Check if a number can be represented as sum of two positive perfect biquadrates

Given a positive integer N, the task is to check if N can be written as the sum of two perfect biquadrates or not i.e., (N = X⁴ + Y⁴), where X and Y are non-negative integers. If it is possible, then print Yes. Otherwise, print No.

Examples:

Input: N = 97
Output: Yes
Explanation: Summation of the biquadrates of 2 and 3 is 97, i.e. 24 + 34 = 16 + 81 = 97(= N).

Input: N = 7857
Output: Yes

Approach 1: Naïve

The simplest approach to solve the given problem is to consider all possible combinations of integers X and Y and check if the sum of their biquadrates equals N or not. If found to be true, then print Yes. Otherwise, print No.

Time Complexity: O(sqrt(N))
Auxiliary Space: O(1)

Approach 2: Binary Search

we can use binary search to optimize the search for pairs of biquadrates that sum up to the given number. In this program, we first create a vector of all possible biquadrates up to the square root of the square root of N. Then, we iterate through this vector and for each biquadrate, we check if its complement (i.e., N minus the biquadrate) exists in the vector using binary search. If we find such a complement, we return “Yes”. If we don’t find any such pair, we return “No”. 

C++

#include <bits/stdc++.h>
using namespace std;

// Function to check if the given number
// N can be expressed as the sum of two
// biquadrates or not
string sumOfTwoBiquadrates(int N)
{
    // Base Case
    if (N == 0)
        return "Yes";

    // Find the range to traverse
    int j = floor(sqrt(sqrt(N)));
    int i = 1;

    // Store all possible biquadrates in a vector
    vector<int> biquadrates;
    for (int x = 1; x <= j; x++) {
        int biquadrate = x * x * x * x;
        biquadrates.push_back(biquadrate);
    }

    // Use binary search to find the pair of biquadrates
    // that sum up to N
    for (int i = 0; i < biquadrates.size(); i++) {
        int complement = N - biquadrates[i];
        if (binary_search(biquadrates.begin(),
                          biquadrates.end(), complement)) {
            return "Yes";
        }
    }

    // If no such pairs exist
    return "No";
}

// Driver Code
int main()
{
    int N = 7857;
    cout << sumOfTwoBiquadrates(N);

    return 0;
}

Java

import java.util.*;

public class Main {
    // Function to check if the given number
    // N can be expressed as the sum of two
    // biquadrates or not
    public static String sumOfTwoBiquadrates(int N)
    {
        // Base Case
        if (N == 0)
            return "Yes";

        // Find the range to traverse
        int j = (int)Math.floor(Math.sqrt(Math.sqrt(N)));
        int i = 1;

        // Store all possible biquadrates in a vector
        List<Integer> biquadrates
            = new ArrayList<Integer>();
        for (int x = 1; x <= j; x++) {
            int biquadrate = x * x * x * x;
            biquadrates.add(biquadrate);
        }

        // Use binary search to find the pair of biquadrates
        // that sum up to N
        for (i = 0; i < biquadrates.size(); i++) {
            int complement = N - biquadrates.get(i);
            if (Collections.binarySearch(biquadrates,
                                         complement)
                >= 0) {
                return "Yes";
            }
        }

        // If no such pairs exist
        return "No";
    }

    // Driver Code
    public static void main(String[] args)
    {
        int N = 7857;
        System.out.println(sumOfTwoBiquadrates(N));
    }
}

Python3

import math

# Function to check if the given number
# N can be expressed as the sum of two
# biquadrates or not
def sumOfTwoBiquadrates(N):
    # Base Case
    if N == 0:
        return "Yes"

    # Find the range to traverse
    j = math.floor(math.sqrt(math.sqrt(N)))
    i = 1

    # Store all possible biquadrates in a list
    biquadrates = []
    for x in range(1, j+1):
        biquadrate = x * x * x * x
        biquadrates.append(biquadrate)

    # Use binary search to find the pair of biquadrates
    # that sum up to N
    for i in range(len(biquadrates)):
        complement = N - biquadrates[i]
        if complement in biquadrates:
            return "Yes"

