 Open in App
Not now

# Range Minimum Query (Square Root Decomposition and Sparse Table)

• Difficulty Level : Hard
• Last Updated : 26 May, 2022

We have an array arr[0 . . . n-1]. We should be able to efficiently find the minimum value from index L (query start) to R (query end) where 0 <= L <= R <= n-1. Consider a situation when there are many range queries.
Example:

```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```

A simple solution is to run a loop from L to R and find the minimum element in the given range. This solution takes O(n) time to query in the worst case.
Another approach is to use Segment tree. With segment tree, preprocessing time is O(n) and time to for range minimum query is O(Logn). The extra space required is O(n) to store the segment tree. Segment tree allows updates also in O(Log n) time.

### Can we do better if we know that the array is static?

How to optimize query time when there are no update operations and there are many range minimum queries?
Below are different methods.

Method 1 (Simple Solution)
A Simple Solution is to create a 2D array lookup[][] where an entry lookup[i][j] stores the minimum value in range arr[i..j]. The minimum of a given range can now be calculated in O(1) time. ## C++

 `// C++ program to do range``// minimum query in O(1) time with``// O(n*n) extra space and O(n*n)``// preprocessing time.``#include ``using` `namespace` `std;``#define MAX 500` `// lookup[i][j] is going to store``// index of minimum value in``// arr[i..j]``int` `lookup[MAX][MAX];` `// Structure to represent a query range``struct` `Query {``    ``int` `L, R;``};` `// Fills lookup array lookup[n][n]``// for all possible values``// of query ranges``void` `preprocess(``int` `arr[], ``int` `n)``{``    ``// Initialize lookup[][] for the``    ``// intervals with length 1``    ``for` `(``int` `i = 0; i < n; i++)``        ``lookup[i][i] = i;` `    ``// Fill rest of the entries in bottom up manner``    ``for` `(``int` `i = 0; i < n; i++) {``        ``for` `(``int` `j = i + 1; j < n; j++)` `            ``// To find minimum of [0,4],``            ``// we compare minimum``            ``// of arr[lookup] with arr.``            ``if` `(arr[lookup[i][j - 1]] < arr[j])``                ``lookup[i][j] = lookup[i][j - 1];``            ``else``                ``lookup[i][j] = j;``    ``}``}` `// Prints minimum of given m``// query ranges in arr[0..n-1]``void` `RMQ(``int` `arr[], ``int` `n, Query q[], ``int` `m)``{``    ``// Fill lookup table for``    ``// all possible input queries``    ``preprocess(arr, n);` `    ``// One by one compute sum of all queries``    ``for` `(``int` `i = 0; i < m; i++)``    ``{``        ``// Left and right boundaries``        ``// of current range``        ``int` `L = q[i].L, R = q[i].R;` `        ``// Print sum of current query range``        ``cout << ``"Minimum of ["` `<< L``             ``<< ``", "` `<< R << ``"] is "``             ``<< arr[lookup[L][R]] << endl;``    ``}``}` `// Driver code``int` `main()``{``    ``int` `a[] = { 7, 2, 3, 0, 5, 10, 3, 12, 18 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);``    ``Query q[] = { { 0, 4 }, { 4, 7 }, { 7, 8 } };``    ``int` `m = ``sizeof``(q) / ``sizeof``(q);``    ``RMQ(a, n, q, m);``    ``return` `0;``}`

