Find ways an Integer can be expressed as sum of n-th power of unique natural numbers

Given two numbers x and n, find a number of ways x can be expressed as sum of n-th power of unique natural numbers.

Examples :

Input  : x = 10, n = 2
Output : 1
Explanation: 10 = 1² + 3², Hence total 1 possibility

Input  : x = 100, n = 2
Output : 3
Explanation: 
100 = 10² OR 6² + 8² OR 1² + 3² + 4² + 5² + 7² Hence total 3 possibilities

The idea is simple. We iterate through all number starting from 1. For every number, we recursively try all greater numbers and if we are able to find sum, we increment result

C++

// C++ program to count number of ways any
// given integer x can be expressed as n-th
// power of unique natural numbers.
#include <bits/stdc++.h>
using namespace std;

// Function to calculate and return the
// power of any given number
int power(int num, unsigned int n)
{
    if (n == 0)
        return 1;
    else if (n % 2 == 0)
        return power(num, n / 2) * power(num, n / 2);
    else
        return num * power(num, n / 2) * power(num, n / 2);
}

// Function to check power representations recursively
int checkRecursive(int x, int n, int curr_num = 1,
                   int curr_sum = 0)
{
    // Initialize number of ways to express
    // x as n-th powers of different natural
    // numbers
    int results = 0;

    // Calling power of 'i' raised to 'n'
    int p = power(curr_num, n);
    while (p + curr_sum < x) {
        // Recursively check all greater values of i
        results += checkRecursive(x, n, curr_num + 1,
                                  p + curr_sum);
        curr_num++;
        p = power(curr_num, n);
    }

    // If sum of powers is equal to x
    // then increase the value of result.
    if (p + curr_sum == x)
        results++;

    // Return the final result
    return results;
}

// Driver Code.
int main()
{
    int x = 10, n = 2;
    cout << checkRecursive(x, n);
    return 0;
}

Java

// Java program to count number of ways any
// given integer x can be expressed as n-th
// power of unique natural numbers.

class GFG {

    // Function to calculate and return the
    // power of any given number
    static int power(int num, int n)
    {
        if (n == 0)
            return 1;
        else if (n % 2 == 0)
            return power(num, n / 2) * power(num, n / 2);
        else
            return num * power(num, n / 2)
                * power(num, n / 2);
    }

    // Function to check power representations recursively
    static int checkRecursive(int x, int n, int curr_num,
                              int curr_sum)
    {
        // Initialize number of ways to express
        // x as n-th powers of different natural
        // numbers
        int results = 0;

        // Calling power of 'i' raised to 'n'
        int p = power(curr_num, n);
        while (p + curr_sum < x) {
            // Recursively check all greater values of i
            results += checkRecursive(x, n, curr_num + 1,
                                      p + curr_sum);
            curr_num++;
            p = power(curr_num, n);
        }

        // If sum of powers is equal to x
        // then increase the value of result.
        if (p + curr_sum == x)
            results++;

        // Return the final result
        return results;
    }

    // Driver Code.
    public static void main(String[] args)
    {
        int x = 10, n = 2;
        System.out.println(checkRecursive(x, n, 1, 0));
    }
}

// This code is contributed by mits

Python3

# Python3 program to count number of ways any
# given integer x can be expressed as n-th
# power of unique natural numbers.

# Function to calculate and return the
# power of any given number


def power(num, n):

    if(n == 0):
        return 1
    elif(n % 2 == 0):
        return power(num, n // 2) * power(num, n // 2)
    else:
        return num * power(num, n // 2) * power(num, n // 2)

# Function to check power representations recursively


def checkRecursive(x, n, curr_num=1, curr_sum=0):

    # Initialize number of ways to express
    # x as n-th powers of different natural
    # numbers
    results = 0

    # Calling power of 'i' raised to 'n'
    p = power(curr_num, n)
    while(p + curr_sum < x):

        # Recursively check all greater values of i
        results += checkRecursive(x, n, curr_num + 1, p + curr_sum)
        curr_num = curr_num + 1
        p = power(curr_num, n)

    # If sum of powers is equal to x
    # then increase the value of result.
    if(p + curr_sum == x):
        results = results + 1

