Find sum of all odd perfect squares in the range [L, R]

Given two integers L and R. The task is to find the sum of all the odd numbers which are perfect square in the range [L, R].

Examples:

Input: L = 1, R = 9
Output: 10
Explanation: The odd Numbers in the range are 1, 3, 5, 7, 9 and only 1, 9 are perfect squares of 1, 3. So, 1 + 9 = 10.

Input: L = 50, R = 10,000
Output: 166566

Naive Approach: The basic idea to solve this problem is to traverse the numbers in the range L to R, and for each odd number, check whether it is a perfect square as well.

Time Complexity: O(R-L)
Auxiliary Space: O(1)

Efficient Approach: The approach of the solution is based on the mathematical concept of sequence. The idea is to use sum of square of first N odd numbers.

Squares of first n odd natural numbers = \sum (2n-1)^{2} = \frac{n(2n+1)(2n-1)}{3}

Follow the steps below to solve the problem.:

Check count of perfect squares between 1 and the perfect squared odd number just greater or equal to L.
Check count of odd perfect squares in range [1, L).
Calculate sum (sum1) of odd perfect squares in range [1, L).
Check count of perfect squares in range [1, R].
Check count of odd perfect squares in range [1, R].
Calculate sum (sum2) of odd perfect squares in range [1, R].
Subtract sum1 from sum2 to get the sum of odd numbers which are perfect squares in range [L, R].

Below is the implementation of the above approach:

C++

// C++ implementation for the above approach
#include <cmath>
#include <iostream>
using namespace std;

// Function to find sum of all the odd
// numbers,which are perfect squares
// in range [L, R]
int findSum(int L, int R)
{
    // If L > R or both less than 0 
    if (L < 0 || R < 0 || L > R)
        return -1;

    int l, r, n1, n2, s1, s2;

    // Check count of numbers 
    // which are perfect squares between  
    // 1 & perfect squared odd number
    // just greater or equal to L
    l = ceil(sqrt(L));
    if (!(l & 1))
        l++;

    // Check count of numbers which
    // are perfect squares in range [1, R]
    r = floor(sqrt(R));
    if (!(r & 1))
        r--;

    // Check count of odd numbers which
    // are perfect squares in range [1, L)
    n1 = floor((float)l / 2);

    // Check count of odd numbers which
    // are perfect squares in range [1, R]
    n2 = ceil((float)r / 2);

    // Calculate sum of odd numbers which
    // are perfect squares in range [1, L)
    s1 = n1 * ((4 * n1 * n1) - 1) / 3;

    // Calculate sum of odd numbers which
    // are perfect squares in range [1, R]
    s2 = n2 * ((4 * n2 * n2) - 1) / 3;

    // Return sum of odd numbers which
    // are perfect squares in range [L, R]
    return s2 - s1;
}

// Driver Code
int main()
{
    int L = 1;
    int R = 9;

    cout << findSum(L, R);
    return 0;
}

Java

// Java implementation for the above approach
import java.util.*;
public class GFG
{
// Function to find sum of all the odd
// numbers,which are perfect squares
// in range [L, R]
static int findSum(int L, int R)
{
    // If L > R or both less than 0 
    if (L < 0 || R < 0 || L > R)
        return -1;

    int l, r, n1, n2, s1, s2;

    // Check count of numbers 
    // which are perfect squares between  
    // 1 & perfect squared odd number
    // just greater or equal to L
    l = (int)Math.ceil(Math.sqrt(L));
    if ((l & 1) == 0)
        l++;

    // Check count of numbers which
    // are perfect squares in range [1, R]
    r = (int)Math.floor(Math.sqrt(R));
    if ((r & 1) == 0)
        r--;

    // Check count of odd numbers which
    // are perfect squares in range [1, L)
    n1 = (int)Math.floor((float)l / 2);

    // Check count of odd numbers which
    // are perfect squares in range [1, R]
    n2 = (int)Math.ceil((float)r / 2);

    // Calculate sum of odd numbers which
    // are perfect squares in range [1, L)
    s1 = n1 * ((4 * n1 * n1) - 1) / 3;

    // Calculate sum of odd numbers which
    // are perfect squares in range [1, R]
    s2 = n2 * ((4 * n2 * n2) - 1) / 3;

    // Return sum of odd numbers which
    // are perfect squares in range [L, R]
    return s2 - s1;
}

// Driver Code
public static void main(String args[])
{
    int L = 1;
    int R = 9;

    System.out.println(findSum(L, R));
}
}

// This code is contributed by Samim Hossain Mondal.

