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 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 = 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 –

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.

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