Find two numbers whose difference of fourth power is equal to N
Given an integer N, the task is to find two non-negative integers X and Y such that X4 – Y4 = N. If no such pair exists, print -1.
Examples:
Input: N = 15
Output: X = 2, Y = 1
Explanation:
X4 – Y4 = (2)4 – (1)4 = (16) – (1) = 15
Input: N = 10
Output: -1
Explanation :
No such value of X and Y are there which satisfy the condition.
Approach:
To solve the problem mentioned above, we have to observe that we need to find the minimum and the maximum values of x and y that is possible to satisfy the equation.
- The minimum value for the two integers can be 0 since X & Y are non-negative.
- The maximum value of X and Y can be ceil(N(1/4)).
- Hence, iterate over the range [0, ceil(N(1/4))] and find any suitable pair of X and Y that satisfies the condition.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void solve( int n)
{
int upper_limit = ceil ( pow (
n, 1.0 / 4));
for ( int x = 0; x <= upper_limit; x++) {
for ( int y = 0; y <= upper_limit; y++) {
int num1 = x * x * x * x;
int num2 = y * y * y * y;
if (num1 - num2 == n) {
cout << "x = " << x
<< ", y = " << y;
return ;
}
}
}
cout << -1 << endl;
}
int main()
{
int n = 15;
solve(n);
return 0;
}
|
Java
import java.util.*;
class GFG{
static void solve( int n)
{
int upper_limit = ( int ) (Math.ceil
(Math.pow(n, 1.0 / 4 )));
for ( int x = 0 ; x <= upper_limit; x++)
{
for ( int y = 0 ; y <= upper_limit; y++)
{
int num1 = x * x * x * x;
int num2 = y * y * y * y;
if (num1 - num2 == n)
{
System.out.print( "x = " + x +
", y = " + y);
return ;
}
}
}
System.out.print(- 1 );
}
public static void main(String[] args)
{
int n = 15 ;
solve(n);
}
}
|
Python3
from math import pow , ceil
def solve(n) :
upper_limit = ceil( pow (n, 1.0 / 4 ));
for x in range (upper_limit + 1 ) :
for y in range (upper_limit + 1 ) :
num1 = x * x * x * x;
num2 = y * y * y * y;
if (num1 - num2 = = n) :
print ( "x =" , x, ", y =" , y);
return ;
print ( - 1 ) ;
if __name__ = = "__main__" :
n = 15 ;
solve(n);
|
C#
using System;
class GFG{
static void solve( int n)
{
int upper_limit = ( int ) (Math.Ceiling
(Math.Pow(n, 1.0 / 4)));
for ( int x = 0; x <= upper_limit; x++)
{
for ( int y = 0; y <= upper_limit; y++)
{
int num1 = x * x * x * x;
int num2 = y * y * y * y;
if (num1 - num2 == n)
{
Console.Write( "x = " + x +
", y = " + y);
return ;
}
}
}
Console.Write(-1);
}
public static void Main(String[] args)
{
int n = 15;
solve(n);
}
}
|
Javascript
<script>
function solve(n)
{
let upper_limit = Math.ceil(Math.pow(n, 1.0 / 4));
for (let x = 0; x <= upper_limit; x++) {
for (let y = 0; y <= upper_limit; y++) {
let num1 = x * x * x * x;
let num2 = y * y * y * y;
if (num1 - num2 == n) {
document.write( "x = " + x + ", y = " + y);
return ;
}
}
}
document.write(-1);
}
let n = 15;
solve(n);
</script>
|
Time Complexity: O(sqrt(N))
Auxiliary space: O(1)
Another Approach:
In this implementation, we create an unordered map (hash table) to store the values of x^4 – n for each x from 0 to the upper limit. Then, we iterate through the possible values of y and check if y^4 – n is already in the hash table. If it is, we retrieve the corresponding value of x and print the solution. If no such pair exists, we print -1.
C++
#include <bits/stdc++.h>
using namespace std;
void solve( int n) {
unordered_map< int , int > hashTable;
int upper_limit = ceil ( pow (n, 1.0/4));
for ( int x = 0; x <= upper_limit; x++) {
int x4 = x * x * x * x;
hashTable[x4 - n] = x;
}
for ( int y = 0; y <= upper_limit; y++) {
int y4 = y * y * y * y;
if (hashTable.find(y4) != hashTable.end()) {
int x = hashTable[y4];
cout << "x = " << x << ", y = " << y << endl;
return ;
}
}
cout << -1 << endl;
}
int main() {
int n = 15;
solve(n);
return 0;
}
|
Java
import java.util.HashMap;
public class EquationSolver {
static void solve( int n) {
HashMap<Integer, Integer> hashTable = new HashMap<>();
int upperLimit = ( int ) Math.ceil(Math.pow(n, 1.0 / 4 ));
for ( int x = 0 ; x <= upperLimit; x++) {
int x4 = ( int ) Math.pow(x, 4 );
hashTable.put(x4 - n, x);
}
for ( int y = 0 ; y <= upperLimit; y++) {
int y4 = ( int ) Math.pow(y, 4 );
if (hashTable.containsKey(y4)) {
int x = hashTable.get(y4);
System.out.println( "x = " + x + ", y = " + y);
return ;
}
}
System.out.println(- 1 );
}
public static void main(String[] args) {
int n = 15 ;
solve(n);
}
}
|
Python3
import math
def solve(n):
hash_table = {}
upper_limit = math.ceil(n * * ( 1.0 / 4 ))
for x in range (upper_limit + 1 ):
x4 = x * * 4
hash_table[x4 - n] = x
for y in range (upper_limit + 1 ):
y4 = y * * 4
if y4 in hash_table:
x = hash_table[y4]
print (f "x = {x}, y = {y}" )
return
print ( - 1 )
def main():
n = 15
solve(n)
if __name__ = = "__main__" :
main()
|
C#
using System;
using System.Collections.Generic;
class Program
{
static void Solve( int n)
{
Dictionary< int , int > hashTable = new Dictionary< int , int >();
int upperLimit = ( int )Math.Ceiling(Math.Pow(n, 1.0 / 4));
for ( int x = 0; x <= upperLimit; x++)
{
int x4 = x * x * x * x;
hashTable[x4 - n] = x;
}
for ( int y = 0; y <= upperLimit; y++)
{
int y4 = y * y * y * y;
if (hashTable.ContainsKey(y4))
{
int x = hashTable[y4];
Console.WriteLine( "x = " + x + ", y = " + y);
return ;
}
}
Console.WriteLine(-1);
}
static void Main()
{
int n = 15;
Solve(n);
}
}
|
Javascript
function solve(n) {
const hashTable = {};
const upperLimit = Math.ceil(Math.pow(n, 1/4));
for (let x = 0; x <= upperLimit; x++) {
const x4 = Math.pow(x, 4);
hashTable[x4 - n] = x;
}
for (let y = 0; y <= upperLimit; y++) {
const y4 = Math.pow(y, 4);
if (hashTable[y4] !== undefined) {
const x = hashTable[y4];
console.log(`x = ${x}, y = ${y}`);
return ;
}
}
console.log(-1);
}
const n = 15;
solve(n);
|
Time Complexity: O(N^(1/4))
Auxiliary space: O(N^(1/4))
Last Updated :
08 Oct, 2023
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