Given an array arr[], the task is to answer queries to find the maximum of all the elements in the index range arr[L…R].
Examples:
Input: arr[] = {6, 7, 4, 5, 1, 3}, q[][] = {{0, 5}, {3, 5}, {2, 4}}
Output:
7
5
5
Input: arr[] = {3, 34, 1}, q[][] = {{1, 2}}
Output:
34
Approach: A similar problem to answer range minimum queries has been discussed here. The same approach can be modified to answer range maximum queries. Below is the modification:
// Maximum of single element subarrays is same // as the only element lookup[i][0] = arr[i] // If lookup[0][2] ? lookup[4][2], // then lookup[0][3] = lookup[0][2] If lookup[i][j-1] ? lookup[i+2j-1-1][j-1] lookup[i][j] = lookup[i][j-1] // If lookup[0][2] < lookup[4][2], // then lookup[0][3] = lookup[4][2] Else lookup[i][j] = lookup[i+2j-1-1][j-1]
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define MAX 500// lookup[i][j] is going to store maximum// value in arr[i..j]. Ideally lookup table// size should not be fixed and should be// determined using n Log n. It is kept// constant to keep code simple.int lookup[MAX][MAX];// Fills lookup array lookup[][] in bottom up mannervoid buildSparseTable(int arr[], int n){ // Initialize M for the intervals with length 1 for (int i = 0; i < n; i++) lookup[i][0] = arr[i]; // Compute values from smaller to bigger intervals for (int j = 1; (1 << j) <= n; j++) { // Compute maximum value for all intervals with // size 2^j for (int i = 0; (i + (1 << j) - 1) < n; i++) { // For arr[2][10], we compare arr[lookup[0][7]] // and arr[lookup[3][10]] if (lookup[i][j - 1] > lookup[i + (1 << (j - 1))][j - 1]) lookup[i][j] = lookup[i][j - 1]; else lookup[i][j] = lookup[i + (1 << (j - 1))][j - 1]; } }}// Returns maximum of arr[L..R]int query(int L, int R){ // Find highest power of 2 that is smaller // than or equal to count of elements in given // range // For [2, 10], j = 3 int j = (int)log2(R - L + 1); // Compute maximum of last 2^j elements with first // 2^j elements in range // For [2, 10], we compare arr[lookup[0][3]] and // arr[lookup[3][3]] if (lookup[L][j] >= lookup[R - (1 << j) + 1][j]) return lookup[L][j]; else return lookup[R - (1 << j) + 1][j];}// Driver programint main(){ int a[] = { 7, 2, 3, 0, 5, 10, 3, 12, 18 }; int n = sizeof(a) / sizeof(a[0]); buildSparseTable(a, n); cout << query(0, 4) << endl; cout << query(4, 7) << endl; cout << query(7, 8) << endl; return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG { static final int MAX = 500; // lookup[i][j] is going to store maximum // value in arr[i..j]. Ideally lookup table // size should not be fixed and should be // determined using n Log n. It is kept // constant to keep code simple. static int lookup[][] = new int[MAX][MAX]; // Fills lookup array lookup[][] in bottom up manner static void buildSparseTable(int arr[], int n) { // Initialize M for the intervals with length 1 for (int i = 0; i < n; i++) lookup[i][0] = arr[i]; // Compute values from smaller to bigger intervals for (int j = 1; (1 << j) <= n; j++) { // Compute maximum value for all intervals with // size 2^j for (int i = 0; (i + (1 << j) - 1) < n; i++) { // For arr[2][10], we compare arr[lookup[0][7]] // and arr[lookup[3][10]] if (lookup[i][j - 1] > lookup[i + (1 << (j - 1))][j - 1]) lookup[i][j] = lookup[i][j - 1]; else lookup[i][j] = lookup[i + (1 << (j - 1))][j - 1]; } } } // Returns maximum of arr[L..R] static int query(int L, int R) { // Find highest power of 2 that is smaller // than or equal to count of elements in given // range // For [2, 10], j = 3 int j = (int)Math.log(R - L + 1); // Compute maximum of last 2^j elements with first // 2^j elements in range // For [2, 10], we compare arr[lookup[0][3]] and // arr[lookup[3][3]] if (lookup[L][j] >= lookup[R - (1 << j) + 1][j]) return lookup[L][j]; else return lookup[R - (1 << j) + 1][j]; } // Driver program public static void main(String args[]) { int a[] = { 7, 2, 3, 0, 5, 10, 3, 12, 18 }; int n = a.length; buildSparseTable(a, n); System.out.println(query(0, 4)); System.out.println(query(4, 7)); System.out.println(query(7, 8)); }} |
Python3
