Given a number N, the task is to find the sum of alternating sign squares of first N natural numbers, i.e.,
12 – 22 + 32 – 42 + 52 – 62 + ….
Examples:
Input: N = 2 Output: 5 Explanation: Required sum = 12 - 22 = -1 Input: N = 8 Output: 36 Explanation: Required sum = 12 - 22 + 32 - 42 + 52 - 62 + 72 - 82 = 36
Naive approach: O(N)
The Naive or Brute force approach to solve this problem states to find the square of each number from 1 to N and add them with alternating sign in order to get the required sum.
- For each number in 1 to N, find its square
- Add these squares with alternating sign
- This would give the required sum.
Below is the implementation of the above approach:
C++
// C++ program to find Sum of alternating// sign Squares of first N natural numbers#include <iostream>using namespace std;// Function to calculate// the alternating sign sumint summation(int n){ // Variable to store the sum int sum = 0; // Loop to iterate each number // from 1 to N for (int i = 1; i <= n; i++) { // The alternating sign is put // by checking if the number // is even or odd if (i % 2 == 1) // Add the square with the sign sum += (i * i); else // Add the square with the sign sum -= (i * i); } return sum;}// Driver codeint main(){ int N = 2; cout << summation(N); return 0;} |
Java
// Java program to find Sum of alternating// sign Squares of first N natural numbersclass GFG{ // Function to calculate // the alternating sign sum static int summation(int n) { // Variable to store the sum int sum = 0; // Loop to iterate each number // from 1 to N for (int i = 1; i <= n; i++) { // The alternating sign is put // by checking if the number // is even or odd if (i % 2 == 1) // Add the square with the sign sum += (i * i); else // Add the square with the sign sum -= (i * i); } return sum; } // Driver code public static void main (String[] args) { int N = 2; System.out.println(summation(N)); }}// This code is contributed by AnkitRai01 |
Python3
# Python3 program to find Sum of alternating # sign Squares of first N natural numbers # Function to calculate # the alternating sign sum def summation(n) : # Variable to store the sum sum = 0; # Loop to iterate each number # from 1 to N for i in range(1, n + 1) : # The alternating sign is put # by checking if the number # is even or odd if (i % 2 == 1) : # Add the square with the sign sum += (i * i); else : # Add the square with the sign sum -= (i * i); return sum; # Driver code if __name__ == "__main__" : N = 2; print(summation(N)); # This code is contributed by AnkitRai01 |
C#
// C# program to find Sum of alternating// sign Squares of first N natural numbersusing System;class GFG{ // Function to calculate // the alternating sign sum static int summation(int n) { // Variable to store the sum int sum = 0; // Loop to iterate each number // from 1 to N for (int i = 1; i <= n; i++) { // The alternating sign is put // by checking if the number // is even or odd if (i % 2 == 1) // Add the square with the sign sum += (i * i); else // Add the square with the sign sum -= (i * i); } return sum; } // Driver code public static void Main() { int N = 2; Console.WriteLine(summation(N)); }}// This code is contributed by AnkitRai01 |
Javascript
<script>// JavaScript program to find Sum of alternating // sign Squares of first N natural numbers // Function to calculate // the alternating sign sum function summation(n) { // Variable to store the sum let sum = 0; // Loop to iterate each number // from 1 to N for (let i = 1; i <= n; i++) { // The alternating sign is put // by checking if the number // is even or odd if (i % 2 == 1) // Add the square with the sign sum += (i * i); else // Add the square with the sign sum -= (i * i); } return sum; } // Driver code let N = 2; document.write(summation(N)); // This code is contributed by Surbhi Tyagi</script> |
Output:
-3
Efficient Approach: O(1)
There exists a formula for finding the sum of squares of first n numbers with alternating signs:
![\[\LARGE 1^{2}-2^{2}+3^{2}-4^{2}+... = (-1)^{n+1} \text{ } \frac{n(n+1)}{2}\]](https://geeksforgeeks.org/wp-content/uploads/2023/10/rubhabha_quicklatex.com-c1cb79f89bcaa66b20929ec8fd6c4131_l3.png "Rendered by QuickLaTeX.com")
How does this work?
We can prove this formula using induction. We can easily see that the formula is true for n = 1 and n = 2 as sums are 1 and -3 respectively. Let it be true for n = k-1. So sum of k-1 numbers is (-1)k(k - 1) * k / 2 In the following steps, we show that it is true for k assuming that it is true for k-1. Sum of k numbers =(-1)k (Sum of k-1 numbers + k2) =(-1)k+1 ((k - 1) * k / 2 + k2) =(-1)k+1 (k * (k + 1) / 2), which is true.
Hence inorder to find the sum of alternating sign squares of first N natural numbers, simply compute the formula 
and print the result.
C++
// C++ program to find Sum of alternating// sign Squares of first N natural numbers#include <iostream>using namespace std;// Function to calculate// the alternating sign sumint summation(int n){ // Variable to store the absolute sum int abs_sum = n * (n + 1) / 2; // Variable to store the sign int sign = n + 1 % 2 == 0 ? 1 : -1; // Variable to store the resultant sum int result_sum = sign * abs_sum; return result_sum;}// Driver codeint main(){ int N = 2; cout << summation(N); return 0;} |
Java
// Java program to find Sum of alternating// sign Squares of first N natural numbersclass GFG { // Function to calculate // the alternating sign sum static int summation(int n) { // Variable to store the absolute sum int abs_sum = n * (n + 1) / 2; // Variable to store the sign int sign = n + 1 % 2 == 0 ? 1 : -1; // Variable to store the resultant sum int result_sum = sign * abs_sum; return result_sum; } // Driver code public static void main (String[] args) { int N = 2; System.out.println(summation(N)); }}// This code is contributed by AnkitRai01 |
Python3
# Python3 program to find Sum of alternating # sign Squares of first N natural numbers # Function to calculate # the alternating sign sum def summation(n) : # Variable to store the absolute sum abs_sum = n * (n + 1) // 2; # Variable to store the sign sign = 1 if ((n + 1) % 2 == 0 ) else -1; # Variable to store the resultant sum result_sum = sign * abs_sum; return result_sum; # Driver code if __name__ == "__main__" : N = 2; print(summation(N)); # This code is contributed by AnkitRai01 |
C#
// C# program to find Sum of alternating// sign Squares of first N natural numbersusing System;public class GFG { // Function to calculate // the alternating sign sum static int summation(int n) { // Variable to store the absolute sum int abs_sum = (int)(n * (n + 1) / 2); // Variable to store the sign int sign = n + 1 % 2 == 0 ? 1 : -1; // Variable to store the resultant sum int result_sum = sign * abs_sum; return result_sum; } // Driver code public static void Main() { int N = 2; Console.WriteLine(summation(N)); }}// This code is contributed by AnkitRai01 |
Javascript
<script>// Javascript program to find Sum of alternating// sign Squares of first N natural numbers// Function to calculate// the alternating sign sumfunction summation(n){ // Variable to store the absolute sum var abs_sum = n * (n + 1) / 2; // Variable to store the sign var sign = n + 1 % 2 == 0 ? 1 : -1; // Variable to store the resultant sum var result_sum = sign * abs_sum; return result_sum;}// Driver codevar N = 2;document.write(summation(N));// This code is contributed by rutvik_56.</script> |
Output:
-3
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