    # Return the final result
    return results


# Driver Code.
if __name__ == '__main__':
    x = 10
    n = 2
    print(checkRecursive(x, n))


# This code is contributed by
# Sanjit_Prasad

C#

// C# program to count number of ways any
// given integer x can be expressed as
// n-th power of unique natural numbers.
using System;

class GFG {

    // Function to calculate and return
    // the power of any given number
    static int power(int num, int n)
    {
        if (n == 0)
            return 1;
        else if (n % 2 == 0)
            return power(num, n / 2) * power(num, n / 2);
        else
            return num * power(num, n / 2)
                * power(num, n / 2);
    }

    // Function to check power
    // representations recursively
    static int checkRecursive(int x, int n, int curr_num,
                              int curr_sum)
    {
        // Initialize number of ways to express
        // x as n-th powers of different natural
        // numbers
        int results = 0;

        // Calling power of 'i' raised to 'n'
        int p = power(curr_num, n);
        while (p + curr_sum < x) {
            // Recursively check all greater values of i
            results += checkRecursive(x, n, curr_num + 1,
                                      p + curr_sum);
            curr_num++;
            p = power(curr_num, n);
        }

        // If sum of powers is equal to x
        // then increase the value of result.
        if (p + curr_sum == x)
            results++;

        // Return the final result
        return results;
    }

    // Driver Code.
    public static void Main()
    {
        int x = 10, n = 2;
        System.Console.WriteLine(
            checkRecursive(x, n, 1, 0));
    }
}

// This code is contributed by mits

PHP

<?php
// PHP program to count
// number of ways any
// given integer x can
// be expressed as n-th
// power of unique
// natural numbers.

// Function to calculate and return
// the power of any given number
function power($num, $n)
{

    if ($n == 0)
        return 1;
    else if ($n % 2 == 0)
        return power($num, (int)($n / 2)) *
               power($num, (int)($n / 2));
    else
        return $num * power($num, (int)($n / 2)) *
                      power($num, (int)($n / 2));
}

// Function to check power
// representations recursively
function checkRecursive($x, $n,
                        $curr_num = 1,
                        $curr_sum = 0)
{

    // Initialize number of
    // ways to express
    // x as n-th powers
    // of different natural
    // numbers
    $results = 0;

    // Calling power of 'i'
    // raised to 'n'
    $p = power($curr_num, $n);
    while ($p + $curr_sum < $x)
    {

        // Recursively check all
        // greater values of i
        $results += checkRecursive($x, $n,
                                   $curr_num + 1,
                                   $p + $curr_sum);
        $curr_num++;
        $p = power($curr_num, $n);
    }

    // If sum of powers
    // is equal to x
    // then increase the
    // value of result.
    if ($p + $curr_sum == $x)
        $results++;

    // Return the final result
    return $results;
}

// Driver Code.
$x = 10; $n = 2;
echo(checkRecursive($x, $n));

// This code is contributed by Ajit.
?>

Javascript

<script>
// javascript program to count number of ways any
// given integer x can be expressed as n-th
// power of unique natural numbers.

    // Function to calculate and return the
    // power of any given number
    function power(num , n)
    {
        if (n == 0)
            return 1;
        else if (n % 2 == 0)
            return power(num, parseInt(n / 2)) * power(num, parseInt(n / 2));
        else
            return num * power(num,parseInt(n / 2)) * power(num, parseInt(n / 2));
    }

    // Function to check power representations recursively
    function checkRecursive(x , n , curr_num , curr_sum)
    {

        // Initialize number of ways to express
        // x as n-th powers of different natural
        // numbers
        var results = 0;

        // Calling power of 'i' raised to 'n'
        var p = power(curr_num, n);
        while (p + curr_sum < x)
        {

            // Recursively check all greater values of i
            results += checkRecursive(x, n, curr_num + 1, p + curr_sum);
            curr_num++;
            p = power(curr_num, n);
        }

        // If sum of powers is equal to x
        // then increase the value of result.
        if (p + curr_sum == x)
            results++;

        // Return the final result
        return results;
    }

    // Driver Code.
        var x = 10, n = 2;
        document.write(checkRecursive(x, n, 1, 0));

// This code is contributed by gauravrajput1
</script>

Output

1

Time Complexity: O(x^(1/n)), which is the maximum possible value of curr_num.
Space Complexity: O(log(x)) for the recursion stack.

Alternate Solution :

Below is an alternate simpler solution provided by Shivam Kanodia.

