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 

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 <bits/stdc++.h>
#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[100010];

// 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[][2] = { { 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]);
    }

    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[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];
    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

Javascript

<script>

// Javascript program to find the sum of all
// perfect squares in the given range

// Array to precompute the sum of squares
// from 1 to 100010 so that for every
// query, the answer can be returned in O(1).
var pref= Array(100010).fill(0);

// Function to check if a number is
// a perfect square or not
function isPerfectSquare(x)
{
    // Find floating point value of
    // square root of x.
    var 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.
function compute()
{
    for (var i = 1; i <= 100000; ++i) {
        pref[i] = pref[i - 1]
                  + isPerfectSquare(i);
    }
}

// Function to print the sum for each query
function printSum(L, R)
{
    var sum = pref[R] - pref[L - 1];
    document.write(sum + " ");
}

// Driver code

// To calculate the precompute function
compute();
var Q = 4;
arr = [ [ 1, 10 ],
                 [ 1, 100 ],
                 [ 2, 25 ],
                 [ 4, 50 ] ];

// Calling the printSum function
// for every query
for(var i = 0; i < Q; i++)
    printSum(arr[i][0], arr[i][1]);


</script>

Output:

14 385 54 139

Time Complexity: O(Q + 10000 * x)

Auxiliary Space: O(100010)