Min-Max Range Queries in Array

18 Apr, 2022
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 
Simple Solution : We solve this problem using Tournament Method for each query. Complexity for this approach will be O(queries * n).
Efficient solution : This problem can be solved more efficiently by using Segment Tree. First read given segment tree link then start solving this problem. 
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 rangestruct 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 stvoid 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 program to test above functionsint 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);     // Print minimum and maximum value in arr[qs..qe]    printf("Minimum = %d and Maximum = %d ",                     result.minimum, result.maximum);     return 0;}
Output: 
Minimum = 1 and Maximum = 14 
Time Complexity : O(queries * logn)
Auxiliary Space: O(n)
Can we do better if there are no updates on 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 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 range 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 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 you want to share more information about the topic discussed above. 
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