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 minimum element in given range. This solution takes **O(n) **time to query in 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 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]. 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[0][3]] with arr[4]. 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 program int main() { int a[] = {7, 2, 3, 0, 5, 10, 3, 12, 18}; int n = sizeof(a)/sizeof(a[0]); Query q[] = {{0, 4}, {4, 7}, {7, 8}}; int m = sizeof(q)/sizeof(q[0]); 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 query in** O(1)**, but preprocessing takes **O(n ^{2}) **time. Also, this approach needs

**O(n**extra space which may become huge for large input arrays.

^{2})

**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 ^{j} **where 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 2

^{3})

**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]Ifarr[lookup[i][j-1]] <= arr[lookup[i+2^{j-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]Elselookup[i][j] = lookup[i+2^{j-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(Log_{2}(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]Ifarr[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]ElseRMQ(L, R) = lookup[R-(int)pow(2,j)+1][j]

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

Below is C++ implementation of above idea.

// 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<bits/stdc++.h> 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][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] 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[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] 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 program int main() { int a[] = {7, 2, 3, 0, 5, 10, 3, 12, 18}; int n = sizeof(a)/sizeof(a[0]); Query q[] = {{0, 4}, {4, 7}, {7, 8}}; int m = sizeof(q)/sizeof(q[0]); 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

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