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) = 15Input: 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++
// C++ implementation to find the// values of x and y for the given// equation with integer N#include <bits/stdc++.h>using namespace std;// Function which find required x & yvoid solve(int n){ // Upper limit of x & y, // if such x & y exists 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++) { // num1 stores x^4 int num1 = x * x * x * x; // num2 stores y^4 int num2 = y * y * y * y; // If condition is satisfied // the print and return if (num1 - num2 == n) { cout << "x = " << x << ", y = " << y; return; } } } // If no such pair exists cout << -1 << endl;}// Driver codeint main(){ int n = 15; solve(n); return 0;} |
Java
// Java implementation to find the// values of x and y for the given// equation with integer Nimport java.util.*;class GFG{// Function which find required x & ystatic void solve(int n){ // Upper limit of x & y, // if such x & y exists 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++) { // num1 stores x^4 int num1 = x * x * x * x; // num2 stores y^4 int num2 = y * y * y * y; // If condition is satisfied // the print and return if (num1 - num2 == n) { System.out.print("x = " + x + ", y = " + y); return; } } } // If no such pair exists System.out.print(-1);}// Driver codepublic static void main(String[] args){ int n = 15; solve(n);}}// This code is contributed by shivanisinghss2110 |
Python3
# Python3 implementation to find the# values of x and y for the given# equation with integer Nfrom math import pow, ceil# Function which find required x & ydef solve(n) : # Upper limit of x & y, # if such x & y exists upper_limit = ceil(pow(n, 1.0 / 4)); for x in range(upper_limit + 1) : for y in range(upper_limit + 1) : # num1 stores x^4 num1 = x * x * x * x; # num2 stores y^4 num2 = y * y * y * y; # If condition is satisfied # the print and return if (num1 - num2 == n) : print("x =", x, ", y =" , y); return; # If no such pair exists print(-1) ;# Driver codeif __name__ == "__main__" : n = 15; solve(n); # This code is contributed by AnkitRai01 |
C#
// C# implementation to find the// values of x and y for the given// equation with integer Nusing System;class GFG{// Function which find required x & ystatic void solve(int n){ // Upper limit of x & y, // if such x & y exists 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++) { // num1 stores x^4 int num1 = x * x * x * x; // num2 stores y^4 int num2 = y * y * y * y; // If condition is satisfied // the print and return if (num1 - num2 == n) { Console.Write("x = " + x + ", y = " + y); return; } } } // If no such pair exists Console.Write(-1);}// Driver codepublic static void Main(String[] args){ int n = 15; solve(n);}}// This code is contributed by shivanisinghss2110 |
Javascript
<script> // Javascript implementation to find the // values of x and y for the given // equation with integer N // Function which find required x & y function solve(n) { // Upper limit of x & y, // if such x & y exists 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++) { // num1 stores x^4 let num1 = x * x * x * x; // num2 stores y^4 let num2 = y * y * y * y; // If condition is satisfied // the print and return if (num1 - num2 == n) { document.write("x = " + x + ", y = " + y); return; } } } // If no such pair exists document.write(-1); } let n = 15; solve(n); </script> |
Output
x = 2, y = 1
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; // Calculate upper limit of x & y int upper_limit = ceil(pow(n, 1.0/4)); // Store x^4 - n in hash table for (int x = 0; x <= upper_limit; x++) { int x4 = x * x * x * x; hashTable[x4 - n] = x; } // Check if there is a y such that y^4 - x^4 = n 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; } } // If no such pair exists cout << -1 << endl;}int main() { int n = 15; solve(n); return 0;} |
Java
//Java implementation to find the// values of x and y for the given// equation with integer Nimport java.util.HashMap;public class EquationSolver { // Function which finds required x & y static void solve(int n) { HashMap<Integer, Integer> hashTable = new HashMap<>(); // Calculate upper limit of x & y int upperLimit = (int) Math.ceil(Math.pow(n, 1.0 / 4)); // Store x^4 - n in the hash table for (int x = 0; x <= upperLimit; x++) { int x4 = (int) Math.pow(x, 4); hashTable.put(x4 - n, x); } // Check if there is a y such that y^4 - x^4 = n 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; } } // If no such pair exists System.out.println(-1); } // Driver code public static void main(String[] args) { int n = 15; solve(n); }} |
Python3
# Python3 implementation to find the# values of x and y for the given# equation with integer Nimport math# Function which find required x & ydef solve(n): hash_table = {} # Calculate upper limit of x & y upper_limit = math.ceil(n ** (1.0/4)) # Store x^4 - n in the hash table for x in range(upper_limit + 1): x4 = x ** 4 hash_table[x4 - n] = x # Check if there is a y such that y^4 - x^4 = n 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 # If no such pair exists print(-1)#Driver codedef 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>(); // Calculate upper limit of x & y int upperLimit = (int)Math.Ceiling(Math.Pow(n, 1.0 / 4)); // Store x^4 - n in hash table for (int x = 0; x <= upperLimit; x++) { int x4 = x * x * x * x; hashTable[x4 - n] = x; } // Check if there is a y such that y^4 - x^4 = n 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; } } // If no such pair exists Console.WriteLine(-1); } static void Main() { int n = 15; Solve(n); }} |
Javascript
// Function to find required x & yfunction solve(n) { const hashTable = {}; // Calculate upper limit of x & y const upperLimit = Math.ceil(Math.pow(n, 1/4)); // Store x^4 - n in the hash table for (let x = 0; x <= upperLimit; x++) { const x4 = Math.pow(x, 4); hashTable[x4 - n] = x; } // Check if there is a y such that y^4 - x^4 = n 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; } } // If no such pair exists console.log(-1);}// Driver codeconst n = 15;solve(n); |
Output
x = 2, y = 1
Time Complexity: O(N^(1/4))
Auxiliary space: O(N^(1/4))
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