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

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 n1/3. The idea is very simple. For every distinct pair (x, y) formed by numbers less than the n1/3, 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 = n1/3
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 C++ implementation of above idea –

// 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;
}

Output:

(2, 24) and (18, 20)

Time Complexity of above solution is O(n2/3) which is much less than O(n).

Can we solve the above problem in O(n1/3) time? We will be discussing that in next post.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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