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Min-Max Range Queries in Array

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  • Difficulty Level : Hard
  • Last Updated : 23 Sep, 2022
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Given an array arr[0 . . . n-1]. We need to efficiently find the minimum and maximum value from index qs (query start) to qe (query end) where 0 <= qs <= qe <= n-1. We are given multiple queries

Examples: 

Input: arr[] = {1, 8, 5, 9, 6, 14, 2, 4, 3, 7}, queries = 5

        qs = 0 qe = 4
        qs = 3 qe = 7
        qs = 1 qe = 6
        qs = 2 qe = 5
        qs = 0 qe = 8

Output: Minimum = 1 and Maximum = 9 
              Minimum = 2 and Maximum = 14 
              Minimum = 2 and Maximum = 14
              Minimum = 5 and Maximum = 14
              Minimum = 1 and Maximum = 14 

Input: arr[] = {2, 5, 3, 1, 8}, queries = 2

        qs = 2 qe = 3
        qs = 0 qe = 2

Output: Minimum = 1 and Maximum = 3 
              Minimum = 2 and Maximum = 5 

Naive Approach: To solve the problem follow the below idea:

We solve this problem using the Tournament Method for each query.
The time complexity of this approach will be O(queries * n)

Min-Max Range Queries in Array using segment trees:

To solve the problem follow the below idea:

This problem can be solved more efficiently by using a Segment Tree

Below is the implementation of the above approach:

C++




// C++ program to find minimum and maximum using segment
// tree
#include <bits/stdc++.h>
using namespace std;
 
// Node for storing minimum and maximum value of given range
struct node {
    int minimum;
    int maximum;
};
 
// A utility function to get the middle index from corner
// indexes.
int getMid(int s, int e) { return s + (e - s) / 2; }
 
/*  A recursive function to get the minimum and maximum
   value in a given range of array indexes. The following
   are parameters for this function.
 
    st    --> Pointer to segment tree
    index --> Index of current node in the segment tree.
   Initially 0 is passed as root is always at index 0 ss &
   se  --> Starting and ending indexes of the segment
                  represented  by current node, i.e.,
   st[index] qs & qe  --> Starting and ending indexes of
   query range */
struct node MaxMinUntill(struct node* st, int ss, int se,
                         int qs, int qe, int index)
{
    // If segment of this node is a part of given range,
    // then return
    //  the minimum and maximum node of the segment
    struct node tmp, left, right;
    if (qs <= ss && qe >= se)
        return st[index];
 
    // If segment of this node is outside the given range
    if (se < qs || ss > qe) {
        tmp.minimum = INT_MAX;
        tmp.maximum = INT_MIN;
        return tmp;
    }
 
    // If a part of this segment overlaps with the given
    // range
    int mid = getMid(ss, se);
    left = MaxMinUntill(st, ss, mid, qs, qe, 2 * index + 1);
    right = MaxMinUntill(st, mid + 1, se, qs, qe,
                         2 * index + 2);
    tmp.minimum = min(left.minimum, right.minimum);
    tmp.maximum = max(left.maximum, right.maximum);
    return tmp;
}
 
// Return minimum and maximum of elements in range from
// index qs (query start) to qe (query end).  It mainly uses
// MaxMinUtill()
struct node MaxMin(struct node* st, int n, int qs, int qe)
{
    struct node tmp;
 
    // Check for erroneous input values
    if (qs < 0 || qe > n - 1 || qs > qe) {
        printf("Invalid Input");
        tmp.minimum = INT_MIN;
        tmp.minimum = INT_MAX;
        return tmp;
    }
 
    return MaxMinUntill(st, 0, n - 1, qs, qe, 0);
}
 
// A recursive function that constructs Segment Tree for
// array[ss..se]. si is index of current node in segment
// tree st
void constructSTUtil(int arr[], int ss, int se,
                     struct node* st, int si)
{
    // If there is one element in array, store it in current
    // node of segment tree and return
    if (ss == se) {
        st[si].minimum = arr[ss];
        st[si].maximum = arr[ss];
        return;
    }
 
    // If there are more than one elements, then recur for
    // left and right subtrees and store the minimum and
    // maximum of two values in this node
    int mid = getMid(ss, se);
    constructSTUtil(arr, ss, mid, st, si * 2 + 1);
    constructSTUtil(arr, mid + 1, se, st, si * 2 + 2);
 
    st[si].minimum = min(st[si * 2 + 1].minimum,
                         st[si * 2 + 2].minimum);
    st[si].maximum = max(st[si * 2 + 1].maximum,
                         st[si * 2 + 2].maximum);
}
 
/* Function to construct segment tree from given array. This
   function allocates memory for segment tree and calls
   constructSTUtil() to fill the allocated memory */
struct node* constructST(int arr[], int n)
{
    // Allocate memory for segment tree
 
    // Height of segment tree
    int x = (int)(ceil(log2(n)));
 
    // Maximum size of segment tree
    int max_size = 2 * (int)pow(2, x) - 1;
 
    struct node* st = new struct node[max_size];
 
    // Fill the allocated memory st
    constructSTUtil(arr, 0, n - 1, st, 0);
 
    // Return the constructed segment tree
    return st;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 8, 5, 9, 6, 14, 2, 4, 3, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Build segment tree from given array
    struct node* st = constructST(arr, n);
 
    int qs = 0; // Starting index of query range
    int qe = 8; // Ending index of query range
    struct node result = MaxMin(st, n, qs, qe);
 
    // Function call
    printf("Minimum = %d and Maximum = %d ", result.minimum,
           result.maximum);
 
    return 0;
}


Output

Minimum = 1 and Maximum = 14 

Time Complexity: O(queries * log N)
Auxiliary Space: O(N)

Can we do better if there are no updates on the array? 

The above segment tree-based solution also allows array updates also to happen in O(Log n) time. Assume a situation when there are no updates (or the array is static). We can actually process all queries in O(1) time with some preprocessing. One simple solution is to make a 2D table of nodes that stores all ranges minimum and maximum. This solution requires O(1) query time but requires O(N2) preprocessing time and O(N2) extra space which can be a problem for large N. We can solve this problem in O(1) query time, O(n Log n) space and O(n Log n) preprocessing time using the Sparse Table.

This article is contributed by Shashank Mishra. This article is reviewed by team GeeksForGeeks. 
Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.

 


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