Related Articles
Find Cube Pairs | Set 2 (A n^(1/3) Solution)
• Difficulty Level : Medium
• Last Updated : 16 Apr, 2021

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```

We have discussed a O(n2/3) solution in below set 1.
Find Cube Pairs | Set 1 (A n^(2/3) Solution)
In this post, a O(n1/3) solution is discussed.
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 to create an auxiliary array of size n1/3. Each index i in the array will store value equal to cube of that index i.e. arr[i] = i^3. Now the problem reduces to finding pair of elements in an sorted array whose sum is equal to given number n. The problem is discussed in detail here.
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 ``#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 a array of size of size 'cubeRoot'``    ``int` `cube[cubeRoot + 1];` `    ``// for index i, cube[i] will contain i^3``    ``for` `(``int` `i = 1; i <= cubeRoot; i++)``        ``cube[i] = i*i*i;` `    ``// Find all pairs in above sorted``    ``// array cube[] whose sum is equal to n``    ``int` `l = 1;``    ``int` `r = cubeRoot;` `    ``while` `(l < r)``    ``{``        ``if` `(cube[l] + cube[r] < n)``            ``l++;``        ``else` `if``(cube[l] + cube[r] > n)``            ``r--;``        ``else` `{``            ``cout << ``"("` `<< l <<  ``", "`  `<< r``                 ``<< ``")"` `<< endl;``            ``l++; r--;``        ``}``    ``}``}` `// Driver function``int` `main()``{``    ``int` `n = 20683;``    ``findPairs(n);` `    ``return` `0;``}`

## Java

 `// Java program to find pairs``// that can represent the given``// number as sum of two cubes``import` `java.io.*;` `class` `GFG``{``    ` `// 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 a array of``    ``// size of size 'cubeRoot'``    ``int` `cube[] = ``new` `int``[cubeRoot + ``1``];` `    ``// for index i, cube[i]``    ``// will contain i^3``    ``for` `(``int` `i = ``1``; i <= cubeRoot; i++)``        ``cube[i] = i * i * i;` `    ``// Find all pairs in above``    ``// sorted array cube[]``    ``// whose sum is equal to n``    ``int` `l = ``1``;``    ``int` `r = cubeRoot;` `    ``while` `(l < r)``    ``{``        ``if` `(cube[l] + cube[r] < n)``            ``l++;``        ``else` `if``(cube[l] + cube[r] > n)``            ``r--;``        ``else` `{``            ``System.out.println(``"("` `+ l +``                              ``", "` `+ r +``                                  ``")"` `);``            ``l++; r--;``        ``}``    ``}``}` `// Driver Code``public` `static` `void` `main (String[] args)``{``int` `n = ``20683``;``findPairs(n);``}``}` `// This code is contributed by anuj_67.`

## Python3

 `# Python3 program to find pairs that``# can represent the given number``# as sum of two cubes``import` `math` `# Function to find pairs that can``# represent the given number as``# sum of two cubes``def` `findPairs( n):``    ` `    ``# find cube root of n``    ``cubeRoot ``=` `int``(math.``pow``(n, ``1.0` `/` `3.0``));` `    ``# create a array of``    ``# size of size 'cubeRoot'``    ``cube ``=` `[``0``] ``*` `(cubeRoot ``+` `1``);` `    ``# for index i, cube[i]``    ``# will contain i^3``    ``for` `i ``in` `range``(``1``, cubeRoot ``+` `1``):``        ``cube[i] ``=` `i ``*` `i ``*` `i;` `    ``# Find all pairs in above sorted``    ``# array cube[] whose sum``    ``# is equal to n``    ``l ``=` `1``;``    ``r ``=` `cubeRoot;` `    ``while` `(l < r):``        ``if` `(cube[l] ``+` `cube[r] < n):``            ``l ``+``=` `1``;``        ``elif``(cube[l] ``+` `cube[r] > n):``            ``r ``-``=` `1``;``        ``else``:``            ``print``(``"("` `, l , ``", "` `, math.floor(r),``                                  ``")"``, end ``=` `"");``            ``print``();``            ``l ``+``=` `1``;``            ``r ``-``=` `1``;` `# Driver code``n ``=` `20683``;``findPairs(n);` `# This code is contributed by mits`

## C#

 `// C# program to find pairs``// that can represent the given``// number as sum of two cubes``using` `System;` `class` `GFG``{``    ` `// 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 a array of``    ``// size of size 'cubeRoot'``    ``int` `[]cube = ``new` `int``[cubeRoot + 1];` `    ``// for index i, cube[i]``    ``// will contain i^3``    ``for` `(``int` `i = 1; i <= cubeRoot; i++)``        ``cube[i] = i * i * i;` `    ``// Find all pairs in above``    ``// sorted array cube[]``    ``// whose sum is equal to n``    ``int` `l = 1;``    ``int` `r = cubeRoot;` `    ``while` `(l < r)``    ``{``        ``if` `(cube[l] + cube[r] < n)``            ``l++;``        ``else` `if``(cube[l] + cube[r] > n)``            ``r--;``        ``else` `{``            ``Console.WriteLine(``"("` `+ l +``                              ``", "` `+ r +``                                  ``")"` `);``            ``l++; r--;``        ``}``    ``}``}` `// Driver Code``public` `static` `void` `Main ()``{``    ``int` `n = 20683;``    ``findPairs(n);``}``}` `// This code is contributed by anuj_67.`

## PHP

 ` ``\$n``)``            ``\$r``--;``        ``else``        ``{``            ``echo` `"("` `, ``\$l` `, ``", "` `, ``floor``(``\$r``)``                ``, ``")"` `;``                ``echo` `"\n"``;``            ``\$l``++;``\$r``--;``        ``}``    ``}``}` `// Driver code``\$n` `= 20683;``findPairs(``\$n``);` `// This code is contributed by anuj_67.``?>`

## Javascript

 ``

Output:

```(10, 27)
(19, 24)```

Time Complexity of above solution is O(n^(1/3)).