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

• Difficulty Level : Easy
• Last Updated : 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.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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++

// 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 = N
int 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 program
int 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

// 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

# Python 3 program to count pairs
# whose sum cubes is N
import math

# Function to count the pairs
# satisfying a ^ 3 + b ^ 3 = N
def 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 10
for i in range(1, 11):
print('For n = ', i, ', ', countPairs(i),
' pair exists')

# This code is contributed by Smitha.

C#

// 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
// PHP program to count pairs
// whose sum cubes is N

// Function to count the pairs
// satisfying a ^ 3 + b ^ 3 = N
function 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 10
for (\$i = 1; \$i<= 10; \$i++)
echo "For n = " , \$i , ", ",
countPairs(\$i) ," pair exists\n";

// This code is contributed by jit_t
?>

Javascript

<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=5954491572551680
This 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.