Check if N can be converted to the form K power K by the given operation

Given a positive number N, we have to find whether N can be converted to the form K^K where K is also a positive integer, using the following operation any number of times :

• Choose any digit less than the current value of N, say d.
• N = N – d², change N each time

If it is possible to express the number in the required form then print “Yes” otherwise print “No”.
Examples:

Input: N = 13 
Output: Yes 
Explanation:
For integer 13 choose d = 3 : N = 13 – 3² = 4, 4 is of the form 2². Hence, the output is 4.

Input: N = 90 
Output: No 
Explanation:
It is not possible to express the number 90 in required form.

Naive Approach:
To solve the problem mentioned above we will use Recursion. In each recursive step, traverse through all the digits of the current value of N, and choose it as d. This way all the search spaces will be explored and if in any of them N comes out to be of the form K^K stop the recursion and return true. To check whether the number is of the given form, pre-store all such numbers in a set. This method takes O(D^N), where D is the number of digits in N time and can be further optimized.

Below is the implementation of the given approach:

C++14

// C++ implementation to Check whether a given
// number N can be converted to the form K
// power K by the given operation
#include <bits/stdc++.h>
using namespace std;

unordered_set<int> kPowKform;

// Function to check if N can
// be converted to K power K
int func(int n)
{
    if (n <= 0)
        return 0;

    // Check if n is of the form k^k
    if (kPowKform.count(n))
        return 1;

    int answer = 0;
    int x = n;

    // Iterate through each digit of n
    while (x > 0) {
        int d = x % 10;

        if (d != 0) {
            // Check if it is possible to
            // obtain number of given form
            if (func(n - d * d)) {
                answer = 1;
                break;
            }
        }

        // Reduce the number each time
        x /= 10;
    }

    // Return the result
    return answer;
}

// Function to check the above method
void canBeConverted(int n)
{

    // Check if conversion if possible
    if (func(n))
        cout << "Yes";

    else
        cout << "No";
}

// Driver code
int main()
{
    int N = 90;

    // Pre store K power K form of numbers
    // Loop till 8, because 8^8 > 10^7

    for (int i = 1; i <= 8; i++) {
        int val = 1;
        for (int j = 1; j <= i; j++)
            val *= i;

        kPowKform.insert(val);
    }

    canBeConverted(N);

    return 0;
}

Java

// Java implementation to
// Check whether a given
// number N can be converted
// to the form K power K by
// the given operation
import java.util.*;
class GFG{

static HashSet<Integer> kPowKform =
       new HashSet<Integer>();

// Function to check if N can
// be converted to K power K
static int func(int n)
{
  if (n <= 0)
    return 0;

  // Check if n is of the form k^k
  if (kPowKform.contains(n))
    return 1;

  int answer = 0;
  int x = n;

  // Iterate through
  // each digit of n
  while (x > 0)
  {
    int d = x % 10;

    if (d != 0)
    {
      // Check if it is possible to
      // obtain number of given form
      if (func(n - d * d) == 1)
      {
        answer = 1;
        break;
      }
    }

    // Reduce the number each time
    x /= 10;
  }

  // Return the result
  return answer;
}

// Function to check the above method
static void canBeConverted(int n)
{
  // Check if conversion if possible
  if (func(n) == 1)
    System.out.print("Yes");
  else
    System.out.print("No");
}

// Driver code
public static void main(String[] args)
{
  int N = 90;

  // Pre store K power K form of numbers
  // Loop till 8, because 8^8 > 10^7
  for (int i = 1; i <= 8; i++)
  {
    int val = 1;
    for (int j = 1; j <= i; j++)
      val *= i;

    kPowKform.add(val);
  }
  canBeConverted(N);
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation to Check whether a given
# number N can be converted to the form K
# power K by the given operation

kPowKform=dict()

# Function to check if N can
# be converted to K power K
def func(n):
    global kPowKform
    if (n <= 0):
        return 0

    # Check if n is of the form k^k
    if (n in kPowKform):
        return 1

    answer = 0
    x = n

    # Iterate through each digit of n
    while (x > 0):
        d = x % 10

        if (d != 0):
            # Check if it is possible to
            # obtain number of given form
            if (func(n - d * d)):
                answer = 1
                break


        # Reduce the number each time
        x //= 10

    # Return the result
    return answer

# Function to check the above method
def canBeConverted(n):

    # Check if conversion if possible
    if (func(n)):
        print("Yes")

    else:
        print("No")