## Java

 `// Java program to do range minimum query``// in O(1) time with O(n*n) extra space``// and O(n*n) preprocessing time.``import` `java.util.*;` `class` `GFG {``    ``static` `int` `MAX = ``500``;` `    ``// lookup[i][j] is going to store index of``    ``// minimum value in arr[i..j]``    ``static` `int``[][] lookup = ``new` `int``[MAX][MAX];` `    ``// Structure to represent a query range``    ``static` `class` `Query {``        ``int` `L, R;` `        ``public` `Query(``int` `L, ``int` `R)``        ``{``            ``this``.L = L;``            ``this``.R = R;``        ``}``    ``};` `    ``// Fills lookup array lookup[n][n] for``    ``// all possible values of query ranges``    ``static` `void` `preprocess(``int` `arr[], ``int` `n)``    ``{``        ``// Initialize lookup[][] for``        ``// the intervals with length 1``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``lookup[i][i] = i;` `        ``// Fill rest of the entries in bottom up manner``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``for` `(``int` `j = i + ``1``; j < n; j++)` `                ``// To find minimum of [0,4],``                ``// we compare minimum of``                ``// arr[lookup] with arr.``                ``if` `(arr[lookup[i][j - ``1``]] < arr[j])``                    ``lookup[i][j] = lookup[i][j - ``1``];``                ``else``                    ``lookup[i][j] = j;``        ``}``    ``}` `    ``// Prints minimum of given m query``    ``// ranges in arr[0..n-1]``    ``static` `void` `RMQ(``int` `arr[], ``int` `n, Query q[], ``int` `m)``    ``{``        ``// Fill lookup table for``        ``// all possible input queries``        ``preprocess(arr, n);` `        ``// One by one compute sum of all queries``        ``for` `(``int` `i = ``0``; i < m; i++) {``            ``// Left and right boundaries``            ``// of current range``            ``int` `L = q[i].L, R = q[i].R;` `            ``// Print sum of current query range``            ``System.out.println(``"Minimum of ["` `+ L + ``", "` `+ R``                               ``+ ``"] is "``                               ``+ arr[lookup[L][R]]);``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `a[] = { ``7``, ``2``, ``3``, ``0``, ``5``, ``10``, ``3``, ``12``, ``18` `};``        ``int` `n = a.length;``        ``Query q[] = { ``new` `Query(``0``, ``4``), ``new` `Query(``4``, ``7``),``                      ``new` `Query(``7``, ``8``) };``        ``int` `m = q.length;``        ``RMQ(a, n, q, m);``    ``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program to do range``# minimum query in O(1) time with``# O(n*n) extra space and O(n*n)``# preprocessing time.``MAX` `=` `500`` ` `# lookup[i][j] is going to store``# index of minimum value in``# arr[i..j]``lookup ``=` `[[``0` `for` `j ``in` `range``(``MAX``)]``             ``for` `i ``in` `range``(``MAX``)]`` ` `# Structure to represent``# a query range``class` `Query:``    ` `    ``def` `__init__(``self``, L, R):``        ` `        ``self``.L ``=` `L``        ``self``.R ``=` `R`` ` `# Fills lookup array lookup[n][n]``# for all possible values``# of query ranges``def` `preprocess(arr, n):` `    ``# Initialize lookup[][] for the``    ``# intervals with length 1``    ``for` `i ``in` `range``(n):``        ``lookup[i][i] ``=` `i;`` ` `    ``# Fill rest of the entries in``    ``# bottom up manner``    ``for` `i ``in` `range``(n):``        ``for` `j ``in` `range``(i ``+` `1``, n):`` ` `            ``# To find minimum of [0,4],``            ``# we compare minimum``            ``# of arr[lookup] with arr.``            ``if` `(arr[lookup[i][j ``-` `1``]] < arr[j]):``                ``lookup[i][j] ``=` `lookup[i][j ``-` `1``];``            ``else``:``                ``lookup[i][j] ``=` `j;   `` ` `# Prints minimum of given m``# query ranges in arr[0..n-1]``def` `RMQ(arr, n, q, m):` `    ``# Fill lookup table for``    ``# all possible input queries``    ``preprocess(arr, n);`` ` `    ``# One by one compute sum of``    ``# all queries``    ``for` `i ``in` `range``(m):` `        ``# Left and right boundaries``        ``# of current range``        ``L ``=` `q[i].L``        ``R ``=` `q[i].R;`` ` `        ``# Print sum of current query range``        ``print``(``"Minimum of ["` `+` `str``(L) ``+` `", "` `+``               ``str``(R) ``+` `"] is "` `+``               ``str``(arr[lookup[L][R]]))` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``a ``=` `[``7``, ``2``, ``3``, ``0``, ``5``,``         ``10``, ``3``, ``12``, ``18``]``    ``n ``=` `len``(a)   ``    ``q ``=` `[Query(``0``, ``4``),``         ``Query(``4``, ``7``),``         ``Query(``7``, ``8``)]   ``    ``m ``=` `len``(q)   ``    ``RMQ(a, n, q, m);`` ` `# This code is contributed by Rutvik_56`