Python3

# Python3 implementation for the above approach
import math

# Function to find sum of all the odd
# numbers,which are perfect squares
# in range [L, R]
def findSum(L, R):
    
    # If L > R or both less than 0
    if (L < 0 or R < 0 or L > R):
        return -1
        
    # Check count of numbers which are
    # perfect squares between 1 & perfect
    # squared odd number just greater or 
    # equal to L
    l = math.ceil(math.sqrt(L))
    if (not (l & 1)):
        l += 1

    # Check count of numbers which
    # are perfect squares in range [1, R]
    r = math.floor(math.sqrt(R))
    if (not (r & 1)):
        r -= 1

    # Check count of odd numbers which
    # are perfect squares in range [1, L)
    n1 = math.floor(l / 2)

    # Check count of odd numbers which
    # are perfect squares in range [1, R]
    n2 = math.ceil(r / 2)

    # Calculate sum of odd numbers which
    # are perfect squares in range [1, L)
    s1 = int(n1 * ((4 * n1 * n1) - 1) / 3)

    # Calculate sum of odd numbers which
    # are perfect squares in range [1, R]
    s2 = int(n2 * ((4 * n2 * n2) - 1) / 3)

    # Return sum of odd numbers which
    # are perfect squares in range [L, R]
    return s2 - s1

# Driver Code
if __name__ == "__main__":

    L = 1
    R = 9

    print(findSum(L, R))

# This code is contributed by rakeshsahni

// C# implementation for the above approach
using System;
class GFG
{
  
// Function to find sum of all the odd
// numbers,which are perfect squares
// in range [L, R]
static int findSum(int L, int R)
{
  
    // If L > R or both less than 0 
    if (L < 0 || R < 0 || L > R)
        return -1;

    int l, r, n1, n2, s1, s2;

    // Check count of numbers 
    // which are perfect squares between  
    // 1 & perfect squared odd number
    // just greater or equal to L
    l = (int)Math.Ceiling(Math.Sqrt(L));
    if ((l & 1) == 0)
        l++;

    // Check count of numbers which
    // are perfect squares in range [1, R]
    r = (int)Math.Floor(Math.Sqrt(R));
    if ((r & 1) == 0)
        r--;

    // Check count of odd numbers which
    // are perfect squares in range [1, L)
    n1 = (int)Math.Floor((float)l / 2);

    // Check count of odd numbers which
    // are perfect squares in range [1, R]
    n2 = (int)Math.Ceiling((float)r / 2);

    // Calculate sum of odd numbers which
    // are perfect squares in range [1, L)
    s1 = n1 * ((4 * n1 * n1) - 1) / 3;

    // Calculate sum of odd numbers which
    // are perfect squares in range [1, R]
    s2 = n2 * ((4 * n2 * n2) - 1) / 3;

    // Return sum of odd numbers which
    // are perfect squares in range [L, R]
    return s2 - s1;
}

// Driver Code
public static void Main()
{
    int L = 1;
    int R = 9;

    Console.Write(findSum(L, R));
}
}

// This code is contributed by Samim Hossain Mondal.

JavaScript

<script>

// JavaScript implementation for the above approach 

// Function to find sum of all the odd
// numbers,which are perfect squares
// in range [L, R]
function findSum(L, R) 
{
    
    // If L > R or both less than 0 
    if (L < 0 || R < 0 || L > R)
        return -1;

    let l, r, n1, n2, s1, s2;

    // Check count of numbers 
    // which are perfect squares between  
    // 1 & perfect squared odd number
    // just greater or equal to L
    l = Math.ceil(Math.sqrt(L));
    if (!(l & 1))
        l++;

    // Check count of numbers which
    // are perfect squares in range [1, R]
    r = Math.floor(Math.sqrt(R));
    if (!(r & 1))
        r--;

    // Check count of odd numbers which
    // are perfect squares in range [1, L)
    n1 = Math.floor(l / 2);

    // Check count of odd numbers which
    // are perfect squares in range [1, R]
    n2 = Math.ceil(r / 2);

    // Calculate sum of odd numbers which
    // are perfect squares in range [1, L)
    s1 = n1 * ((4 * n1 * n1) - 1) / 3;

    // Calculate sum of odd numbers which
    // are perfect squares in range [1, R]
    s2 = n2 * ((4 * n2 * n2) - 1) / 3;

    // Return sum of odd numbers which
    // are perfect squares in range [L, R]
    return s2 - s1;
}

// Driver Code
let L = 1;
let R = 9;

document.write(findSum(L, R));

// This code is contributed by Potta Lokesh

</script>

Output

Time Complexity: O(1)
Auxiliary Space: O(1)

Find sum of all odd perfect squares in the range [L, R]

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