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# Min-Max Range Queries in Array

• Difficulty Level : Hard
• Last Updated : 20 Aug, 2021

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:

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```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``using` `namespace` `std;` `// Node for storing minimum nd 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 program to test above functions``int` `main()``{``    ``int` `arr[] = {1, 8, 5, 9, 6, 14, 2, 4, 3, 7};``    ``int` `n = ``sizeof``(arr)/``sizeof``(arr);` `    ``// 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)

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.