Given two integers 
and 
and an array arr[] every element of which at index 
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]](https://geeksforgeeks.org/wp-content/uploads/2023/11/rubhabha_quicklatex.com-d2b6a2931a24fda60602c832f27503d4_l3-1.png "Rendered by QuickLaTeX.com")
.
Examples:
Input: L = 1 , R = 5
Output: -3
Sum = (-1) + 2 + (-3) + 4 + (-5) = -3Input: 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++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// function to return the odd sumlong 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 sumlong 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 Programint main(){ int L = 1, R = 5; // function call to print answer cout << sumLtoR(L, R); return 0;} |
Java
// Java implementation of above approachimport java.io.*;class GFG { // function to return the odd sumstatic 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 sumstatic 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.. |
Python3
# Python3 implementation of above approach# function to return the odd sumdef 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 sumdef Even_Sum(n): # total even elements upto n total = n//2 # sum of even elements upto n even = total*(total+1) return evendef 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 codeL =1; R = 5print(sumLtoR(L,R))# This code is contributed by Shrikant13 |
C#
// C# implementation of above approachclass GFG{ // function to return the odd sumstatic 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 sumstatic 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 Codepublic 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 |
PHP
<?php// PHP implementation of above approach// function to return the odd sumfunction 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 $odd;}// function to return the even sumfunction Even_Sum($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 ;$R = 5;// function call to print answerecho sumLtoR($L, $R);// This code is contributed by ANKITRAI1?> |
Javascript
<script>// Javascript implementation of above approach// Function to return the odd sumfunction Odd_Sum(n){ // Total odd elements upto n var total = parseInt((n + 1) / 2); // Sum of odd elements upto n var odd = total * total; return odd;}// Function to return the even sumfunction Even_Sum(n){ // Total even elements upto n var total = parseInt((n) / 2); // Sum of even elements upto n var even = total * (total + 1); return even;}// Function to find sum from L to R.function sumLtoR(L, R){ var 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 codevar L = 1, R = 5;// Function call to print answerdocument.write(sumLtoR(L, R));// This code is contributed by SoumikMondal</script> |
Output
-3
Complexity Analysis:
- Time Complexity: O(1)
- Auxiliary Space: O(1), since no extra space has been taken.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
