Given an integer X, the task is to find a pair A and B such that their difference of fifth power is X, i.e., A5 – B5 = X. If there is no such pair print “Not Possible”.
Input: X = 33
Output: 1 -2
Explanation:
Input: N = 211
Output: -2 -3
Explanation:
Naive Approach: A simple solution is to use two for loops, one for A and one for B, ranging from -109 to 109.
Efficient Approach: The idea is to narrow down the range of A and B using mathematical techniques.
Since A5 – B5 = X => A5 = X + B5. For A to be as high as possible, B also has to be as high as possible, as it is evident from the inequality.
Consider A = N and B = N – 1
=> N5 – (N – 1)5 = X.
By binomial expansion, we know
(p + 1)yp <= (y + 1)p+1 – yp+1 <= (p+1)(y+1)p
So we can say that the maximum value of LHS is 4N4.
Hence 4N5 <= X
=> N <= (X/5)1/5.
=> This gives us N ~ 120.
Since A and B can also be negative, we simply extrapolate the range and the final range we get is [-120, 120].
Below is the implementation of the above approach:
// C++ implementation to find a pair // of integers A & B such that // difference of fifth power is // equal to the given number X #include <bits/stdc++.h> using namespace std;
// Function to find a pair // of integers A & B such that // difference of fifth power is // equal to the given number X void findPair( int x)
{ int lim = 120;
// Loop to choose every possible
// pair with in the range
for ( int i = -lim; i <= lim; i++) {
for ( int j = -lim; j <= lim; j++) {
// Check if equation holds
if ( pow (i, 5) - pow (j, 5) == x) {
cout << i << ' ' << j << endl;
return ;
}
}
}
cout << "-1" ;
} // Driver Code signed main()
{ int X = 33;
// Function Call
findPair(X);
return 0;
} |
// Java implementation to find a // pair of integers A & B such // that difference of fifth power // is equal to the given number X class GFG{
// Function to find a pair // of integers A & B such that // difference of fifth power is // equal to the given number X static void findPair( int x)
{ int lim = 120 ;
// Loop to choose every possible
// pair with in the range
for ( int i = -lim; i <= lim; i++)
{
for ( int j = -lim; j <= lim; j++)
{
// Check if equation holds
if (Math.pow(i, 5 ) -
Math.pow(j, 5 ) == x)
{
System.out.print(i + " " +
j + "\n" );
return ;
}
}
}
System.out.print( "-1" );
} // Driver Code public static void main(String[] args)
{ int X = 33 ;
// Function Call
findPair(X);
} } // This code is contributed by PrinciRaj1992 |
# Python3 implementation to find # a pair of integers A & B such # that difference of fifth power # is equal to the given number X import math
# Function to find a pair # of integers A & B such that # difference of fifth power is # equal to the given number X def findPair(x):
lim = 120
# Loop to choose every possible
# pair with in the range
for i in range ( - lim, lim + 1 ):
for j in range ( - lim, lim + 1 ):
# Check if equation holds
if (math. pow (i, 5 ) -
math. pow (j, 5 ) = = x):
print (i, end = ' ' )
print (j, end = '\n' )
return
print ( "-1" )
# Driver Code X = 33
# Function Call findPair(X) # This code is contributed by PratikBasu |
// C# implementation to find a // pair of integers A & B such // that difference of fifth power // is equal to the given number X using System;
class GFG{
// Function to find a pair of // integers A & B such that // difference of fifth power is // equal to the given number X static void findPair( int x)
{ int lim = 120;
// Loop to choose every possible
// pair with in the range
for ( int i = -lim; i <= lim; i++)
{
for ( int j = -lim; j <= lim; j++)
{
// Check if equation holds
if (Math.Pow(i, 5) -
Math.Pow(j, 5) == x)
{
Console.Write(i + " " +
j + "\n" );
return ;
}
}
}
Console.Write( "-1" );
} // Driver code public static void Main(String[] args)
{ int X = 33;
// Function call
findPair(X);
} } // This code is contributed by 29AjayKumar |
<script> // JavaScript implementation to find a // pair of integers A & B such // that difference of fifth power // is equal to the given number X // Function to find a pair // of integers A & B such that // difference of fifth power is // equal to the given number X function findPair(x)
{ let lim = 120;
// Loop to choose every possible
// pair with in the range
for (let i = -lim; i <= lim; i++)
for (let j = -lim; j <= lim; j++)
// Check if equation holds
if (Math.pow(i, 5) -Math.pow(j, 5) == x)
{
document.write(i + " " + j);
return ;
}
document.write( "-1" );
} // Driver Code let X = 33;
// Function Call
findPair(X);
// This code is contributed by mohan </script> |
Output:
1 -2
Time Complexity: O(240*240)