Range Min Using Sparse Table

Approach

As the title of this article itself suggests we will be using the concept of a sparse table to solve the problem. Let’s first see how a sparse table is constructed from a given array.

Construction of Sparse Table

The main intuition behind sparse table representation comes from the binary representation of an integer. As any integer can have its unique binary representation, it can be expressed as the sum of powers of 2. So, any interval or range can be expressed as the sum of decreasing powers of 2.

We will precompute the minimum of the ranges having the length of a power of 2 from each index of the given array. To answer queries efficiently we will use these precomputed values.

To store these precomputed values we will use a 2D array known as the sparse table of size n*(k+1) where n is the length of the given array and k=logn (of base 2).

For any sparse table st, st[i][j] will store the minimum of the range [ i to  i+2j-1] that is of length 2j.

range [i,  i+2j-1] can be broken into two ranges [i, i+2j-1-1] and [ i+2j-1,  i+2j-1] of length  2j-1.So, st[i][j] can be expressed as 
st[ i][ j] = min(st[ i ][ j-1] ,  st[ i+2j-1] [  j-1])

//constructing the sparse table

int st[n][k+1];

for(int i=0;i<n;i++)
{
  st[i][0]=arr[i];
}

for(int j=1;j<=k;j++)
{
	for(int i=0;i<n;i++)
	{
  	st[i][j] = min(st[i][j-1] , st[ i+1<<(j-1)][j-1]);
	}
}

For answering the given range min query [l,r] we will try to split the range into lengths of powers of 2 starting from the largest possible value k. If the first split is the range [l, l+2j-1] i.e. st[ l, j], we will consider the remaining range [r-2j +1, r] i.e.

st[r-2j +1, j].

//answering range min query (l,r)
int    query(int l,int r)
{
  int minimum=INT_MAX;
  int j=log(r-l+1)/log(2);

  minimum=min(st[ l][ j], st[r-(1<<j)+1][ j]);


 return minimum;
}    

Code

//C++ implementation of constrcuting Sparse Table #include<bits/stdc++.h> using namespace std; //driver code int main() { int n=6; int arr[] = { 2, 4, 3, 5, 7, 9 }; //constructing Sparse Table int k=log(n)/log(2); int st[n][k+1]; for(int i=0;i<n;i++) { st[i][0]=arr[i]; } for(int j=1;j<=k;j++) { for(int i=0;i<n;i++) { st[i][j] = min( st[i][j-1] , st[ i+1<<(j-1)][j-1] ); } } //answering query (l,r)     int l=0,r=2;    int minimum=INT_MAX; int j=log(r-l+1)/log(2); minimum=min(st[l][j],st[r-(1<<j)+1][j]); cout<<" min of the given range for the given query: "<<minimum; } You can also try this code with Online C++ Compiler Run Code

Output

min of the given range for the given query: 2 You can also try this code with Online C++ Compiler Run Code

Complexity analysis

The Time Complexity of constructing a sparse table is O(n*k) ≅ O(n*logn) and answering a single query takes O(1)where n is the size of the given array.

Space complexity is O(n*logn).

FAQs

1. What is the Sparse Table?
   Sparse Table is a special kind of data structure to answer range queries on an array efficiently. It is implemented as a 2d array.
    
2. In which type of problem Sparse Table can be used?
   To perform range queries on any static data set, the sparse table is used.

Key Takeaways

This article covered the construction of a sparse table and how to find min range using the sparse table.

Also, check out the article on the range sum using sparse table to clear the concepts about sparse table data structure.
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