Check whether a number can be represented as difference of two consecutive cubes

Given a number N, the task is to check if the number N can be represented as the difference of two consecutive cubes or not. If Yes then print those numbers else print No.

Examples:

Input: N = 19 
Output:
Yes  
2 3
Explanation:
3³ – 2³ = 19

Input: N = 10 
Output: No 

Approach: The key observation in the problem is that a number can be represented as difference of two consecutive cubes if and only if:

=> N = (K+1)³ – K³
=> N = 3*K² + 3*K + 1 
=> 12*N = 36*K² + 36*K + 12 
=> 12*N = (6*K + 3)² + 3 
=> 12*N – 3 = (6*K + 3)²
which means (12*N – 3) must be a perfect square to break N into difference of two consecutive cubes.

Therefore, if the above condition holds true then we will print the numbers using a for a loop by check that for which value of i if (i+1)³ – i³ = N and print the number i and i + 1.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to print the two consecutive
// numbers whose difference is N
void print(int N)
{
    // Iterate in the range [0, 10^5]
    for (int i = 0; i < 100000; i++) {

        if (pow(i + 1, 3)
                - pow(i, 3)
            == N) {

            cout << i << ' ' << i + 1;
            return;
        }
    }
}

// Function to check if N is a
// perfect cube
bool isPerfectSquare(long double x)
{
    // Find floating point value of
    // square root of x.
    long double sr = sqrt(x);

    // If square root is an integer
    return ((sr - floor(sr)) == 0);
}

// Function to check whether a number
// can be represented as difference
// of two consecutive cubes
bool diffCube(int N)
{
    // Check if 12 * N - 3 is a
    // perfect square or not
    return isPerfectSquare(12 * N - 3);
}

// Driver Code
int main()
{
    // Given Number N
    int N = 19;
    if (diffCube(N)) {
        cout << "Yes\n";
        print(N);
    }
    else {
        cout << "No\n";
    }
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{

// Function to print the two consecutive
// numbers whose difference is N
static void print(int N)
{
    // Iterate in the range [0, 10^5]
    for (int i = 0; i < 100000; i++)
    {

        if (Math.pow(i + 1, 3) - Math.pow(i, 3) == N)
        {
            int j = i + 1;
            System.out.println(i + " " + j);
            return;
        }
    }
}

// Function to check if N is a
// perfect cube
static boolean isPerfectSquare(double x)
{
    // Find floating point value of
    // square root of x.
    double sr = Math.sqrt(x);

    // If square root is an integer
    return ((sr - Math.floor(sr)) == 0);
}

// Function to check whether a number
// can be represented as difference
// of two consecutive cubes
static boolean diffCube(int N)
{
    // Check if 12 * N - 3 is a
    // perfect square or not
    return isPerfectSquare(12 * N - 3);
}

// Driver Code
public static void main(String[] args)
{
    // Given Number N
    int N = 19;
    if (diffCube(N))
    {
        System.out.println("Yes");
        print(N);
    }
    else
    {
        System.out.println("No");
    }
}
}

// This code is contributed by rock_cool

Python3

# Python3 program for the above approach
import math

# Function to print the two consecutive
# numbers whose difference is N
def printt(N):

    # Iterate in the range [0, 10^5]
    for i in range(100000):
        if (pow(i + 1, 3) - pow(i, 3) == N):
            print(i, '', i + 1)
            return

# Function to check if N is a
# perfect cube
def isPerfectSquare(x):

    # Find floating povalue of
    # square root of x.
    sr = math.sqrt(x)

    # If square root is an integer
    return ((sr - math.floor(sr)) == 0)

# Function to check whether a number
# can be represented as difference
# of two consecutive cubes
def diffCube(N):

    # Check if 12 * N - 3 is a
    # perfect square or not
    return isPerfectSquare(12 * N - 3)

# Driver Code

# Given number N
N = 19

if (diffCube(N)):
    print("Yes")
    printt(N)

else:
    print("No")

# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;
class GFG{

// Function to print the two consecutive
// numbers whose difference is N
static void print(int N)
{
    // Iterate in the range [0, 10^5]
    for (int i = 0; i < 100000; i++)
    {

        if (Math.Pow(i + 1, 3) - Math.Pow(i, 3) == N)
        {
            int j = i + 1;
            Console.WriteLine(i + " " + j);
            return;
        }
    }
}

// Function to check if N is a
// perfect cube
static bool isPerfectSquare(double x)
{
    // Find floating point value of
    // square root of x.
    double sr = Math.Sqrt(x);

    // If square root is an integer
    return ((sr - Math.Floor(sr)) == 0);
}

// Function to check whether a number
// can be represented as difference
// of two consecutive cubes
static bool diffCube(int N)
{
    // Check if 12 * N - 3 is a
    // perfect square or not
    return isPerfectSquare(12 * N - 3);
}

// Driver Code
public static void Main(String[] args)
{
    // Given Number N
    int N = 19;
    if (diffCube(N))
    {
        Console.WriteLine("Yes");
        print(N);
    }
    else
    {
        Console.WriteLine("No");
    }
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// Javascript program for the above approach

// Function to print the two consecutive
// numbers whose difference is N
function print(N)
{

    // Iterate in the range [0, 10^5]
    for(let i = 0; i < 100000; i++)
    {
        if (parseInt(Math.pow(i + 1, 3), 10) -
            parseInt(Math.pow(i, 3), 10) == N)
        {
            document.write(i + " " + (i + 1));
            return;
        }
    }
}

