Sum of the elements from index L to R in an array when arr[i] = i * (-1)^i

Given two integers L and R and an array arr[] every element of which at index i is calculated as arr[i] = i * (-1)i. The task is to find the sum of these elements of the array within the index range [L, R].

Examples:

Input: L = 1 , R = 5
Output: -3
Sum = (-1) + 2 + (-3) + 4 + (-5) = -3

Input: L = 5 , R = 100000000
Output: 49999998

Naive Approach: According to the definition of array elements, each odd element of the array is negative and even element is positive. So, to find the sum run a for loop from (L to R) and maintain the sum of all the odd(negative) and even(positive) numbers. Finally, return the sum.

Efficient Approach: It can be noted that the sum of all the odd elements of this series will always be equal to (totalOdd)2 where totalOdd = total number of odd elements and sum of the even numbers will be totalEven * (totalEven + 1). Now, all we have to do is to find the sum of all the odd elements upto L and upto R. Store the difference of both to get the sum of all the odd elements between L and R. Do this same for even numbers and finally return the difference of even and odd sums.

Below is the implementation of the above approach:

C++

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// CPP implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// function to return the odd sum
long int Odd_Sum(int n)
{
  
    // total odd elements upto n
    long int total = (n + 1) / 2;
  
    // sum of odd elements upto n
    long int odd = total * total;
  
    return odd;
}
  
// function to return the even sum
long int Even_Sum(int n)
{
  
    // total even elements upto n
    long int total = (n) / 2;
  
    // sum of even elements upto n
    long int even = total * (total + 1);
  
    return even;
}
  
// Function to find sum from L to R.
int sumLtoR(int L, int R)
{
  
    long int odd_sum, even_sum;
  
    odd_sum = Odd_Sum(R) - Odd_Sum(L - 1);
  
    even_sum = Even_Sum(R) - Even_Sum(L - 1);
  
    // return final sum from L to R
    return even_sum - odd_sum;
}
  
// Driver Program
int main()
{
  
    int L = 1, R = 5;
  
    // function call to print answer
    cout << sumLtoR(L, R);
  
    return 0;
}

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Java

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// Java implementation of above approach
  
import java.io.*;
  
class GFG {
      
  
  
// function to return the odd sum
static long  Odd_Sum(int n)
{
  
    // total odd elements upto n
    long  total = (n + 1) / 2;
  
    // sum of odd elements upto n
    long  odd = total * total;
  
    return odd;
}
  
// function to return the even sum
static long  Even_Sum(int n)
{
  
    // total even elements upto n
    long  total = (n) / 2;
  
    // sum of even elements upto n
    long  even = total * (total + 1);
  
    return even;
}
  
// Function to find sum from L to R.
static long sumLtoR(int L, int R)
{
  
    long  odd_sum, even_sum;
  
    odd_sum = Odd_Sum(R) - Odd_Sum(L - 1);
  
    even_sum = Even_Sum(R) - Even_Sum(L - 1);
  
    // return final sum from L to R
    return even_sum - odd_sum;
}
  
// Driver Program
  
    public static void main (String[] args) {
        int L = 1, R = 5;
  
    // function call to print answer
    System.out.println( sumLtoR(L, R));
    }
}
// This code is contributed by shs..

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Python3

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# Python3 implementation of above approach
  
# function to return the odd sum
def Odd_Sum(n):
  
    # total odd elements upto n
    total =(n+1)//2
  
    # sum of odd elements upto n
    odd = total*total
    return odd
  
# function to return the even sum
def Even_Sum(n):
  
    # total even elements upto n
    total = n//2
  
    # sum of even elements upto n
    even = total*(total+1)
    return even
  
def sumLtoR(L,R):
    odd_sum = Odd_Sum(R)-Odd_Sum(L-1)
    even_sum = Even_Sum(R)- Even_Sum(L-1)
  
    # return final sum from L to R
    return even_sum-odd_sum
  
  
# Driver code
L =1; R = 5
print(sumLtoR(L,R))
  
# This code is contributed by Shrikant13

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C#

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// C# implementation of above approach
class GFG
{
      
// function to return the odd sum
static long Odd_Sum(int n)
{
  
    // total odd elements upto n
    long total = (n + 1) / 2;
  
    // sum of odd elements upto n
    long odd = total * total;
  
    return odd;
}
  
// function to return the even sum
static long Even_Sum(int n)
{
  
    // total even elements upto n
    long total = (n) / 2;
  
    // sum of even elements upto n
    long even = total * (total + 1);
  
    return even;
}
  
// Function to find sum from L to R.
static long sumLtoR(int L, int R)
{
    long odd_sum, even_sum;
  
    odd_sum = Odd_Sum(R) - Odd_Sum(L - 1);
  
    even_sum = Even_Sum(R) - Even_Sum(L - 1);
  
    // return final sum from L to R
    return even_sum - odd_sum;
}
  
// Driver Code
public static void Main () 
{
    int L = 1, R = 5;
  
    // function call to print answer
    System.Console.WriteLine(sumLtoR(L, R));
}
}
  
// This code is contributed by mits

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PHP

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<?php
// PHP implementation of above approach
  
// function to return the odd sum
function Odd_Sum($n)
{
  
    // for total odd elements upto n
    // divide by 2
    $total = ($n + 1) >> 1;
  
    // sum of odd elements upto n
    $odd = $total * $total;
  
    return $odd;
}
  
// function to return the even sum
function Even_Sum($n)
{
  
    // for total even elements upto n
    // divide by 2
    $total = $n >> 1;
      
    // sum of even elements upto n
    $even = $total * ($total + 1);
  
    return $even;
}
  
// Function to find sum from L to R.
function sumLtoR($L, $R)
{
    $odd_sum = Odd_Sum($R) - 
               Odd_Sum($L - 1);
  
    $even_sum = Even_Sum($R) - 
                Even_Sum($L - 1);
  
      
    // print final sum from L to R
    return $even_sum - $odd_sum ;
}
  
// Driver Code
$L = 1 ;
$R = 5;
  
// function call to print answer
echo sumLtoR($L, $R);
  
// This code is contributed by ANKITRAI1
?>

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Output:

-3


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