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:
// 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 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 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# 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 |
<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> |
14 385 54 139
Time Complexity: O(Q + 10000 * x)
Auxiliary Space: O(100010)