# Sum of all Perfect Squares lying in the range [L, R] for Q queries

Given Q queries in the form of 2D array arr[][] whose every row consists of two numbers L and R which signifies the range [L, R], the task is to find the sum of all perfect squares lying in this range.

Examples:

Input: Q = 2, arr[][] = {{4, 9}, {4, 16}}
Output: 13 29
Explanation:
From 4 to 9: only 4 and 9 are perfect squares. Therefore, 4 + 9 = 13.
From 4 to 16: 4, 9 and 16 are the perfect squares. Therefore, 4 + 9 + 16 = 29.

Input: Q = 4, arr[][] = {{1, 10}, {1, 100}, {2, 25}, {4, 50}}
Output: 14 385 54 139

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

Approach: The idea is to use a prefix sum array. The sum all squares are precomputed and stored in an array pref[] so that every query can be answered in O(1) time. Every ‘i’th index in the pref[] array represents the sum of perfect squares from 1 to that number. Therefore, the sum of perfect squares from the given range ‘L’ to ‘R’ can be found as follows:

`sum = pref[R] - pref[L - 1]`

Below is the implementation of the above approach:

## CPP

 `// C++ program to find the sum of all ` `// perfect squares in the given range ` ` `  `#include ` `#define ll int ` `using` `namespace` `std; ` ` `  `// Array to precompute the sum of squares ` `// from 1 to 100010 so that for every ` `// query, the answer can be returned in O(1). ` `long` `long` `pref; ` ` `  `// Function to check if a number is ` `// a perfect square or not ` `int` `isPerfectSquare(``long` `long` `int` `x) ` `{ ` `    ``// Find floating point value of ` `    ``// square root of x. ` `    ``long` `double` `sr = ``sqrt``(x); ` ` `  `    ``// If square root is an integer ` `    ``return` `((sr - ``floor``(sr)) == 0) ? x : 0; ` `} ` ` `  `// Function to precompute the perfect ` `// squares upto 100000. ` `void` `compute() ` `{ ` `    ``for` `(``int` `i = 1; i <= 100000; ++i) { ` `        ``pref[i] = pref[i - 1] ` `                  ``+ isPerfectSquare(i); ` `    ``} ` `} ` ` `  `// Function to print the sum for each query ` `void` `printSum(``int` `L, ``int` `R) ` `{ ` `    ``int` `sum = pref[R] - pref[L - 1]; ` `    ``cout << sum << ``" "``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// To calculate the precompute function ` `    ``compute(); ` ` `  `    ``int` `Q = 4; ` `    ``int` `arr[] = { { 1, 10 }, ` `                     ``{ 1, 100 }, ` `                     ``{ 2, 25 }, ` `                     ``{ 4, 50 } }; ` ` `  `    ``// Calling the printSum function ` `    ``// for every query ` `    ``for` `(``int` `i = 0; i < Q; i++) { ` `        ``printSum(arr[i], arr[i]); ` `    ``} ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find the sum of all ` `// perfect squares in the given range ` `class` `GFG ` `{ ` ` `  `// Array to precompute the sum of squares ` `// from 1 to 100010 so that for every ` `// query, the answer can be returned in O(1). ` `static` `int` `[]pref = ``new` `int``[``100010``]; ` ` `  `// Function to check if a number is ` `// a perfect square or not ` `static` `int` `isPerfectSquare(``int` `x) ` `{ ` `    ``// Find floating point value of ` `    ``// square root of x. ` `    ``double` `sr = Math.sqrt(x); ` ` `  `    ``// If square root is an integer ` `    ``return` `((sr - Math.floor(sr)) == ``0``) ? x : ``0``; ` `} ` ` `  `// Function to precompute the perfect ` `// squares upto 100000. ` `static` `void` `compute() ` `{ ` `    ``for` `(``int` `i = ``1``; i <= ``100000``; ++i)  ` `    ``{ ` `        ``pref[i] = pref[i - ``1``] ` `                ``+ isPerfectSquare(i); ` `    ``} ` `} ` ` `  `// Function to print the sum for each query ` `static` `void` `printSum(``int` `L, ``int` `R) ` `{ ` `    ``int` `sum = pref[R] - pref[L - ``1``]; ` `    ``System.out.print(sum+ ``" "``); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``// To