the other the other equal elements and continues traversal with the next different element. This algorithm is evidently useful only when there are consecutive equal elements in large amounts.

Algorithm :  

1. Merge all the consecutive equal elements in one group. 
2. While processing a query, start from R. Let index = R. 
3. Compare a[index] with X. If they are equal, then print “Yes” and break out of traversing the rest of the range. Else, skip all the consecutive elements belonging to the group of a[index]. Index becomes equal to one less than the index of the root of this group. 
4. Continue the above step either till X is found or till index becomes less than L. 
5. If index becomes less than L, print “No”. 

Below is the implementation of the above idea.  

2 does not exist between [0, 2] 
1 exists between [1, 4] 
5 exists between [2, 4] 

Efficient Approach(Using Mo’s Algorithm) : 

Mo’s algorithm is one of the finest applications for square root decomposition. 

It is based on the basic idea of using the answer to the previous query to compute the answer for the current query. This is made possible because the Mo’s algorithm is constructed in such a way that if F([L, R]) is known, then F([L + 1, R]), F([L – 1, R]), F([L, R + 1]) and F([L, R – 1]) can be computed easily, each in O(F) time.
Answering queries in the order they are asked, then the time complexity is not improved to what is needed to be. To reduce the time complexity considerably, the queries are divided into blocks and then sorted. 

The exact algorithm to sort the queries is as follows : 

• Denote BLOCK_SIZE = sqrt(N)
• All the queries with the same L/BLOCK_SIZE are put in the same block
• Within a block, the queries are sorted based on their R values
• The sort function thus compares two queries, Q1 and Q2 as follows: 
  Q1 must come before Q2 if: 
  1. L1/BLOCK_SIZE<L2/BLOCK_SIZE 
  2. L1/BLOCK_SIZE=L2/BLOCK_SIZE and R1<R2

After sorting the queries, the next step is to compute the answer to the first query and consequently answer rest of the queries. To determine if a particular element exists or not, check the frequency of the element in that range. A non zero frequency confirms the existence of the element in that range. 
To store the frequency of the elements, STL map has been used in the following code. 

In the example given, first query after sorting the array of queries is {0, 2, 2}. Hash the frequencies of the elements in [0, 2] and then check the frequency of the element 2 from the map. Since, 2 occurs 0 times, print “No”. 

While processing the next query, which is {1, 4, 1} in this case, decrement the frequencies of the elements in the range [0, 1) and increment the frequencies of the elements in range [3, 4]. This step gives the frequencies of elements in [1, 4] and it can easily be seen from the map that 1 exists in this range.

Time complexity : 

The pre-processing part, that is sorting the queries takes O(m Log m) time. 
The index variable for R changes at most times throughout the run and that for L changes its value at most times. Hence, processing all queries takes + = time.

Space complexity:

The space complexity of the given program is O((m+n) * sqrt(n)) where n is the size of the input array a[] and m is the number of queries. 

Below is the C++ implementation of the above idea : 

2 does not exist between [0, 2] 
1 exists between [1, 4] 
5 exists between [2, 4] 

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