# Driver code
if __name__ == '__main__':
    N = 90

    # Pre store K power K form of numbers
    # Loop till 8, because 8^8 > 10^7

    for i in range(1,9):
        val = 1
        for j in range(1,i+1):
            val *= i

        kPowKform[val]=1


    canBeConverted(N)

# This code is contributed by mohit kumar 29

C#

// C# implementation to check whether a given
// number N can be converted to the form K
// power K by the given operation
using System;
using System.Collections.Generic;

class GFG{

static SortedSet<int> kPowKform = new SortedSet<int>();

// Function to check if N can
// be converted to K power K
static int func(int n)
{
    if (n <= 0)
        return 0;

    // Check if n is of the form k^k
    if (kPowKform.Contains(n))
        return 1;

    int answer = 0;
    int x = n;

    // Iterate through each digit of n
    while (x > 0)
    {
        int d = x % 10;

        if (d != 0)
        {

            // Check if it is possible to
            // obtain number of given form
            if (func(n - d * d) == 1)
            {
                answer = 1;
                break;
            }
        }

        // Reduce the number each time
        x /= 10;
    }

    // Return the result
    return answer;
}

// Function to check the above method
static void canBeConverted(int n)
{

    // Check if conversion if possible
    if (func(n) == 1)
        Console.Write("Yes");
    else
        Console.Write("No");
}

// Driver code
public static void Main()
{
    int N = 90;

    // Pre store K power K form of numbers
    // Loop till 8, because 8^8 > 10^7
    for(int i = 1; i <= 8; i++)
    {
        int val = 1;
        for(int j = 1; j <= i; j++)
            val *= i;

        kPowKform.Add(val);
    }
    canBeConverted(N);
}
}

// This code is contributed by sanjoy_62

Javascript

<script>

// Javascript implementation to Check whether a given
// number N can be converted to the form K
// power K by the given operation
var kPowKform = new Set();

// Function to check if N can
// be converted to K power K
function func(n)
{
    if (n <= 0)
        return 0;

    // Check if n is of the form k^k
    if (kPowKform.has(n))
        return 1;

    var answer = 0;
    var x = n;

    // Iterate through each digit of n
    while (x > 0)
    {
        var d = x % 10;

        if (d != 0)
        {

            // Check if it is possible to
            // obtain number of given form
            if (func(n - d * d)) {
                answer = 1;
                break;
            }
        }

        // Reduce the number each time
        x = parseInt(x/10);
    }

    // Return the result
    return answer;
}

// Function to check the above method
function canBeConverted(n)
{

    // Check if conversion if possible
    if (func(n))
        document.write( "Yes");

    else
        document.write( "No");
}

// Driver code
var N = 90;

// Pre store K power K form of numbers
// Loop till 8, because 8^8 > 10^7
for (var i = 1; i <= 8; i++) {
    var val = 1;
    for (var j = 1; j <= i; j++)
        val *= i;
    kPowKform.add(val);
}
canBeConverted(N);

// This code is contributed by noob2000.
</script>

Output:

No

Efficient Approach:
In the recursive approach, we are solving the same subproblem multiple times i.e there are Overlapping Subproblems. So we can use Dynamic Programming and memorize the recursive approach using a cache or memorization table.

Below is the implementation of the above approach:

C++

// C++ implementation to Check whether a given
// number N can be converted to the form K
// power K by the given operation
#include <bits/stdc++.h>
using namespace std;

unordered_set<int> kPowKform;
int dp[100005];

// Function to check if a number is converatable
int func(int n)
{
    if (n <= 0)
        return 0;

    // Check if n is of the form k^k
    if (kPowKform.count(n))
        return 1;

    // Check if the subproblem has been solved before
    if (dp[n] != -1)
        return dp[n];

    int answer = 0;
    int x = n;

    // Iterate through each digit of n
    while (x > 0) {
        int d = x % 10;
        if (d != 0) {
            // Check if it is possible to
            // obtain number of given form
            if (func(n - d * d)) {
                answer = 1;
                break;
            }
        }

        // Reduce the number each time
        x /= 10;
    }

    // Store and return the
    // answer to this subproblem
    return dp[n] = answer;
}

// Function to check the above method
void canBeConverted(int n)
{

    // Initialise the dp table
    memset(dp, -1, sizeof(dp));

