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++ program to find minimum and maximum using segment tree` `#include<bits/stdc++.h>` `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 (quey 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[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)

**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(n^{2}) preprocessing time and O(n^{2}) 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.

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