## C#

 `// C# program to do range minimum query``// in O(1) time with O(n*n) extra space``// and O(n*n) preprocessing time.``using` `System;` `class` `GFG {``    ``static` `int` `MAX = 500;` `    ``// lookup[i][j] is going to store index of``    ``// minimum value in arr[i..j]``    ``static` `int``[, ] lookup = ``new` `int``[MAX, MAX];` `    ``// Structure to represent a query range``    ``public` `class` `Query {``        ``public` `int` `L, R;` `        ``public` `Query(``int` `L, ``int` `R)``        ``{``            ``this``.L = L;``            ``this``.R = R;``        ``}``    ``};` `    ``// Fills lookup array lookup[n][n] for``    ``// all possible values of query ranges``    ``static` `void` `preprocess(``int``[] arr, ``int` `n)``    ``{``        ``// Initialize lookup[][] for``        ``// the intervals with length 1``        ``for` `(``int` `i = 0; i < n; i++)``            ``lookup[i, i] = i;` `        ``// Fill rest of the entries in bottom up manner``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``for` `(``int` `j = i + 1; j < n; j++)` `                ``// To find minimum of [0,4],``                ``// we compare minimum of``                ``// arr[lookup] with arr.``                ``if` `(arr[lookup[i, j - 1]] < arr[j])``                    ``lookup[i, j] = lookup[i, j - 1];``                ``else``                    ``lookup[i, j] = j;``        ``}``    ``}` `    ``// Prints minimum of given m query``    ``// ranges in arr[0..n-1]``    ``static` `void` `RMQ(``int``[] arr, ``int` `n, Query[] q, ``int` `m)``    ``{``        ``// Fill lookup table for``        ``// all possible input queries``        ``preprocess(arr, n);` `        ``// One by one compute sum of all queries``        ``for` `(``int` `i = 0; i < m; i++) {``            ``// Left and right boundaries``            ``// of current range``            ``int` `L = q[i].L, R = q[i].R;` `            ``// Print sum of current query range``            ``Console.WriteLine(``"Minimum of ["` `+ L + ``", "` `+ R``                              ``+ ``"] is "``                              ``+ arr[lookup[L, R]]);``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] a = { 7, 2, 3, 0, 5, 10, 3, 12, 18 };``        ``int` `n = a.Length;``        ``Query[] q = { ``new` `Query(0, 4), ``new` `Query(4, 7),``                      ``new` `Query(7, 8) };``        ``int` `m = q.Length;``        ``RMQ(a, n, q, m);``    ``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output:

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

This approach supports queries 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 the above method.
Preprocessing:
1) Divide the range [0, n-1] into different blocks of √n each.
2) Compute the 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 a minimum of all blocks that lie in this range. For left and right corner blocks which may partially overlap with the given range, we linearly scan them to find the minimum.
The time complexity of the query is O(√n). Note that we have a minimum of the middle block directly accessible and there can be at most O(√n) middle blocks. There can be at most two corner blocks that we may have to scan, so we may have to scan 2*O(√n) elements of corner blocks. Therefore, the overall time complexity is O(√n).
Refer to 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. The sparse table method supports query time O(1) with extra space O(n Log n).