    // Check if conversion if possible
    if (func(n))
        cout << "Yes";

    else
        cout << "No";
}

// Driver code
int main()
{
    int N = 13;

    // Pre store K power K form of numbers
    // Loop till 8, because 8^8 > 10^7
    for (int i = 1; i <= 8; i++) {
        int val = 1;

        for (int j = 1; j <= i; j++)
            val *= i;

        kPowKform.insert(val);
    }

    canBeConverted(N);

    return 0;
}

Java

// Java implementation to
// Check whether a given
// number N can be converted
// to the form K power K by
// the given operation
import java.util.*;
class GFG{

static HashSet<Integer> kPowKform =
       new HashSet<>();
static int []dp = new int[100005];

// Function to check if
// a number is converatable
static int func(int n)
{
  if (n <= 0)
    return 0;

  // Check if n is of the form k^k
  if (kPowKform.contains(n))
    return 1;

  // Check if the subproblem
  // has been solved before
  if (dp[n] != -1)
    return dp[n];

  int answer = 0;
  int x = n;

  // Iterate through each digit of n
  while (x > 0)
  {
    int d = x % 10;
    if (d != 0)
    {
      // Check if it is possible to
      // obtain number of given form
      if (func(n - d * d) != 0)
      {
        answer = 1;
        break;
      }
    }

    // Reduce the number
    // each time
    x /= 10;
  }

  // Store and return the
  // answer to this subproblem
  return dp[n] = answer;
}

// Function to check the above method
static void canBeConverted(int n)
{
  // Initialise the dp table
  for (int i = 0; i < n; i++)
    dp[i] = -1;

  // Check if conversion if possible
  if (func(n) == 0)
    System.out.print("Yes");
  else
    System.out.print("No");
}

// Driver code
public static void main(String[] args)
{
  int N = 13;

  // Pre store K power K form of numbers
  // Loop till 8, because 8^8 > 10^7
  for (int i = 1; i <= 8; i++)
  {
    int val = 1;

    for (int j = 1; j <= i; j++)
      val *= i;

    kPowKform.add(val);
  }
  canBeConverted(N);
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation to check whether
# a given number N can be converted to
# the form K power K by the given operation
kPowKform = dict()

# Function to check if N can
# be converted to K power K
def func(n, dp):

    global kPowKform
    if (n <= 0):
        return 0

    # Check if n is of the form k^k
    if (n in kPowKform):
        return 1

    if (dp[n] != -1):
        return dp[n]

    answer = 0
    x = n

    # Iterate through each digit of n
    while (x > 0):
        d = x % 10

        if (d != 0):

            # Check if it is possible to
            # obtain number of given form
            if (func(n - d * d, dp)):
                answer = 1
                break

        # Reduce the number each time
        x //= 10

    dp[n] = answer

    # Return the result
    return answer

# Function to check the above method
def canBeConverted(n):

    dp = [-1 for i in range(10001)]

    # Check if conversion if possible
    if (func(n, dp)):
        print("Yes")
    else:
        print("No")

# Driver code
if __name__ == '__main__':

    N = 13

    # Pre store K power K form of
    # numbers Loop till 8, because
    # 8^8 > 10^7
    for i in range(1, 9):
        val = 1

        for j in range(1, i + 1):
            val *= i

        kPowKform[val] = 1

    canBeConverted(N)

# This code is contributed by grand_master

C#

// C# implementation to check whether a given
// number N can be converted to the form K
// power K by the given operation
using System;
using System.Collections;
using System.Collections.Generic;

class GFG{

static HashSet<int> kPowKform = new HashSet<int>();
static int []dp = new int[100005];

// Function to check if a number
// is converatable
static int func(int n)
{
    if (n <= 0)
        return 0;

    // Check if n is of the form k^k
    if (kPowKform.Contains(n))
        return 1;

    // Check if the subproblem has
    // been solved before
    if (dp[n] != -1)
        return dp[n];

    int answer = 0;
    int x = n;

    // Iterate through each digit of n
    while (x > 0)
    {
        int d = x % 10;

        if (d != 0)
        {

            // Check if it is possible to
            // obtain number of given form
            if (func(n - d * d) != 0)
            {
                answer = 1;
                break;
            }
        }

        // Reduce the number each time
        x /= 10;
    }

    // Store and return the
    // answer to this subproblem
    dp[n] = answer;
    return answer;
}

// Function to check the above method
static void canBeConverted(int n)
{

    // Initialise the dp table
    Array.Fill(dp, -1);

