Range Minimum Query(GeeskForGeeks)

本文介绍区间最小值查询的三种高效算法:简单解法、平方根分解及稀疏表算法。这些算法可在静态数组中实现快速查询,适用于大量查询场景。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

http://www.geeksforgeeks.org/range-minimum-query-for-static-array/


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 minimum element in given range. This solution takes O(n) time to query in 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 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]. Minimum of a given range can now be calculated in O(1) time.

rmqsimple

// C++ program to do range minimum query in O(1) time with O(n*n)
// extra space and O(n*n) preprocessing time.
#include<bits/stdc++.h>
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 range
struct Query
{
     int L, R;
};
 
// Fills lookup array lookup[n][n] for all possible values of
// query ranges
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]
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 program
int 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;
}

Output:

Minimum of [0, 4] is 0
Minimum of [4, 7] is 3
Minimum of [7, 8] is 12

This approach supports query 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 above method.

Preprocessing:
1) Divide the range [0, n-1] into different blocks of √n each.
2) Compute minimum of every block of size √n and store the results.

Preprocessing takes O(√n * √n) = O(n) time and O(√n) space.

rmq3

Query:
1) To query a range [L, R], we take minimum of all blocks that lie in this range. For left and right corner blocks which may partially overlap with given range, we linearly scan them to find minimum.

Time complexity of query is O(√n). Note that we have minimum of middle block directly accessible and there can be at most O(√n) middle blocks. There can be atmost two corner blocks that we may have to scan, so we may have to scan 2*O(√n) elements of corner blocks. Therefore, overall time complexity is O(√n).

Refer 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. Sparse table method supports query time O(1) with extra space O(n Log n).

The idea is to precompute 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 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)

Preprocessing:
How to fill this lookup table? The idea is simple, fill in bottom up manner using previously computed values.

For example, to find minimum of range [0, 7], we can use minimum of following two.
a) Minimum of range [0, 3]
b) Minimum of range [4, 7]

Based on above example, below is 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-1-1][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-1-1][j-1] 

rmqsparsetable

Query:
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 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[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 in O(1) time with
// O(n Log n) extra space and O(n Log n) preprocessing time
#include<bits/stdc++.h>
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 range
struct 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 program
int 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;
}

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.

This article is contributed by Ruchir Garg. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值