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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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));
    }
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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));
    }
}

chevron_right


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.



My Personal Notes arrow_drop_up

Engineering student who loves competitive programming too much

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : mohit kumar 29