# Range Minimum Query

Hello Everyone,

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++ program to do range`

`// minimum query in O(1) time with`

`// O(n*n) extra space and O(n*n)`

`// preprocessing time.`

`#include <bits/stdc++.h>`

`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;`

`}`

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.

Move to the more efficient method.

Thankyou.