The idea is to precompute a minimum of all subarrays of size 2j where j varies from 0 to Log n. Like method 1, we make a lookup table. Here lookup[i][j] contains a minimum of range starting from i and of size 2j. For example lookup contains a 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 a bottom-up manner using previously computed values.
For example, to find a minimum of range [0, 7], we can use a minimum of the following two.
a) Minimum of range [0, 3]
b) Minimum of range [4, 7]
Based on the above example, below is the formula,

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

// If arr[lookup] >  arr[lookup],
// then lookup = lookup
Else
lookup[i][j] = lookup[i+2j-1][j-1] ``` Query:
For any arbitrary range [l, R], we need to use ranges that are in powers of 2. The idea is to use the closest power of 2. We always need to do at most one comparison (compare a minimum of two ranges which are powers of 2). One range starts with L 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 the given range is (2, 10), we compare a minimum of two ranges (2, 9) and (3, 10).
Based on the above example, below is the formula,

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

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

// If arr[lookup] >  arr[lookup],
// then RMQ(2,10) = lookup
Else
RMQ(L, R) = lookup[R-(int)pow(2,j)+1][j]```

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

Below is the implementation of the above idea.

## C++

 `// C++ program to do range minimum``// query in O(1) time with``// O(n Log n) extra space and``// O(n Log n) preprocessing time``#include ``using` `namespace` `std;``#define MAX 500` `// lookup[i][j] is going to``// store index of minimum value in``// arr[i..j]. Ideally lookup``// table size should not be fixed``// and should be determined using``// n Log n. It is kept``// constant to keep code simple.``int` `lookup[MAX][MAX];` `// Structure to represent a query range``struct` `Query {``    ``int` `L, R;``};` `// Fills lookup array``// lookup[][] in bottom up manner.``void` `preprocess(``int` `arr[], ``int` `n)``{``    ``// Initialize M for the``    ``// intervals with length 1``    ``for` `(``int` `i = 0; i < n; i++)``        ``lookup[i] = i;` `    ``// Compute values from smaller``    ``// to bigger intervals``    ``for` `(``int` `j = 1; (1 << j) <= n; j++)``    ``{``        ``// Compute minimum value for``        ``// all intervals with size``        ``// 2^j``        ``for` `(``int` `i = 0; (i + (1 << j) - 1) < n; i++)``        ``{``            ``// For arr, we``            ``// compare arr[lookup]``            ``// and arr[lookup]``            ``if` `(arr[lookup[i][j - 1]]``                ``< arr[lookup[i + (1 << (j - 1))][j - 1]])``                ``lookup[i][j] = lookup[i][j - 1];``            ``else``                ``lookup[i][j]``                    ``= lookup[i + (1 << (j - 1))][j - 1];``        ``}``    ``}``}` `// Returns minimum of arr[L..R]``int` `query(``int` `arr[], ``int` `L, ``int` `R)``{``    ``// For [2,10], j = 3``    ``int` `j = (``int``)log2(R - L + 1);` `    ``// For [2,10], we compare arr[lookup] and``    ``// arr[lookup],``    ``if` `(arr[lookup[L][j]]``        ``<= arr[lookup[R - (1 << j) + 1][j]])``        ``return` `arr[lookup[L][j]];` `    ``else``        ``return` `arr[lookup[R - (1 << j) + 1][j]];``}` `// Prints minimum of given``// m query ranges in arr[0..n-1]``void` `RMQ(``int` `arr[], ``int` `n, Query q[], ``int` `m)``{``    ``// Fills table lookup[n][Log n]``    ``preprocess(arr, n);` `    ``// One by one compute sum of all queries``    ``for` `(``int` `i = 0; i < m; i++)``    ``{``        ``// Left and right boundaries``        ``// of current range``        ``int` `L = q[i].L, R = q[i].R;` `        ``// Print sum of current query range``        ``cout << ``"Minimum of ["` `<< L << ``", "``             ``<< R << ``"] is "``             ``<< query(arr, L, R) << endl;``    ``}``}` `// Driver code``int` `main()``{``    ``int` `a[] = { 7, 2, 3, 0, 5, 10, 3, 12, 18 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);``    ``Query q[] = { { 0, 4 }, { 4, 7 }, { 7, 8 } };``    ``int` `m = ``sizeof``(q) / ``sizeof``(q);``    ``RMQ(a, n, q, m);``    ``return` `0;``}`

## Java

 `// Java program to do range minimum query``// in O(1) time with O(n Log n) extra space``// and O(n Log n) preprocessing time``import` `java.util.*;` `class` `GFG {` `    ``static` `int` `MAX = ``500``;` `    ``// lookup[i][j] is going to store index``    ``// of minimum value in arr[i..j].``    ``// Ideally lookup table size should not be fixed``    ``// and should be determined using n Log n.``    ``// It is kept constant to keep code simple.``    ``static` `int``[][] lookup = ``new` `int``[MAX][MAX];` `    ``// Structure to represent a query range``    ``static` `class` `Query {``        ``int` `L, R;` `        ``public` `Query(``int` `L, ``int` `R)``        ``{``            ``this``.L = L;``            ``this``.R = R;``        ``}``    ``};` `    ``// Fills lookup array lookup[][]``    ``// in bottom up manner.``    ``static` `void` `preprocess(``int` `arr[], ``int` `n)``    ``{``        ``// Initialize M for the intervals``        ``// with length 1``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``lookup[i][``0``] = i;` `        ``// Compute values from smaller``        ``// to bigger intervals``        ``for` `(``int` `j = ``1``; (``1` `<< j) <= n; j++)``        ``{``            ``// Compute minimum value for``            ``// all intervals with size 2^j``            ``for` `(``int` `i = ``0``;``                 ``(i + (``1` `<< j) - ``1``) < n;``                 ``i++)``            ``{``                ``// For arr, we compare``                ``// arr[lookup]``                ``// and arr[lookup]``                ``if` `(arr[lookup[i][j - ``1``]]``                    ``< arr[lookup[i + (``1` `<< (j - ``1``))]``                                ``[j - ``1``]])``                    ``lookup[i][j] = lookup[i][j - ``1``];``                ``else``                    ``lookup[i][j]``                        ``= lookup[i + (``1` `<< (j - ``1``))][j - ``1``];``            ``}``        ``}``    ``}` `    ``// Returns minimum of arr[L..R]``    ``static` `int` `query(``int` `arr[], ``int` `L, ``int` `R)``    ``{``        ``// For [2,10], j = 3``        ``int` `j = (``int``)Math.log(R - L + ``1``);` `        ``// For [2,10], we compare``        ``// arr[lookup]``        ``// and arr[lookup],``        ``if` `(arr[lookup[L][j]]``            ``<= arr[lookup[R - (``1` `<< j) + ``1``][j]])``            ``return` `arr[lookup[L][j]];` `        ``else``            ``return` `arr[lookup[R - (``1` `<< j) + ``1``][j]];``    ``}` `    ``// Prints minimum of given m``    ``// query ranges in arr[0..n-1]``    ``static` `void` `RMQ(``int` `arr[], ``int` `n,``                    ``Query q[], ``int` `m)``    ``{``        ``// Fills table lookup[n][Log n]``        ``preprocess(arr, n);` `        ``// One by one compute sum of all queries``        ``for` `(``int` `i = ``0``; i < m; i++)``        ``{``            ``// Left and right boundaries``            ``// of current range``            ``int` `L = q[i].L, R = q[i].R;` `            ``// Print