C++

// C++ program to find number of ways to express
// a number as sum of n-th powers of numbers.
#include<bits/stdc++.h>
using namespace std;

int res = 0;
int checkRecursive(int num, int x, int k, int n)
{
    if (x == 0)
        res++;

    int r = (int)floor(pow(num, 1.0 / n));

    for (int i = k + 1; i <= r; i++)
    {
        int a = x - (int)pow(i, n);
        if (a >= 0)
            checkRecursive(num, x -
                          (int)pow(i, n), i, n);
    }
    return res;
}

// Wrapper over checkRecursive()
int check(int x, int n)
{
    return checkRecursive(x, x, 0, n);
}

// Driver Code
int main()
{
    cout << (check(10, 2));
    return 0;
}

// This code is contributed by mits

C

// C program to find number of ways to express
// a number as sum of n-th powers of numbers.
#include <math.h>
#include <stdio.h>

int res = 0;
int checkRecursive(int num, int x, int k, int n)
{
    if (x == 0)
        res++;

    int r = (int)floor(pow(num, 1.0 / n));

    for (int i = k + 1; i <= r; i++) {
        int a = x - (int)pow(i, n);
        if (a >= 0)
            checkRecursive(num, x - (int)pow(i, n), i, n);
    }
    return res;
}

// Wrapper over checkRecursive()
int check(int x, int n)
{
    return checkRecursive(x, x, 0, n);
}

// Driver Code
int main()
{
    printf("%d", (check(10, 2)));
    return 0;
}

// This code is contributed by Rohit Pradhan

Java

// Java program to find number of ways to express a
// number as sum of n-th powers of numbers.
import java.io.*;
import java.util.*;

public class Solution {

    static int res = 0;
    static int checkRecursive(int num, int x, int k, int n)
    {
        if (x == 0)
            res++;

        int r = (int)Math.floor(Math.pow(num, 1.0 / n));

        for (int i = k + 1; i <= r; i++) {
            int a = x - (int)Math.pow(i, n);
          if (a >= 0)
            checkRecursive(num,
                   x - (int)Math.pow(i, n), i, n);
        }
        return res;
    }

    // Wrapper over checkRecursive()
    static int check(int x, int n)
    {
        return checkRecursive(x, x, 0, n);
    }

    public static void main(String[] args)
    {
        System.out.println(check(10, 2));
    }
}

Python3

# Python 3 program to find number of ways to express
# a number as sum of n-th powers of numbers.


def checkRecursive(num, rem_num, next_int, n, ans=0):

    if (rem_num == 0):
        ans += 1

    r = int(num**(1 / n))

    for i in range(next_int + 1, r + 1):
        a = rem_num - int(i**n)
        if a >= 0:
            ans += checkRecursive(num, rem_num - int(i**n), i, n, 0)
    return ans

# Wrapper over checkRecursive()


def check(x, n):
    return checkRecursive(x, x, 0, n)


# Driver Code
if __name__ == '__main__':
    print(check(10, 2))

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find number of
// ways to express a number as sum
// of n-th powers of numbers.
using System;

class Solution {

    static int res = 0;
    static int checkRecursive(int num, int x,
                                int k, int n)
    {
        if (x == 0)
            res++;

        int r = (int)Math.Floor(Math.Pow(num, 1.0 / n));

        for (int i = k + 1; i <= r; i++)
        {
            int a = x - (int)Math.Pow(i, n);
        if (a >= 0)
            checkRecursive(num, x -
                          (int)Math.Pow(i, n), i, n);
        }
        return res;
    }

    // Wrapper over checkRecursive()
    static int check(int x, int n)
    {
        return checkRecursive(x, x, 0, n);
    }

    // Driver code
    public static void Main()
    {
        Console.WriteLine(check(10, 2));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to find number
// of ways to express a number
// as sum of n-th powers of numbers.
$res = 0;

function checkRecursive($num, $x,
                        $k, $n)
{
    global $res;
    if ($x == 0)
        $res++;

    $r = (int)floor(pow($num,
                        1.0 / $n));

    for ($i = $k + 1;
         $i <= $r; $i++)
    {
        $a = $x - (int)pow($i, $n);
        if ($a >= 0)
            checkRecursive($num, $x -
                    (int)pow($i, $n),
                             $i, $n);
    }
    return $res;
}

// Wrapper over
// checkRecursive()
function check($x, $n)
{
    return checkRecursive($x, $x,
                          0, $n);
}

// Driver Code
echo (check(10, 2));

// This code is contributed by ajit
?>

Javascript

<script>

// JavaScript program for the above approach

    let res = 0;
    function checkRecursive(num, x, k, n)
    {
        if (x == 0)
            res++;

        let r = Math.floor(Math.pow(num, 1.0 / n));

        for (let i = k + 1; i <= r; i++) {
            let a = x - Math.pow(i, n);
          if (a >= 0)
            checkRecursive(num,
                   x - Math.pow(i, n), i, n);
        }
        return res;
    }

    // Wrapper over checkRecursive()
    function check(x, n)
    {
        return checkRecursive(x, x, 0, n);
    }

// Driver Code
      document.write(check(10, 2));

// This code is contributed by splevel62.
</script>

Output

1

Time Complexity: O(x^(1/n)),
Space Complexity: O(log(x)) 

Simple Recursive Solution:

contributed by Ram Jondhale.