// Function to check if N is a
// perfect cube
function isPerfectSquare(x)
{

    // Find floating point value of
    // square root of x.
    let sr = Math.sqrt(x);

    // If square root is an integer
    return ((sr - Math.floor(sr)) == 0);
}

// Function to check whether a number
// can be represented as difference
// of two consecutive cubes
function diffCube(N)
{

    // Check if 12 * N - 3 is a
    // perfect square or not
    return isPerfectSquare(12 * N - 3);
}

// Driver code

// Given Number N
let N = 19;

if (diffCube(N) != 0)
{
    document.write("Yes" + "</br>");
    print(N);
}
else
{
    document.write("No");
}

// This code is contributed by divyeshrabadiya07

</script>

Output:

Yes
2 3

Time Complexity: O(N), where N is in the range 10⁵
Auxiliary Space: O(1)

Approach 2: Binary Search:

The given program is used to determine whether a given number N can be expressed as the difference of two consecutive cubes.

•  The approach used in the program is based on the fact that the difference of two consecutive cubes can be expressed as (j^3 – i^3), where j = i + 1.
• The program checks all possible values of i from 0 to 100000. For each value of i, it performs a binary search to find the value of j such that (j^3 – i^3) is equal to N. The search space for j is from i+1 to 100000 since j should always be greater than i. The binary search is performed by maintaining two pointers, lo and hi, that point to the left and right endpoints of the search space, respectively. The mid pointer is calculated as the average of lo and hi, and the difference (j^3 – i^3) is calculated for the mid value. If the difference is equal to N, the values of i and j are printed, and the function returns true. If the difference is less than N, the lo pointer is moved to mid+1 to search for a larger value of j. If the difference is greater than N, the hi pointer is moved to mid-1 to search for a smaller value of j. This process continues until either the pair (i, j) is found, or the search space is exhausted.
• If no such pair (i,j) is found for any value of i, the function returns false indicating that N cannot be expressed as the difference of two consecutive cubes.
• In the main function, a value of N=19 is given as input to the diffCube function. The function diffCube is called to determine if 19 can be expressed as the difference of two consecutive cubes. If such a pair exists, the program prints “Yes”. Otherwise, it prints “No”.

Here is the code of above approach:

C++

#include <bits/stdc++.h>
using namespace std;

// Function to check whether a number
// can be represented as difference
// of two consecutive cubes
bool diffCube(int N)
{
    // Iterate over the possible values of i
    for (int i = 0; i <= 100000; i++) {
        int lo = i + 1, hi = 100000, mid;

        // Binary search for j
    a    while (lo <= hi) {
            mid = lo + (hi - lo) / 2;

            // Check if the difference is N
            if (pow(mid, 3) - pow(i, 3) == N) {
                cout << i << " " << mid << endl;
                return true;
            }

            // If the difference is less than N,
            // move the left pointer
            if (pow(mid, 3) - pow(i, 3) < N)
                lo = mid + 1;
            // If the difference is greater than N,
            // move the right pointer
            else
                hi = mid - 1;
        }
    }

    // If no such pair is found, return false
    return false;
}

// Driver Code
int main()
{
    // Given Number N
    int N = 19;
    if (diffCube(N))
        cout << "Yes\n";
    else
        cout << "No\n";
    return 0;
}

Java

import java.util.*;

public class Main {
  // Function to check whether a number
// can be represented as difference
// of two consecutive cubes
static boolean diffCube(int N) {
    // Iterate over the possible values of i
    for (int i = 0; i <= 100000; i++) {
        int lo = i + 1, hi = 100000, mid;

        // Binary search for j
        while (lo <= hi) {
            mid = lo + (hi - lo) / 2;

            // Check if the difference is N
            if (Math.pow(mid, 3) - Math.pow(i, 3) == N) {
                System.out.println(i + " " + mid);
                return true;
            }

            // If the difference is less than N,
            // move the left pointer
            if (Math.pow(mid, 3) - Math.pow(i, 3) < N)
                lo = mid + 1;
            // If the difference is greater than N,
            // move the right pointer
            else
                hi = mid - 1;
        }
    }

    // If no such pair is found, return false
    return false;
}

// Driver Code
public static void main(String[] args) {
    // Given Number N
    int N = 19;
    if (diffCube(N))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}

Python3

import math

# Function to check whether a number
# can be represented as difference
# of two consecutive cubes
def diff_cube(N):
    # Iterate over the possible values of i
    for i in range(0, 100001):
        lo, hi = i + 1, 100000

        # Binary search for j
        while lo <= hi:
            mid = lo + (hi - lo) // 2

            # Check if the difference is N
            if (mid**3 - i**3) == N:
                print(i, mid)
                return True

            # If the difference is less than N,
            # move the left pointer
            if (mid**3 - i**3) < N:
                lo = mid + 1
            # If the difference is greater than N,
            # move the right pointer
            else:
                hi = mid - 1

    # If no such pair is found, return false
    return False

# Driver Code
if __name__ == '__main__':
    # Given Number N
    N = 19
    if diff_cube(N):
        print("Yes")
    else:
        print("No")

Output

2 3
Yes

Time Complexity: O(N log^2(N)), where N is the given input number
Auxiliary Space: O(1)