sum of current query range``            ``System.out.println(``"Minimum of ["``                               ` `                               ``+ L + ``", "` `+ R``                               ``+ ``"] is "``                               ``+ query(arr, L, R));``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `a[] = { ``7``, ``2``, ``3``, ``0``, ``5``, ``10``, ``3``, ``12``, ``18` `};``        ``int` `n = a.length;``        ``Query q[] = { ``new` `Query(``0``, ``4``), ``new` `Query(``4``, ``7``),``                      ``new` `Query(``7``, ``8``) };``        ``int` `m = q.length;``        ``RMQ(a, n, q, m);``    ``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 program to do range minimum query``# in O(1) time with O(n Log n) extra space``# and O(n Log n) preprocessing time``from` `math ``import` `log2` `MAX` `=` `500` `# lookup[i][j] is going to store index of``# minimum value in arr[i..j].``# Ideally lookup table size should``# not be fixed and should be determined``# using n Log n. It is kept constant``# to keep code simple.``lookup ``=` `[[``0` `for` `i ``in` `range``(``500``)]``          ``for` `j ``in` `range``(``500``)]` `# Structure to represent a query range`  `class` `Query:``    ``def` `__init__(``self``, l, r):``        ``self``.L ``=` `l``        ``self``.R ``=` `r` `# Fills lookup array lookup[][]``# in bottom up manner.`  `def` `preprocess(arr: ``list``, n: ``int``):``    ``global` `lookup` `    ``# Initialize M for the``    ``# intervals with length 1``    ``for` `i ``in` `range``(n):``        ``lookup[i][``0``] ``=` `i` `    ``# Compute values from``    ``# smaller to bigger intervals``    ``j ``=` `1``    ``while` `(``1` `<< j) <``=` `n:` `        ``# Compute minimum value for``        ``# all intervals with size 2^j``        ``i ``=` `0``        ``while` `i ``+` `(``1` `<< j) ``-` `1` `< n:` `            ``# For arr, we compare``            ``# arr[lookup] and``            ``# arr[lookup]``            ``if` `(arr[lookup[i][j ``-` `1``]] <``                    ``arr[lookup[i ``+` `(``1` `<< (j ``-` `1``))][j ``-` `1``]]):``                ``lookup[i][j] ``=` `lookup[i][j ``-` `1``]``            ``else``:``                ``lookup[i][j] ``=` `lookup[i ``+``                                      ``(``1` `<< (j ``-` `1``))][j ``-` `1``]` `            ``i ``+``=` `1``        ``j ``+``=` `1` `# Returns minimum of arr[L..R]`  `def` `query(arr: ``list``, L: ``int``, R: ``int``) ``-``> ``int``:``    ``global` `lookup` `    ``# For [2,10], j = 3``    ``j ``=` `int``(log2(R ``-` `L ``+` `1``))` `    ``# For [2,10], we compare``    ``# arr[lookup] and``    ``# arr[lookup],``    ``if` `(arr[lookup[L][j]] <``=``            ``arr[lookup[R ``-` `(``1` `<< j) ``+` `1``][j]]):``        ``return` `arr[lookup[L][j]]``    ``else``:``        ``return` `arr[lookup[R ``-` `(``1` `<< j) ``+` `1``][j]]` `# Prints minimum of given``# m query ranges in arr[0..n-1]`  `def` `RMQ(arr: ``list``, n: ``int``, q: ``list``, m: ``int``):` `    ``# Fills table lookup[n][Log n]``    ``preprocess(arr, n)` `    ``# One by one compute sum of all queries``    ``for` `i ``in` `range``(m):` `        ``# Left and right boundaries``        ``# of current range``        ``L ``=` `q[i].L``        ``R ``=` `q[i].R` `        ``# Print sum of current query range``        ``print``(``"Minimum of [%d, %d] is %d"` `%``              ``(L, R, query(arr, L, R)))`  `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ``a ``=` `[``7``, ``2``, ``3``, ``0``, ``5``, ``10``, ``3``, ``12``, ``18``]``    ``n ``=` `len``(a)``    ``q ``=` `[Query(``0``, ``4``), Query(``4``, ``7``),``         ``Query(``7``, ``8``)]``    ``m ``=` `len``(q)` `    ``RMQ(a, n, q, m)` `# This code is contributed by``# sanjeev2552`

