# 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 = 1729Output:(1, 12) and (9, 10)Explanation:1729 = 1^3 + 12^3 = 9^3 + 10^3Input:N = 4104Output:(2, 16) and (9, 15)Explanation:4104 = 2^3 + 16^3 = 9^3 + 15^3Input:N = 13832Output:(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*, if their sum (x

^{1/3}^{3}+ y

^{3}) 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 = x^{3}+ y^{3}; 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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