Sparse Table

4.3

We have briefly discussed sparse table in Range Minimum Query (Square Root Decomposition and Sparse Table)

Sparse table concept is used for fast queries on a set of static data (elements do not change). It does preprocessing so that the queries can answered efficiently.

 

Example Problem 1 : Range Minimum Query


We have an array arr[0 . . . n-1]. We need 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

The idea is to precompute minimum of all subarrays of size 2j where j varies from 0 to Log n. We make a table lookup[i][j] such that lookup[i][j] contains minimum of range starting from i and of size 2j. For example lookup[0][3] contains minimum of range [0, 7] (starting with 0 and of size 23)

How to fill this lookup or sparse table?
The idea is simple, fill in bottom up manner using previously computed values. We compute ranges with current power of 2 using values of lower power of two. For example, to find minimum of range [0, 7] (Range size is power of 3), we can use minimum of following two.
a) Minimum of range [0, 3] (Range size is power of 2)
b) Minimum of range [4, 7] (Range size is power of 2)

Based on above example, below is formula,

// Minimum of single element subarrays is same
// as the only element.
lookup[i][i] = 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] 

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(Log2(10-2+1)) = 3
j = floor(Log(R-L+1))

// If lookup[0][3] <=  lookup[3][3], 
// then min(2, 10) = lookup[0][3]
If lookup[L][j] <= lookup[R-(int)pow(2, j)+1][j]
   min(L, R) = lookup[L][j]

// If lookup[0][3] >  arr[lookup[3][3], 
// then min(2, 10) = lookup[3][3]
Else 
   min(L, R) = lookup[i+2j-1-1][j-1]

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
// using sparse table
#include <bits/stdc++.h>
using namespace std;
#define MAX 500

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

// 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 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][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 minimum 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 minimum 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;
}

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.

 

Example Problem 2 : Range GCD Query


We have an array arr[0 . . . n-1]. We need to find the greatest common divisor in the range L and R where 0 <= L <= R <= n-1. Consider a situation when there are many range queries
Examples:

Input : arr[] = {2, 3, 5, 4, 6, 8}
        queries[] = {(0, 2), (3, 5), (2, 3)}
Output : 1
         2
         1

We use below properties of GCD:

  • GCD function is associative [ GCD(a, b, c) = GCD(GCD(a, b), c) = GCD(a, GCD(b, c))], we can compute GCD of a range using GCDs of subranges.
  • If we take GCD of an overlapping range more than once, then it does not change answer. For example GCD(a, b, c) = GCD(GCD(a, b), GCD(b, c)). Therefore like minimum range query problem, we need to do only one comparison to find GCD of given range.

We build sparse table using same logic as above. After building sparse table, we can find all GCDs by breaking given range in powers of 2 and add GCD of every piece to current answer.

// C++ program to do range minimum query
// using sparse table
#include <bits/stdc++.h>
using namespace std;
#define MAX 500

// lookup[i][j] is going to store GCD of
// 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 table[MAX][MAX];

// it builds sparse table.
void buildSparseTable(int arr[], int n)
{
    // GCD of single element is element itself
    for (int i = 0; i < n; i++)
        table[i][0] = arr[i];

    // Build sparse table
    for (int j = 1; j <= k; j++)
        for (int i = 0; i <= n - (1 << j); i++)
            table[i][j] = __gcd(table[i][j - 1],
                    table[i + (1 << (j - 1))][j - 1]);
}

// Returns GCD 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 GCD of last 2^j elements with first
    // 2^j elements in range.
    // For [2, 10], we find GCD of arr[lookup[0][3]] and
    // arr[lookup[3][3]],
    return __gcd(table[L][j], table[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, 2) << endl;
    cout << query(1, 3) << endl;
    cout << query(4, 5) << endl;
    return 0;
}

Output:

1
1
5

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