Min-Max Range Queries in Array
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 = 8Output: Minimum = 1 and Maximum = 9
Minimum = 2 and Maximum = 14
Minimum = 2 and Maximum = 14
Minimum = 5 and Maximum = 14
Minimum = 1 and Maximum = 14Input: arr[] = {2, 5, 3, 1, 8}, queries = 2
qs = 2 qe = 3
qs = 0 qe = 2Output: 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++
- Java
- Python3
- C#
- Javascript
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_MAX; tmp.maximum = INT_MIN; 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 codeint 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;} |
Java
Python3
C#
Javascript
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 reviewed by team GeeksForGeeks.
