Find Cube Pairs | Set 1 (A n^(2/3) Solution)

Last Updated : 05 Oct, 2020

Given a number n, find two pairs that can represent the number as sum of two cubes. In other words, find two pairs (a, b) and (c, d) such that given number n can be expressed as

n = a^3 + b^3 = c^3 + d^3

where a, b, c and d are four distinct numbers.

Examples:

Input: N = 1729
Output: (1, 12) and (9, 10)
Explanation: 
1729 = 1^3 + 12^3 = 9^3 + 10^3

Input: N = 4104
Output: (2, 16) and (9, 15)
Explanation: 
4104 = 2^3 + 16^3 = 9^3 + 15^3

Input: N = 13832
Output: (2, 24) and (18, 20)
Explanation: 
13832 = 2^3 + 24^3 = 18^3 + 20^3

Any number n that satisfies the constraint will have two distinct pairs (a, b) and (c, d) such that a, b, c and d are all less than n^1/3. The idea is very simple. For every distinct pair (x, y) formed by numbers less than the n^1/3, if their sum (x³ + y³) is equal to given number, we store them in a hash table using sum as a key. If pairs with sum equal to given number appears again, we simply print both pairs.

1) Create an empty hash map, say s.
2) cubeRoot = n^1/3
3) for (int x = 1; x < cubeRoot; x++)
     for (int y = x + 1; y <= cubeRoot; y++)
       int sum = x³ + y³;
       if (sum != n) continue;
       if sum exists in s,
         we found two pairs with sum, print the pairs
       else
         insert pair(x, y) in s using sum as key

Below is the implementation of above idea –

C++

// C++ program to find pairs that can represent
// the given number as sum of two cubes
#include <bits/stdc++.h>
using namespace std;
 
// Function to find pairs that can represent
// the given number as sum of two cubes
void findPairs(int n)
{
    // find cube root of n
    int cubeRoot = pow(n, 1.0/3.0);
 
    // create an empty map
    unordered_map<int, pair<int, int> > s;
 
    // Consider all pairs such with values less
    // than cuberoot
    for (int x = 1; x < cubeRoot; x++)
    {
        for (int y = x + 1; y <= cubeRoot; y++)
        {
            // find sum of current pair (x, y)
            int sum = x*x*x + y*y*y;
 
            // do nothing if sum is not equal to
            // given number
            if (sum != n)
                continue;
 
            // if sum is seen before, we found two pairs
            if (s.find(sum) != s.end())
            {
                cout << "(" << s[sum].first << ", "
                     << s[sum].second << ") and ("
                    << x << ", " << y << ")" << endl;
            }
            else
                // if sum is seen for the first time
                s[sum] = make_pair(x, y);
        }
    }
}
 
// Driver function
int main()
{
    int n = 13832;
    findPairs(n);
    return 0;
}

Java

// Java program to find pairs that can represent
// the given number as sum of two cubes
import java.util.*;
 
class GFG
{
    static class pair
    { 
        int first, second; 
        public pair(int first, int second) 
        { 
            this.first = first; 
            this.second = second; 
        } 
    }
     
// Function to find pairs that can represent
// the given number as sum of two cubes
static void findPairs(int n)
{
    // find cube root of n
    int cubeRoot = (int) Math.pow(n, 1.0/3.0);
 
    // create an empty map
    HashMap<Integer, pair> s = new HashMap<Integer, pair>();
 
    // Consider all pairs such with values less
    // than cuberoot
    for (int x = 1; x < cubeRoot; x++)
    {
        for (int y = x + 1; y <= cubeRoot; y++)
        {
            // find sum of current pair (x, y)
            int sum = x*x*x + y*y*y;
 
            // do nothing if sum is not equal to
            // given number
            if (sum != n)
                continue;
 
            // if sum is seen before, we found two pairs
            if (s.containsKey(sum))
            {
                System.out.print("(" + s.get(sum).first+ ", "
                    + s.get(sum).second+ ") and ("
                    + x+ ", " + y+ ")" +"\n");
            }
            else
                // if sum is seen for the first time
                s.put(sum, new pair(x, y));
        }
    }
}
 
// Driver code
public static void main(String[] args)
{
    int n = 13832;
    findPairs(n);
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to find pairs
# that can represent the given 
# number as sum of two cubes
 
# Function to find pairs that 
# can represent the given number
# as sum of two cubes 
def findPairs(n):
 
    # Find cube root of n 
    cubeRoot = pow(n, 1.0 / 3.0); 
   
    # Create an empty map 
    s = {} 
   
    # Consider all pairs such with 
    # values less than cuberoot 
    for x in range(int(cubeRoot)):
        for y in range(x + 1, 
           int(cubeRoot) + 1):
             
            # Find sum of current pair (x, y) 
            sum = x * x * x + y * y * y; 
   
            # Do nothing if sum is not equal to 
            # given number 
            if (sum != n):
                continue; 
   
            # If sum is seen before, we
            # found two pairs 
            if sum in s.keys():
                print("(" + str(s[sum][0]) +
                     ", " + str(s[sum][1]) +
                        ") and (" + str(x) +
                             ", " + str(y) +
                              ")" + "\n")
                     
            else:
                 
                # If sum is seen for the first time 
                s[sum] = [x, y] 
 
# Driver code
if __name__=="__main__":
     
    n = 13832
     
    findPairs(n)
     
# This code is contributed by rutvik_56

C#

// C# program to find pairs that can represent
// the given number as sum of two cubes
using System;
using System.Collections.Generic;
 
class GFG
{
    class pair
    { 
        public int first, second; 
        public pair(int first, int second) 
        { 
            this.first = first; 
            this.second = second; 
        } 
    }
      
// Function to find pairs that can represent
// the given number as sum of two cubes
static void findPairs(int n)
{
    // find cube root of n
    int cubeRoot = (int) Math.Pow(n, 1.0/3.0);
  
    // create an empty map
    Dictionary<int, pair> s = new Dictionary<int, pair>();
  
    // Consider all pairs such with values less
    // than cuberoot
    for (int x = 1; x < cubeRoot; x++)
    {
        for (int y = x + 1; y <= cubeRoot; y++)
        {
            // find sum of current pair (x, y)
            int sum = x*x*x + y*y*y;
  
            // do nothing if sum is not equal to
            // given number
            if (sum != n)
                continue;
  
            // if sum is seen before, we found two pairs
            if (s.ContainsKey(sum))
            {
                Console.Write("(" + s[sum].first+ ", "
                    + s[sum].second+ ") and ("
                    + x+ ", " + y+ ")" +"\n");
            }
            else
                // if sum is seen for the first time
                s.Add(sum, new pair(x, y));
        }
    }
}
  
// Driver code
public static void Main(String[] args)
{
    int n = 13832;
    findPairs(n);
}
}
 
// This code is contributed by PrinciRaj1992

Output:

(2, 24) and (18, 20)

Time Complexity of above solution is O(n^2/3) which is much less than O(n).

Can we solve the above problem in O(n^1/3) time? We will be discussing that in next post.

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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