calculate the precompute function ` `    ``compute(); ` ` `  `    ``int` `Q = ``4``; ` `    ``int` `arr[][] = { { ``1``, ``10` `}, ` `                    ``{ ``1``, ``100` `}, ` `                    ``{ ``2``, ``25` `}, ` `                    ``{ ``4``, ``50` `} }; ` ` `  `    ``// Calling the printSum function ` `    ``// for every query ` `    ``for` `(``int` `i = ``0``; i < Q; i++) ` `    ``{ ` `        ``printSum(arr[i][``0``], arr[i][``1``]); ` `    ``} ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python3 program to find the sum of all  ` `# perfect squares in the given range  ` `from` `math ``import` `sqrt, floor ` ` `  `# Array to precompute the sum of squares  ` `# from 1 to 100010 so that for every  ` `# query, the answer can be returned in O(1).  ` `pref ``=` `[``0``]``*``100010``;  ` ` `  `# Function to check if a number is  ` `# a perfect square or not  ` `def` `isPerfectSquare(x) : ` `     `  `    ``# Find floating point value of ` `    ``# square root of x. ` `    ``sr ``=` `sqrt(x); ` `     `  `    ``# If square root is an integer ` `    ``rslt ``=` `x ``if` `(sr ``-` `floor(sr) ``=``=` `0``) ``else` `0``; ` `    ``return` `rslt;  ` ` `  `# Function to precompute the perfect  ` `# squares upto 100000.  ` `def` `compute() : ` ` `  `    ``for` `i ``in` `range``(``1` `, ``100001``) : ` `        ``pref[i] ``=` `pref[i ``-` `1``] ``+` `isPerfectSquare(i);  ` ` `  `# Function to print the sum for each query  ` `def` `printSum( L, R) :  ` ` `  `    ``sum` `=` `pref[R] ``-` `pref[L ``-` `1``];  ` `    ``print``(``sum` `,end``=` `" "``);  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``# To calculate the precompute function  ` `    ``compute();  ` ` `  `    ``Q ``=` `4``;  ` `    ``arr ``=` `[ [ ``1``, ``10` `],  ` `            ``[ ``1``, ``100` `],  ` `            ``[ ``2``, ``25` `],  ` `            ``[ ``4``, ``50` `] ];  ` ` `  `    ``# Calling the printSum function  ` `    ``# for every query  ` `    ``for` `i ``in` `range``(Q) : ` `        ``printSum(arr[i][``0``], arr[i][``1``]);  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# program to find the sum of all ` `// perfect squares in the given range ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Array to precompute the sum of squares ` `// from 1 to 100010 so that for every ` `// query, the answer can be returned in O(1). ` `static` `int` `[]pref = ``new` `int``; ` ` `  `// Function to check if a number is ` `// a perfect square or not ` `static` `int` `isPerfectSquare(``int` `x) ` `{ ` `    ``// Find floating point value of ` `    ``// square root of x. ` `    ``double` `sr = Math.Sqrt(x); ` ` `  `    ``// If square root is an integer ` `    ``return` `((sr - Math.Floor(sr)) == 0) ? x : 0; ` `} ` ` `  `// Function to precompute the perfect ` `// squares upto 100000. ` `static` `void` `compute() ` `{ ` `    ``for` `(``int` `i = 1; i <= 100000; ++i)  ` `    ``{ ` `        ``pref[i] = pref[i - 1] ` `                ``+ isPerfectSquare(i); ` `    ``} ` `} ` ` `  `// Function to print the sum for each query ` `static` `void` `printSum(``int` `L, ``int` `R) ` `{ ` `    ``int` `sum = pref[R] - pref[L - 1]; ` `    ``Console.Write(sum+ ``" "``); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``// To calculate the precompute function ` `    ``compute(); ` ` `  `    ``int` `Q = 4; ` `    ``int` `[,]arr = { { 1, 10 }, ` `                    ``{ 1, 100 }, ` `                    ``{ 2, 25 }, ` `                    ``{ 4, 50 } }; ` ` `  `    ``// Calling the printSum function ` `    ``// for every query ` `    ``for` `(``int` `i = 0; i < Q; i++) ` `    ``{ ` `        ``printSum(arr[i, 0], arr[i, 1]); ` `    ``} ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```14 385 54 139
```

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Improved By : princiraj1992, AnkitRai01