http://www.geeksforgeeks.org/range-minimum-query-for-static-array/

Range Minimum Query (Square Root Decomposition and Sparse Table)

Input:  arr[]   = {7, 2, 3, 0, 5, 10, 3, 12, 18};
        query[] = [0, 4], [4, 7], [7, 8]

Output: Minimum of [0, 4] is 0
        Minimum of [4, 7] is 3
        Minimum of [7, 8] is 12

Output:

Minimum of [0, 4] is 0
Minimum of [4, 7] is 3
Minimum of [7, 8] is 12

This approach supports query in O(1), but preprocessing takes O(n²) time. Also, this approach needs O(n²) extra space which may become huge for large input arrays.

Method 2 (Square Root Decomposition)
We can use Square Root Decompositions to reduce space required in above method.

Preprocessing:
1) Divide the range [0, n-1] into different blocks of √n each.
2) Compute minimum of every block of size √n and store the results.

Preprocessing takes O(√n * √n) = O(n) time and O(√n) space.

Query:
1) To query a range [L, R], we take minimum of all blocks that lie in this range. For left and right corner blocks which may partially overlap with given range, we linearly scan them to find minimum.

Time complexity of query is O(√n). Note that we have minimum of middle block directly accessible and there can be at most O(√n) middle blocks. There can be atmost two corner blocks that we may have to scan, so we may have to scan 2*O(√n) elements of corner blocks. Therefore, overall time complexity is O(√n).

Refer Sqrt (or Square Root) Decomposition Technique | Set 1 (Introduction) for details.

Method 3 (Sparse Table Algorithm)
The above solution requires only O(√n) space, but takes O(√n) time to query. Sparse table method supports query time O(1) with extra space O(n Log n).

The idea is to precompute minimum of all subarrays of size 2^jwhere j varies from 0 to Log n. Like method 1, we make a lookup table. Here lookup[i][j] contains minimum of range starting from i and of size 2^j. For example lookup[0][3] contains minimum of range [0, 7] (starting with 0 and of size 2³)

Preprocessing:
How to fill this lookup table? The idea is simple, fill in bottom up manner using previously computed values.

For example, to find minimum of range [0, 7], we can use minimum of following two.
a) Minimum of range [0, 3]
b) Minimum of range [4, 7]

Based on above example, below is formula,

// If arr[lookup[0][2]] <=  arr[lookup[4][2]], 
// then lookup[0][3] = lookup[0][2]
If arr[lookup[i][j-1]] <= arr[lookup[i+2j-1-1][j-1]]
   lookup[i][j] = lookup[i][j-1]

// If arr[lookup[0][2]] >  arr[lookup[4][2]], 
// then lookup[0][3] = lookup[4][2]
Else 
   lookup[i][j] = lookup[i+2j-1-1][j-1] 

Query:
For any arbitrary range [l, R], we need to use ranges which are in powers of 2. The idea is to use closest power of 2. We always need to do at most one comparison (compare minimum of two ranges which are powers of 2). One range starts with L and and ends with “L + closest-power-of-2”. The other range ends at R and starts with “R – same-closest-power-of-2 + 1”. For example, if given range is (2, 10), we compare minimum of two ranges (2, 9) and (3, 10).

Based on above example, below is formula,

// For (2,10), j = floor(Log2(10-2+1)) = 3
j = floor(Log(R-L+1))

// If arr[lookup[0][3]] <=  arr[lookup[3][3]], 
// then RMQ(2,10) = lookup[0][3]
If arr[lookup[L][j]] <= arr[lookup[R-(int)pow(2,j)+1][j]]
   RMQ(L, R) = lookup[L][j]

// If arr[lookup[0][3]] >  arr[lookup[3][3]], 
// then RMQ(2,10) = lookup[3][3]
Else 
   RMQ(L, R) = lookup[i+2j-1-1][j-1]

Since we do only one comparison, time complexity of query is O(1).

Below is C++ implementation of above idea.

Output:

Minimum of [0, 4] is 0
Minimum of [4, 7] is 3
Minimum of [7, 8] is 12

So sparse table method supports query operation in O(1) time with O(n Log n) preprocessing time and O(n Log n) space.

This article is contributed by Ruchir Garg. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above