## C#

 `// C# program to do range minimum query``// in O(1) time with O(n Log n) extra space``// and O(n Log n) preprocessing time``using` `System;` `class` `GFG {` `    ``static` `int` `MAX = 500;` `    ``// lookup[i,j] is going to store index``    ``// of minimum value in arr[i..j].``    ``// Ideally lookup table size should not be fixed``    ``// and should be determined using n Log n.``    ``// It is kept constant to keep code simple.``    ``static` `int``[, ] lookup = ``new` `int``[MAX, MAX];` `    ``// Structure to represent a query range``    ``public` `class` `Query {``        ``public` `int` `L, R;` `        ``public` `Query(``int` `L, ``int` `R)``        ``{``            ``this``.L = L;``            ``this``.R = R;``        ``}``    ``};` `    ``// Fills lookup array lookup[,]``    ``// in bottom up manner.``    ``static` `void` `preprocess(``int``[] arr, ``int` `n)``    ``{``        ``// Initialize M for the intervals``        ``// with length 1``        ``for` `(``int` `i = 0; i < n; i++)``            ``lookup[i, 0] = i;` `        ``// Compute values from smaller``        ``// to bigger intervals``        ``for` `(``int` `j = 1; (1 << j) <= n; j++)``        ``{``            ``// Compute minimum value for``            ``// all intervals with size 2^j``            ``for` `(``int` `i = 0;``                 ``(i + (1 << j) - 1) < n;``                 ``i++)``            ``{``                ``// For arr[2,10], we compare``                ``// arr[lookup[0,3]] and arr[lookup[3,3]]``                ``if` `(arr[lookup[i, j - 1]]``                    ``< arr[lookup[i + (1 << (j - 1)),``                                 ``j - 1]])``                    ``lookup[i, j] = lookup[i, j - 1];``                ``else``                    ``lookup[i, j]``                        ``= lookup[i + (1 << (j - 1)), j - 1];``            ``}``        ``}``    ``}` `    ``// Returns minimum of arr[L..R]``    ``static` `int` `query(``int``[] arr, ``int` `L, ``int` `R)``    ``{``        ``// For [2,10], j = 3``        ``int` `j = (``int``)Math.Log(R - L + 1);` `        ``// For [2,10], we compare arr[lookup[0,3]]``        ``// and arr[lookup[3,3]],``        ``if` `(arr[lookup[L, j]]``            ``<= arr[lookup[R - (1 << j) + 1, j]])``            ``return` `arr[lookup[L, j]];` `        ``else``            ``return` `arr[lookup[R - (1 << j) + 1, j]];``    ``}` `    ``// Prints minimum of given m``    ``// query ranges in arr[0..n-1]``    ``static` `void` `RMQ(``int``[] arr,``                    ``int` `n, Query[] q, ``int` `m)``    ``{``        ``// Fills table lookup[n,Log n]``        ``preprocess(arr, n);` `        ``// One by one compute sum of all queries``        ``for` `(``int` `i = 0; i < m; i++)``        ``{``            ``// Left and right``            ``// boundaries of current range``            ``int` `L = q[i].L, R = q[i].R;` `            ``// Print sum of current query range``            ``Console.WriteLine(``"Minimum of ["` `+ L + ``", "` `+ R``                              ``+ ``"] is "` `+ query(arr, L, R));``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] a = { 7, 2, 3, 0, 5, 10, 3, 12, 18 };``        ``int` `n = a.Length;``        ``Query[] q = { ``new` `Query(0, 4), ``new` `Query(4, 7),``                      ``new` `Query(7, 8) };``        ``int` `m = q.Length;``        ``RMQ(a, n, q, m);``    ``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``

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.