C++

#include <iostream>
#include<cmath>
using namespace std;

//Helper function
int getAllWaysHelper(int remainingSum, int power, int base){
      //calculate power
    int result = pow(base, power);

    if(remainingSum == result)
        return 1;
    if(remainingSum < result)
        return 0;
      //Two recursive calls one to include current base's power in sum another to exclude
    int x = getAllWaysHelper(remainingSum - result, power, base + 1);
    int y = getAllWaysHelper(remainingSum, power, base+1);
    return x + y;
}

int getAllWays(int sum, int power) {
    return getAllWaysHelper(sum, power, 1);
}

// Driver Code.
int main()
{
    int x = 10, n = 2;
    cout << getAllWays(x, n);
    return 0;
}

Java

// Java program to implement the approach
import java.io.*;

class GFG {

  public static int getAllWaysHelper(int remainingSum,
                                     int power, int base)
  {
    // calculate power
    int result = (int)Math.pow(base, power);

    if (remainingSum == result)
      return 1;
    if (remainingSum < result)
      return 0;
    // Two recursive calls one to include current base's
    // power in sum another to exclude
    int x = getAllWaysHelper(remainingSum - result,
                             power, base + 1);
    int y = getAllWaysHelper(remainingSum, power,
                             base + 1);
    return x + y;
  }

  public static int getAllWays(int sum, int power)
  {
    return getAllWaysHelper(sum, power, 1);
  }

  // Driver Code
  public static void main(String[] args)
  {
    int x = 10, n = 2;
    System.out.print(getAllWays(x, n));
  }
}

// This code is contributed by Rohit Pradhan

Python3

# Helper function
def getAllWaysHelper(remainingSum, power, base):

    # calculate power
    result = pow(base, power)

    if(remainingSum == result):
        return 1
    if(remainingSum < result):
        return 0

    # Two recursive calls one to include
    # current base's power in sum another to exclude
    x = getAllWaysHelper(remainingSum - result, power, base + 1)
    y = getAllWaysHelper(remainingSum, power, base+1)
    return x + y

def getAllWays(sum, power):
    return getAllWaysHelper(sum, power, 1)

# Driver Code
x,n = 10,2
print(getAllWays(x, n))

# This code is contributed by shinjanpatra.

C#

using System;
class GFG {

    // Helper function
    static int getAllWaysHelper(int remainingSum, int power,
                                int bases)
    {
        // calculate power
        int result = (int)Math.Pow(bases, power);

        if (remainingSum == result)
            return 1;
        if (remainingSum < result)
            return 0;
        // Two recursive calls one to include current base's
        // power in sum another to exclude
        int x = getAllWaysHelper(remainingSum - result,
                                 power, bases + 1);
        int y = getAllWaysHelper(remainingSum, power,
                                 bases + 1);
        return x + y;
    }

    static int getAllWays(int sum, int power)
    {
        return getAllWaysHelper(sum, power, 1);
    }

    // Driver Code.
    public static int Main()
    {
        int x = 10, n = 2;
        Console.Write(getAllWays(x, n));
        return 0;
    }
}

// This code is contributed by Taranpreet

Javascript

<script>

// Helper function
function getAllWaysHelper(remainingSum, power, base)
{

    // calculate power
    let result = Math.pow(base, power);

    if(remainingSum == result)
        return 1;
    if(remainingSum < result)
        return 0;

    // Two recursive calls one to include
    // current base's power in sum another to exclude
    let x = getAllWaysHelper(remainingSum - result, power, base + 1);
    let y = getAllWaysHelper(remainingSum, power, base+1);
    return x + y;
}

function getAllWays(sum, power) {
    return getAllWaysHelper(sum, power, 1);
}

// Driver Code.

let x = 10, n = 2;
document.write(getAllWays(x, n));

// This code is contributed by shinjanpatra

</script>

Output

1

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