The sum of two positive numbers is equal to 2 and if the sum of their cubes is least the numbers are

Given an integer N, the task is to check if N can be represented as the sum of two positive perfect cubes or not.

Examples:

Input: N = 28
Output: Yes
Explanation:
Since, 28 = 27 + 1 = 33 + 13.
Therefore, the required answer is Yes.

Input: N = 34
Output: No

Approach: The idea is to store the perfect cubes of all numbers from 1 to cubic root of N in a Map and check if N can be represented as the sum of two numbers present in the Map or not. Follow the steps below to solve the problem:

Initialize an ordered map, say cubes, to store the perfect cubes of first N natural numbers in sorted order.
Traverse the map and check for the pair having a sum equal to N.
If such a pair is found having sum N, then print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach:

void sumOfTwoPerfectCubes(int N)

for (int i = 1; i * i * i <= N; i++)

map<int, int>::iterator itr;

for (itr = cubes.begin();

itr != cubes.end(); itr++) {

int firstNumber = itr->first;

int secondNumber = N - itr->first;

if (cubes.find(secondNumber)

public static void sumOfTwoPerfectCubes(int N)

HashMap<Integer, Integer> cubes = new HashMap<>();

for (int i = 1; i * i * i <= N; i++)

cubes.put((i * i * i), i);

Iterator<Map.Entry<Integer, Integer> > itr

= cubes.entrySet().iterator();

Map.Entry<Integer, Integer> entry = itr.next();

int firstNumber = entry.getKey();

int secondNumber = N - entry.getKey();

if (cubes.containsKey(secondNumber))

System.out.println("True");

System.out.println("False");

public static void main(String[] args)

def sumOfTwoPerfectCubes(N) :

if secondNumber in cubes :

using System.Collections.Generic;

public static void sumOfTwoPerfectCubes(int N)

int> cubes = new Dictionary<int,

for (int i = 1; i * i * i <= N; i++)

cubes.Add((i * i * i), i);

var val = cubes.Keys.ToList();

int firstNumber = cubes[1];

int secondNumber = N - cubes[1];

if (cubes.ContainsKey(secondNumber))

static public void Main()

function sumOfTwoPerfectCubes(N)

for (var i = 1; i * i * i <= N; i++)

cubes.forEach((value, key) => {

var secondNumber = N - value;

if (cubes.has(secondNumber)) {

document.write( "False");

Time Complexity: O(N1/3 * log(N1/3))
Auxiliary Space: O(N1/3)

Approach 2 :

Using two Pointers:

We will declare lo to 1 and hi to cube root of n(the given number), then by (lo<=hi) this condition, if current is smaller than n we will increment the lo and in other hand if it is greater then decrement the hi, where current is (lo*lo*lo + hi*hi*hi)

bool sumOfTwoCubes(int n)

long long int lo = 1, hi = (long long int)cbrt(n);

long long int curr = (lo * lo * lo + hi * hi * hi);

static boolean sumOfTwoCubes(int n)

int lo = 1, hi = (int)Math.cbrt(n);

int curr = (lo * lo * lo + hi * hi * hi);

public static void main (String[] args)

System.out.println("True");

System.out.println("False");

hi = round(math.pow(n, 1 / 3))

curr = (lo * lo * lo + hi * hi * hi)