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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 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 the implementation of above idea –

## C++

 `// C++ program to find pairs that can represent``// the given number as sum of two cubes``#include ``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 s = ``new` `HashMap();` `    ``// 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(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.

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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