# Range maximum query using Sparse Table

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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] = arr[i]

// If lookup ≥  lookup,
// then lookup = lookup
If lookup[i][j-1] ≥ lookup[i+2j-1-1][j-1]
lookup[i][j] = lookup[i][j-1]

// If lookup <  lookup,
// then lookup = lookup
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 ` `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 manner ` `void` `buildSparseTable(``int` `arr[], ``int` `n) ` `{ ` `    ``// Initialize M for the intervals with length 1 ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``lookup[i] = 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, we compare arr[lookup] ` `            ``// and arr[lookup] ` `            ``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] and ` `    ``// arr[lookup] ` `    ``if` `(lookup[L][j] >= lookup[R - (1 << j) + 1][j]) ` `        ``return` `lookup[L][j]; ` ` `  `    ``else` `        ``return` `lookup[R - (1 << j) + 1][j]; ` `} ` ` `  `// Driver program ` `int` `main() ` `{ ` `    ``int` `a[] = { 7, 2, 3, 0, 5, 10, 3, 12, 18 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` ` `  `    ``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 approach ` `import` `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, we compare arr[lookup] ` `                ``// and arr[lookup] ` `                ``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] and ` `        ``// arr[lookup] ` `        ``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 approach ` `from` `math ``import` `log ` `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. ` `lookup ``=` `[[``0` `for` `i ``in` `range``(``MAX``)]  ` `             ``for` `i ``in` `range``(``MAX``)] ` ` `  `# Fills lookup array lookup[][]  ` `# in bottom up manner ` `def` `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, we compare arr[lookup] ` `            ``# and arr[lookup] ` `            ``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] and ` `    ``# arr[lookup] ` `    ``if` `(lookup[L][j] >``=` `lookup[R ``-` `(``1` `<< j) ``+` `1``][j]): ` `        ``return` `lookup[L][j] ` ` `  `    ``else``: ` `        ``return` `lookup[R ``-` `(``1` `<< j) ``+` `1``][j] ` ` `  `# Driver Code ` `a ``=` `[``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 approach ` `using` `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, we compare arr[lookup] ` `                ``// and arr[lookup] ` `                ``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] and ` `        ``// arr[lookup] ` `        ``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)); ` `    ``} ` `} `

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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Improved By : mohit kumar 29