# Python3 implementation of the approachfrom math import logMAX = 500# lookup[i][j] is going to store maximum# value in arr[i..j]. Ideally lookup table# size should not be fixed and should be# determined using n Log n. It is kept# constant to keep code simple.lookup = [[0 for i in range(MAX)] for i in range(MAX)]# Fills lookup array lookup[][] # in bottom up mannerdef buildSparseTable(arr, n): # Initialize M for the intervals # with length 1 for i in range(n): lookup[i][0] = arr[i] # Compute values from smaller # to bigger intervals i, j = 0, 1 while (1 << j) <= n: # Compute maximum value for # all intervals with size 2^j while (i + (1 << j) - 1) < n: # For arr[2][10], we compare arr[lookup[0][7]] # and arr[lookup[3][10]] if (lookup[i][j - 1] > lookup[i + (1 << (j - 1))][j - 1]): lookup[i][j] = lookup[i][j - 1] else: lookup[i][j] = lookup[i + (1 << (j - 1))][j - 1] i += 1 j += 1# Returns maximum of arr[L..R]def query(L, R): # Find highest power of 2 that is smaller # than or equal to count of elements in given # range # For [2, 10], j = 3 j = int(log(R - L + 1)) # Compute maximum of last 2^j elements with first # 2^j elements in range # For [2, 10], we compare arr[lookup[0][3]] and # arr[lookup[3][3]] if (lookup[L][j] >= lookup[R - (1 << j) + 1][j]): return lookup[L][j] else: return lookup[R - (1 << j) + 1][j]# Driver Codea = [7, 2, 3, 0, 5, 10, 3, 12, 18]n = len(a)buildSparseTable(a, n);print(query(0, 4))print(query(4, 7))print(query(7, 8))# This code is contributed by Mohit Kumar |
C#
// Java implementation of the approachusing System;class GFG { static int MAX = 500; // lookup[i][j] is going to store maximum // value in arr[i..j]. Ideally lookup table // size should not be fixed and should be // determined using n Log n. It is kept // constant to keep code simple. static int[, ] lookup = new int[MAX, MAX]; // Fills lookup array lookup[][] in bottom up manner static void buildSparseTable(int[] arr, int n) { // Initialize M for the intervals with length 1 for (int i = 0; i < n; i++) lookup[i, 0] = arr[i]; // Compute values from smaller to bigger intervals for (int j = 1; (1 << j) <= n; j++) { // Compute maximum value for all intervals with // size 2^j for (int i = 0; (i + (1 << j) - 1) < n; i++) { // For arr[2][10], we compare arr[lookup[0][7]] // and arr[lookup[3][10]] if (lookup[i, j - 1] > lookup[i + (1 << (j - 1)), j - 1]) lookup[i, j] = lookup[i, j - 1]; else lookup[i, j] = lookup[i + (1 << (j - 1)), j - 1]; } } } // Returns maximum of arr[L..R] static int query(int L, int R) { // Find highest power of 2 that is smaller // than or equal to count of elements in given // range // For [2, 10], j = 3 int j = (int)Math.Log(R - L + 1); // Compute maximum of last 2^j elements with first // 2^j elements in range // For [2, 10], we compare arr[lookup[0][3]] and // arr[lookup[3][3]] if (lookup[L, j] >= lookup[R - (1 << j) + 1, j]) return lookup[L, j]; else return lookup[R - (1 << j) + 1, j]; } // Driver program public static void Main(String[] args) { int[] a = { 7, 2, 3, 0, 5, 10, 3, 12, 18 }; int n = a.Length; buildSparseTable(a, n); Console.WriteLine(query(0, 4)); Console.WriteLine(query(4, 7)); Console.WriteLine(query(7, 8)); }} |
Javascript
<script>// Javascript implementation of the approachlet MAX = 500;// lookup[i][j] is going to store maximum// value in arr[i..j]. Ideally lookup table// size should not be fixed and should be// determined using n Log n. It is kept// constant to keep code simple.let lookup = new Array();for(let i = 0; i < MAX; i++){ let temp = []; for(let j = 0; j < MAX; j++) { temp.push([]) } lookup.push(temp)}// Fills lookup array lookup[][] in bottom up mannerfunction buildSparseTable(arr, n){ // Initialize M for the letervals with length 1 for (let i = 0; i < n; i++) lookup[i][0] = arr[i]; // Compute values from smaller to bigger letervals for (let j = 1; (1 << j) <= n; j++) { // Compute maximum value for all letervals with // size 2^j for (let i = 0; (i + (1 << j) - 1) < n; i++) { // For arr[2][10], we compare arr[lookup[0][7]] // and arr[lookup[3][10]] if (lookup[i][j - 1] > lookup[i + (1 << (j - 1))][j - 1]) lookup[i][j] = lookup[i][j - 1]; else lookup[i][j] = lookup[i + (1 << (j - 1))][j - 1]; } }}// Returns maximum of arr[L..R]function query(L, R){ // Find highest power of 2 that is smaller // than or equal to count of elements in given // range // For [2, 10], j = 3 let j = Math.floor(Math.log2(R - L + 1)); // Compute maximum of last 2^j elements with first // 2^j elements in range // For [2, 10], we compare arr[lookup[0][3]] and // arr[lookup[3][3]] if (lookup[L][j] >= lookup[R - (1 << j) + 1][j]) return lookup[L][j]; else return lookup[R - (1 << j) + 1][j];}// Driver program let a = [ 7, 2, 3, 0, 5, 10, 3, 12, 18 ]; let n = a.length; buildSparseTable(a, n); document.write(query(0, 4) + "<br>"); document.write(query(4, 7) + "<br>"); document.write(query(7, 8) + "<br>");// This code is contributed by _saurabh_jaiswal.</script> |
Output:
7 12 18
So sparse table method supports query operation in O(1) time with O(n Log n) preprocessing time and O(n Log n) space.
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