Program To Print Sum Of All Cubes In Java Between 1 To N

Finding the sum of consecutive cube numbers is a common problem in programming. In this article, we will look at techniques for calculating sum of all cubes in Java Between 1 to n.

Problem Statement

Given a number n, print sum of all cubes in java between 1 to n.

Examples:

Input: 3

Output: 36

Explanation

Cubes of first three natural numbers is 1^3 + 2^3 + 3^3 =36.

Print Sum Of All Cubes In Java Solution

We will discuss two solutions here:

1. Naive Approach Using Loops
2. Using Mathematical Formula

Naive Approach

The idea here is to iterate from 1 to n using loop and add the cube of current number to the sum.

Java Code Implementation

Let’s see Java Program for cube sum of first n natural numbers using Naive Approach

// Java Program for cube sum of first n natural numbers using Naive Approach

public class SumOfCubes {

    // Method to calculate the sum of cubes of natural numbers up to n
    public static long sumOfCubes(int n) {
        long sum = 0; // Initialize sum to 0
        for (int i = 1; i <= n; i++) {
            sum += i * i * i; // Add the cube of the current number to the sum
        }
        return sum; // Return the computed sum
    }

    public static void main(String[] args) {
        int n = 3; // Example input
        long result = sumOfCubes(n); // Calculate the sum of cubes
        System.out.println("The sum of cubes of first " + n + " natural numbers is: " + result);
    }
}

Output:

The sum of cubes of first 3 natural numbers is: 36

Time Complexity: O(n).

This is because we have to loop through each number from 1 to n and perform a constant-time operation (cubing a number and adding it to the sum).

Space Complexity: O(1).

Using Mathematical Formula

Mathematical Formula to get sum of cube of first n natural number is

To know the proof and derivation of this formula check out this article. We will use this formula to calculate sum of cubes of numbers between 1 to n.

Java Code Implementation

Let’s see Java Program for cube sum of first n natural numbers using Mathematical Formula

// Sum of cube of first n natural number in java using formula

public class SumOfCubes {

    // Function to calculate the sum of cubes of the first n natural numbers
    public static long sumOfCubes(int n) {
        long sum = (long) n * (n + 1) / 2;  // Sum of the first n natural numbers
        sum *= sum;  // Squaring the sum to get the sum of cubes
        return sum;
    }

    public static void main(String[] args) {
        int n = 5;  // Example input
        System.out.println("The sum of cubes of the first " + n + " natural numbers is: " + sumOfCubes(n));
    }
}

Output:

The sum of cubes of the first 5 natural numbers is: 225

Time Complexity: O(1).

This is because the calculations are done in constant time. No matter the size of n, the computational steps remain the same

Space Complexity: O(1).

Key Takeaways

• A naive iterative approach uses a loop to cube each number and add to the sum
• Mathematical formulas can calculate the sum in constant time without loops
• Time complexity is O(n) for iteration vs O(1) for the formula
• Understanding both methods provides a good overview of algorithmic approaches

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