    # If no such pairs exist
    return "No"

# Driver Code
N = 7857
print(sumOfTwoBiquadrates(N))

Javascript

function sumOfTwoBiquadrates(N) {
    // Base Case
    if (N === 0) {
        return "Yes";
    }

    // Find the range to traverse
    const j = Math.floor(Math.sqrt(Math.sqrt(N)));
    let i = 1;

    // Store all possible biquadrates in a vector
    const biquadrates = [];
    for (let x = 1; x <= j; x++) {
        const biquadrate = x * x * x * x;
        biquadrates.push(biquadrate);
    }

    // Use binary search to find the pair of biquadrates
    // that sum up to N
    for (let i = 0; i < biquadrates.length; i++) {
        const complement = N - biquadrates[i];
        if (biquadrates.includes(complement)) {
            return "Yes";
        }
    }

    // If no such pairs exist
    return "No";
}

// Driver Code
const N = 7857;
console.log(sumOfTwoBiquadrates(N));

C#

// Importing required libraries
using System;
using System.Collections.Generic;

namespace ConsoleApp {
class Program {
    // Function to check if the given number
    // N can be expressed as the sum of two
    // biquadrates or not
    static string SumOfTwoBiquadrates(int N)
    {
        // Base Case
        if (N == 0)
            return "Yes";
        // Find the range to traverse
        int j = (int)Math.Floor(Math.Sqrt(Math.Sqrt(N)));
        int i = 1;

        // Store all possible biquadrates in a list
        List<int> biquadrates = new List<int>();
        for (int x = 1; x <= j; x++) {
            int biquadrate = x * x * x * x;
            biquadrates.Add(biquadrate);
        }

        // Use binary search to find the pair of biquadrates
        // that sum up to N
        foreach(int biquadrate in biquadrates)
        {
            int complement = N - biquadrate;
            if (biquadrates.BinarySearch(complement) >= 0) {
                return "Yes";
            }
        }

        // If no such pairs exist
        return "No";
    }

    // Main Function
    static void Main(string[] args)
    {
        int N = 7857;
        Console.WriteLine(SumOfTwoBiquadrates(N));
    }
}
}
// This code is contributed by sarojmcy2e

Output: 

Yes

Time Complexity: O(sqrt(N) * logN), where N is the given number.
Auxiliary Space: O(1)

Approach 3: Two Pointer (Efficient)

The above approach can also be optimized by using the two-pointer approach, the idea is to find the two numbers over the range .  Follow the below step to solve this problem:

1. Initialize two pointers, say i as 0 and j as .
2. Iterate a loop until j becomes less than i, and perform the following steps:
   • If the value of j⁴ is N and i⁴ is N, then print Yes.
   • If the value of (i⁴ + i⁴) or (j⁴ + j⁴) or (i⁴ + j⁴) is N, then print Yes.
   • If the value of (i⁴ + j⁴) is less than N, then increment the pointer i by 1. Otherwise, decrement the pointer j by 1.
3. After completing the above steps, if the loops terminate, then print No as there exists no such pairs satisfying the given criteria.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if the given number
// N can be expressed as the sum of two
// biquadrates or not
string sumOfTwoBiquadrates(int N)
{
    // Base Case
    if (N == 0)
        return "Yes";

    // Find the range to traverse
    int j = floor(sqrt(sqrt(N)));
    int i = 1;

    while (i <= j) {

        // If x & y exists
        if (j * j * j * j == N)
            return "Yes";
        if (i * i * i * i == N)
            return "Yes";
        if (i * i * i * i + j * j * j * j == N)
            return "Yes";

        // If sum of powers of i and j
        // of 4 is less than N, then
        // increment the value of i
        if (i * i * i * i + j * j * j * j < N)
            i++;

        // Otherwise, decrement the
        // value of j
        else
            j--;
    }

    // If no such pairs exist
    return "No";
}

// Driver Code
int main()
{
    int N = 7857;
    cout << sumOfTwoBiquadrates(N);

    return 0;
}

Java

// Java program for the above approach
public class GFG
{

// Function to check if the given number
// N can be expressed as the sum of two
// biquadrates or not
static String sumOfTwoBiquadrates(int N)
{

