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

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 rangestruct Query {    int L, R;}; // Fills lookup array lookup[n][n]// for all possible values// of query rangesvoid 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 codeint 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;}

## 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[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]    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 rangeclass Query:         def __init__(self, L, R):                 self.L = L        self.R = R  # Fills lookup array lookup[n][n]# for all possible values# of query rangesdef 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[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]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 codeif __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[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]    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[0][3] 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[0][2]] <=  arr[lookup[4][2]],
// then lookup[0][3] = lookup[0][2]
If arr[lookup[i][j-1]] <= arr[lookup[i+2j-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][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[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[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 rangestruct 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 codeint 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;}

## 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 timeimport 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[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            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 timefrom 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[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]             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[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]  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 Codeif __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 timeusing 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.