Count pairs (a, b) whose sum of cubes is N (a^3 + b^3 = N) - GeeksforGeeks

04 Aug, 2021
Given N, count all ‘a’ and ‘b’ that satisfy the condition a^3 + b^3 = N. Examples:  
Input : N = 9
Output : 2
1^3 + 2^3 = 9
2^3 + 1^3 = 9

Input : N = 28
Output : 2
 1^3 + 3^3 = 28
 3^3 + 1^3 = 28
Note:- (a, b) and (b, a) are to be considered as two different pairs. Asked in : Adobe 
 
Implementation:
Traverse numbers from 1 to cube root of N. 
  a) Subtract cube of current number from 
    N and check if their difference is a 
    perfect cube or not.
      i) If perfect cube then increment count.

2- Return count.
Below is the implementation of above approach:  
C++
Java
Python 3
C#
PHP
Javascript
// C++ program to count pairs whose sum// cubes is N#include<bits/stdc++.h>using namespace std; // Function to count the pairs satisfying// a ^ 3 + b ^ 3 = Nint countPairs(int N){    int count = 0;     // Check for each number 1 to cbrt(N)    for (int i = 1; i <= cbrt(N); i++)    {        // Store cube of a number        int cb = i*i*i;         // Subtract the cube from given N        int diff = N - cb;         // Check if the difference is also        // a perfect cube        int cbrtDiff = cbrt(diff);         // If yes, then increment count        if (cbrtDiff*cbrtDiff*cbrtDiff == diff)            count++;    }     // Return count    return count;} // Driver programint main(){    // Loop to Count no. of pairs satisfying    // a ^ 3 + b ^ 3 = i for N = 1 to 10    for (int i = 1; i<= 10; i++)        cout << "For n = " << i << ", "             << countPairs(i) <<" pair exists\n";     return 0;}
// Java program to count pairs whose sum// cubes is N class Test{    // method to count the pairs satisfying    // a ^ 3 + b ^ 3 = N    static int countPairs(int N)    {        int count = 0;              // Check for each number 1 to cbrt(N)        for (int i = 1; i <= Math.cbrt(N); i++)        {            // Store cube of a number            int cb = i*i*i;                  // Subtract the cube from given N            int diff = N - cb;                  // Check if the difference is also            // a perfect cube            int cbrtDiff = (int) Math.cbrt(diff);                  // If yes, then increment count            if (cbrtDiff*cbrtDiff*cbrtDiff == diff)                count++;        }              // Return count        return count;    }         // Driver method    public static void main(String args[])    {        // Loop to Count no. of pairs satisfying        // a ^ 3 + b ^ 3 = i for N = 1 to 10        for (int i = 1; i<= 10; i++)            System.out.println("For n = " + i + ", " +                     + countPairs(i) + " pair exists");    }}
# Python 3 program to count pairs# whose sum cubes is Nimport math # Function to count the pairs# satisfying a ^ 3 + b ^ 3 = Ndef countPairs(N):     count = 0     # Check for each number 1    # to cbrt(N)    for i in range(1, int(math.pow(N, 1/3) + 1)):             # Store cube of a number        cb = i * i * i         # Subtract the cube from given N        diff = N - cb         # Check if the difference is also        # a perfect cube        cbrtDiff = int(math.pow(diff, 1/3))         # If yes, then increment count        if (cbrtDiff * cbrtDiff * cbrtDiff == diff):            count += 1         # Return count    return count  # Driver program # Loop to Count no. of pairs satisfying# a ^ 3 + b ^ 3 = i for N = 1 to 10for i in range(1, 11):    print('For n = ', i, ', ', countPairs(i),                                ' pair exists')  # This code is contributed by Smitha.
// C# program to count pairs whose sum// cubes is N  using System;class Test{    // method to count the pairs satisfying    // a ^ 3 + b ^ 3 = N    static int countPairs(int N)    {        int count = 0;               // Check for each number 1 to cbrt(N)        for (int i = 1; i <= Math.Pow(N,(1.0/3.0)); i++)        {            // Store cube of a number            int cb = i*i*i;                   // Subtract the cube from given N            int diff = N - cb;                   // Check if the difference is also            // a perfect cube            int cbrtDiff = (int) Math.Pow(diff,(1.0/3.0));                   // If yes, then increment count            if (cbrtDiff*cbrtDiff*cbrtDiff == diff)                count++;        }               // Return count        return count;    }          // Driver method    public static void Main()    {        // Loop to Count no. of pairs satisfying        // a ^ 3 + b ^ 3 = i for N = 1 to 10        for (int i = 1; i<= 10; i++)            Console.Write("For n = " + i + ", " +                     + countPairs(i) + " pair exists"+"\n");    }}
<?php// PHP program to count pairs// whose sum cubes is N // Function to count the pairs// satisfying a ^ 3 + b ^ 3 = Nfunction countPairs($N){    $count = 0;     // Check for each number    // 1 to cbrt(N)    for ($i = 1;         $i <= (int)pow($N, 1 / 3); $i++)    {        // Store cube of a number        $cb = $i * $i * $i;         // Subtract the cube from        // given N        $diff = ($N - $cb);         // Check if the difference is        // also a perfect cube        $cbrtDiff = (int)pow($diff, 1 / 3);         // If yes, then increment count        if ($cbrtDiff * $cbrtDiff *                        $cbrtDiff == $diff)            $count++;    }     // Return count    return $count;} // Driver Code // Loop to Count no. of pairs// satisfying a ^ 3 + b ^ 3 = i// for N = 1 to 10for ($i = 1; $i<= 10; $i++)    echo "For n = " , $i , ", ",          countPairs($i) ," pair exists\n"; // This code is contributed by jit_t?>
<script>    // Javascript program to count pairs whose sum cubes is N         // method to count the pairs satisfying    // a ^ 3 + b ^ 3 = N    function countPairs(N)    {        let count = 0;                 // Check for each number 1 to cbrt(N)        for (let i = 1; i <= parseInt(Math.pow(N,(1.0/3.0)), 10); i++)        {            // Store cube of a number            let cb = i*i*i;                     // Subtract the cube from given N            let diff = N - cb;                     // Check if the difference is also            // a perfect cube            let cbrtDiff = parseInt(Math.pow(diff,(1.0/3.0)), 10);                     // If yes, then increment count            if (cbrtDiff*cbrtDiff*cbrtDiff == diff)                count++;        }                 // Return count        return count;    }         // Loop to Count no. of pairs satisfying    // a ^ 3 + b ^ 3 = i for N = 1 to 10    for (let i = 1; i<= 10; i++)      document.write("For n = " + i + ", " + countPairs(i) + " pair exists"+"</br>");     </script>
Output:  
For n= 1, 1 pair exists
For n= 2, 1 pair exists
For n= 3, 0 pair exists
For n= 4, 0 pair exists
For n= 5, 0 pair exists
For n= 6, 0 pair exists
For n= 7, 0 pair exists
For n= 8, 1 pair exists
For n= 9, 2 pair exists
For n= 10, 0 pair exists
Reference: https://www.careercup.com/question?id=5954491572551680This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
Smitha Dinesh Semwal
ukasp
jit_t
Akanksha_Rai
divyeshrabadiya07
gabaa406
Adobe
maths-cube
Mathematical
Adobe
Mathematical
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Merge two sorted arrays
Modulo Operator (%) in C/C++ with Examples
Prime Numbers
Print all possible combinations of r elements in a given array of size n
Program for factorial of a number
Operators in C / C++
Find minimum number of coins that make a given value
The Knight's tour problem | Backtracking-1
Program to find sum of elements in a given array
Program for Decimal to Binary Conversion