    // Base Case
    if (N == 0)
        return "Yes";

    // Find the range to traverse
    int j = (int)Math.floor((double)(Math.sqrt((int)(Math.sqrt(N)))));

    int i = 1;

    while (i <= j) {

        // If x & y exists
        if (j * j * j * j == N)
            return "Yes";
        if (i * i * i * i == N)
            return "Yes";
        if (i * i * i * i
                + j * j * j * j
            == N)
            return "Yes";

        // If sum of powers of i and j
        // of 4 is less than N, then
        // increment the value of i
        if (i * i * i * i
                + j * j * j * j
            < N)
            i++;

        // Otherwise, decrement the
        // value of j
        else
            j--;
    }

    // If no such pairs exist
    return "No";
}

// Driver Code
public static void main(String []args)
{
    int N = 7857;
    System.out.println(sumOfTwoBiquadrates(N));


}}

// This code is contributed by AnkThon

Python3

# python program for the above approach
import math

# Function to check if the given number
# N can be expressed as the sum of two
# biquadrates or not


def sumOfTwoBiquadrates(N):

    # Base Case
    if (N == 0):
        return "Yes"

    # Find the range to traverse
    j = int(math.sqrt(math.sqrt(N)))
    i = 1

    while (i <= j):

        # If x & y exists
        if (j * j * j * j == N):
            return "Yes"
        if (i * i * i * i == N):
            return "Yes"
        if (i * i * i * i + j * j * j * j == N):
            return "Yes"

        # If sum of powers of i and j
        # of 4 is less than N, then
        # increment the value of i
        if (i * i * i * i + j * j * j * j < N):
            i += 1

        # Otherwise, decrement the
        # value of j
        else:
            j -= 1

    # If no such pairs exist
    return "No"


# Driver Code
if __name__ == "__main__":

    N = 7857
    print(sumOfTwoBiquadrates(N))

# This code is contributed by rakeshsahni

Javascript

<script>
// Javascript program for the above approach

// Function to check if the given number
// N can be expressed as the sum of two
// biquadrates or not
function sumOfTwoBiquadrates(N)
{

  // Base Case
  if (N == 0) return "Yes";

  // Find the range to traverse
  let j = Math.floor(Math.sqrt(Math.sqrt(N)));
  let i = 1;

  while (i <= j)
  {

    // If x & y exists
    if (j * j * j * j == N) return "Yes";
    if (i * i * i * i == N) return "Yes";
    if (i * i * i * i + j * j * j * j == N) return "Yes";

    // If sum of powers of i and j
    // of 4 is less than N, then
    // increment the value of i
    if (i * i * i * i + j * j * j * j < N) i++;

    // Otherwise, decrement the
    // value of j
    else j--;
  }

  // If no such pairs exist
  return "No";
}

// Driver Code

let N = 7857;
document.write(sumOfTwoBiquadrates(N));

// This code is contributed by gfgking.
</script>

C#

// C# program for the above approach
using System;
class GFG {

    // Function to check if the given number
    // N can be expressed as the sum of two
    // biquadrates or not
    static string sumOfTwoBiquadrates(int N)
    {

        // Base Case
        if (N == 0)
            return "Yes";

        // Find the range to traverse
        int j = (int)Math.Floor(
            (double)(Math.Sqrt((int)(Math.Sqrt(N)))));

        int i = 1;

        while (i <= j) {

            // If x & y exists
            if (j * j * j * j == N)
                return "Yes";
            if (i * i * i * i == N)
                return "Yes";
            if (i * i * i * i + j * j * j * j == N)
                return "Yes";

            // If sum of powers of i and j
            // of 4 is less than N, then
            // increment the value of i
            if (i * i * i * i + j * j * j * j < N)
                i++;

            // Otherwise, decrement the
            // value of j
            else
                j--;
        }

        // If no such pairs exist
        return "No";
    }

    // Driver Code
    public static void Main()
    {
        int N = 7857;
        Console.WriteLine(sumOfTwoBiquadrates(N));
    }
}

// This code is contributed by ukasp.

Output:

Yes

Time Complexity:
Auxiliary Space: O(1)