Range maximum query using Sparse Table

Given an array arr[], the task is to answer queries to find the maximum of all the elements in the index range arr[L…R].

Examples:

Input: arr[] = {6, 7, 4, 5, 1, 3}, q[][] = {{0, 5}, {3, 5}, {2, 4}}
Output:
7
5
5

Input: arr[] = {3, 34, 1}, q[][] = {{1, 2}}
Output:
34

Approach: A similar problem to answer range minimum queries has been discussed here. The same approach can be modified to answer range maximum queries. Below is the modification:

// Maximum of single element subarrays is same
// as the only element
lookup[i][0] = arr[i]

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

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

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define MAX 500
  
// lookup[i][j] is going to store maximum
// 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];
  
// Fills lookup array lookup[][] in bottom up manner
void buildSparseTable(int arr[], int n)
{
    // Initialize M for the intervals with length 1
    for (int i = 0; i < n; i++)
        lookup[i][0] = arr[i];
  
    // Compute values from smaller to bigger intervals
    for (int j = 1; (1 << j) <= n; j++) {
  
        // Compute maximum 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][7]]
            // and arr[lookup[3][10]]
            if (lookup[i][j - 1] > 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 maximum of arr[L..R]
int query(int L, int R)
{
    // Find highest power of 2 that is smaller
    // than or equal to count of elements in given
    // range
    // For [2, 10], j = 3
    int j = (int)log2(R - L + 1);
  
    // Compute maximum of last 2^j elements with first
    // 2^j elements in range
    // For [2, 10], we compare arr[lookup[0][3]] and
    // arr[lookup[3][3]]
    if (lookup[L][j] >= lookup[R - (1 << j) + 1][j])
        return lookup[L][j];
  
    else
        return lookup[R - (1 << j) + 1][j];
}
  
// Driver program
int main()
{
    int a[] = { 7, 2, 3, 0, 5, 10, 3, 12, 18 };
    int n = sizeof(a) / sizeof(a[0]);
  
    buildSparseTable(a, n);
  
    cout << query(0, 4) << endl;
    cout << query(4, 7) << endl;
    cout << query(7, 8) << endl;
  
    return 0;
}

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Java














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// Java implementation of the approach 


import java.util.*; 


class GFG { 


  


    static final int MAX = 500; 


  


    // lookup[i][j] is going to store maximum 


    // 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]; 


  


    // Fills lookup array lookup[][] in bottom up manner 


    static void buildSparseTable(int arr[], int n) 


    { 


        // Initialize M for the intervals with length 1 


        for (int i = 0; i < n; i++) 


            lookup[i][0] = arr[i]; 


  


        // Compute values from smaller to bigger intervals 


        for (int j = 1; (1 << j) <= n; j++) { 


  


            // Compute maximum 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][7]] 


                // and arr[lookup[3][10]] 


                if (lookup[i][j - 1] > 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 maximum of arr[L..R] 


    static int query(int L, int R) 


    { 


        // Find highest power of 2 that is smaller 


        // than or equal to count of elements in given 


        // range 


        // For [2, 10], j = 3 


        int j = (int)Math.log(R - L + 1); 


  


        // Compute maximum of last 2^j elements with first 


        // 2^j elements in range 


        // For [2, 10], we compare arr[lookup[0][3]] and 


        // arr[lookup[3][3]] 


        if (lookup[L][j] >= lookup[R - (1 << j) + 1][j]) 


            return lookup[L][j]; 


  


        else


            return lookup[R - (1 << j) + 1][j]; 


    } 


  


    // Driver program 


    public static void main(String args[]) 


    { 


        int a[] = { 7, 2, 3, 0, 5, 10, 3, 12, 18 }; 


        int n = a.length; 


  


        buildSparseTable(a, n); 


  


        System.out.println(query(0, 4)); 


        System.out.println(query(4, 7)); 


        System.out.println(query(7, 8)); 


    } 


} 



















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Python3














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# Python3 implementation of the approach 


from math import log 


MAX = 500


  


# lookup[i][j] is going to store maximum 


# 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(MAX)]  


             for i in range(MAX)] 


  


# Fills lookup array lookup[][]  


# in bottom up manner 


def buildSparseTable(arr, n): 


  


    # Initialize M for the intervals  


    # with length 1 


    for i in range(n): 


        lookup[i][0] = arr[i] 


  


    # Compute values from smaller  


    # to bigger intervals 


    i, j = 0, 1


    while (1 << j) <= n: 


  


        # Compute maximum value for  


        # all intervals with size 2^j 


        while (i + (1 << j) - 1) < n: 


  


            # For arr[2][10], we compare arr[lookup[0][7]] 


            # and arr[lookup[3][10]] 


            if (lookup[i][j - 1] >  


                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 maximum of arr[L..R] 


def query(L, R): 


      


    # Find highest power of 2 that is smaller 


    # than or equal to count of elements in given 


    # range 


    # For [2, 10], j = 3 


    j = int(log(R - L + 1)) 


  


    # Compute maximum of last 2^j elements with first 


    # 2^j elements in range 


    # For [2, 10], we compare arr[lookup[0][3]] and 


    # arr[lookup[3][3]] 


    if (lookup[L][j] >= lookup[R - (1 << j) + 1][j]): 


        return lookup[L][j] 


  


    else: 


        return lookup[R - (1 << j) + 1][j] 


  


# Driver Code 


a = [7, 2, 3, 0, 5, 10, 3, 12, 18] 


n = len(a) 


  


buildSparseTable(a, n); 


  


print(query(0, 4)) 


print(query(4, 7)) 


print(query(7, 8)) 


  


# This code is contributed by Mohit Kumar 



















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C#














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// Java implementation of the approach 


using System; 


class GFG { 


  


    static int MAX = 500; 


  


    // lookup[i][j] is going to store maximum 


    // 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]; 


  


    // Fills lookup array lookup[][] in bottom up manner 


    static void buildSparseTable(int[] arr, int n) 


    { 


        // Initialize M for the intervals with length 1 


        for (int i = 0; i < n; i++) 


            lookup[i, 0] = arr[i]; 


  


        // Compute values from smaller to bigger intervals 


        for (int j = 1; (1 << j) <= n; j++) { 


  


            // Compute maximum 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][7]] 


                // and arr[lookup[3][10]] 


                if (lookup[i, j - 1] > 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 maximum of arr[L..R] 


    static int query(int L, int R) 


    { 


        // Find highest power of 2 that is smaller 


        // than or equal to count of elements in given 


        // range 


        // For [2, 10], j = 3 


        int j = (int)Math.Log(R - L + 1); 


  


        // Compute maximum of last 2^j elements with first 


        // 2^j elements in range 


        // For [2, 10], we compare arr[lookup[0][3]] and 


        // arr[lookup[3][3]] 


        if (lookup[L, j] >= lookup[R - (1 << j) + 1, j]) 


            return lookup[L, j]; 


  


        else


            return lookup[R - (1 << j) + 1, j]; 


    } 


  


    // Driver program 


    public static void Main(String[] args) 


    { 


        int[] a = { 7, 2, 3, 0, 5, 10, 3, 12, 18 }; 


        int n = a.Length; 


  


        buildSparseTable(a, n); 


  


        Console.WriteLine(query(0, 4)); 


        Console.WriteLine(query(4, 7)); 


        Console.WriteLine(query(7, 8)); 


    } 


} 



















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Output:


7
12
18





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


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