    // Check if conversion if possible
    if (func(n) != 0)
        Console.Write("Yes");
    else
        Console.Write("No");
}

// Driver code
public static void Main(string[] args)
{
   int N = 13;

    // Pre store K power K form of numbers
    // Loop till 8, because 8^8 > 10^7
    for(int i = 1; i <= 8; i++)
    {
        int val = 1;

        for(int j = 1; j <= i; j++)
            val *= i;

        kPowKform.Add(val);
    }
    canBeConverted(N);
}
}

// This code is contributed by rutvik_56

Javascript

<script>

// Javascript implementation to Check whether a given
// number N can be converted to the form K
// power K by the given operation
var kPowKform = new Set();
var dp = Array(100005);

// Function to check if a number is converatable
function func(n)
{
    if (n <= 0)
        return 0;

    // Check if n is of the form k^k
    if (kPowKform.has(n))
        return 1;

    // Check if the subproblem has been solved before
    if (dp[n] != -1)
        return dp[n];

    var answer = 0;
    var x = n;

    // Iterate through each digit of n
    while (x > 0) {
        var d = x % 10;
        if (d != 0) {
            // Check if it is possible to
            // obtain number of given form
            if (func(n - d * d)) {
                answer = 1;
                break;
            }
        }

        // Reduce the number each time
        x /= 10;
    }

    // Store and return the
    // answer to this subproblem
    return dp[n] = answer;
}

// Function to check the above method
function canBeConverted(n)
{

    // Initialise the dp table
    dp = Array(100005).fill(-1);

    // Check if conversion if possible
    if (func(n))
        document.write( "Yes");

    else
        document.write( "No");
}

// Driver code
var N = 13;

// Pre store K power K form of numbers
// Loop till 8, because 8^8 > 10^7
for (var i = 1; i <= 8; i++) {
    var val = 1;
    for (var j = 1; j <= i; j++)
        val *= i;
    kPowKform.add(val);
}
canBeConverted(N);

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

Output:

Yes

Time Complexity: O(D * N), where D is the number of digits in N.

Efficient approach: Using DP Tabulation method ( Iterative approach )

The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memorization(top-down) because memorization method needs extra stack space of recursion calls.

Steps to solve this problem :

• Initialize the dp table with n+1 elements and set all elements to 0 using the memset function.
• Use another loop to solve subproblems in a bottom-up manner. For each value i from 1 to n:
• Check if i is of the form k^k. If yes, set dp[i] to 1 and continue to the next iteration.
• If i is not of the form k^k, iterate through each digit of i using a while loop. For each digit d, check if it is possible to obtain a number of the given form by subtracting d*d from i and checking the corresponding value in the dp table.
• If it is possible to obtain a number of the given form, set found to true and break out of the while loop.
• Store the value of found in dp[i].
• Print the final solution. If dp[n] is equal to 1, print “Yes“, else print “No“.

Implementation :

C++

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

// Function to check if a number is convertible
void canBeConverted(int n)
{
    // Initialize the dp table
    int dp[n+1];
    memset(dp, 0, sizeof(dp));

    // Pre-store K power K form of numbers
    // Loop till 8, because 8^8 > 10^7
    unordered_set<int> kPowKform;
    for (int i = 1; i <= 8; i++) {
        int val = 1;
        for (int j = 1; j <= i; j++)
            val *= i;
        kPowKform.insert(val);
    }

    // Base case
    dp[0] = 0;

    // Solve subproblems in a bottom-up manner
    for (int i = 1; i <= n; i++) {
        // Check if i is of the form k^k
        if (kPowKform.count(i)) {
            dp[i] = 1;
            continue;
        }

        int x = i;
        bool found = false;

        // Iterate through each digit of i
        while (x > 0) {
            int d = x % 10;
            if (d != 0) {
                // Check if it is possible to obtain number of given form
                if (dp[i - d*d] == 1) {
                    found = true;
                    break;
                }
            }

            // Reduce the number each time
            x /= 10;
        }

        // Store the answer to this subproblem
        dp[i] = found;
    }

    // Print the final solution
    if (dp[n] == 1)
        cout << "Yes";
    else
        cout << "No";
}

// Driver code
int main()
{
    int n = 13;
    canBeConverted(n);
    return 0;
}

Java

import java.util.*;

public class Main
{

  // Function to check if a number is convertible
  public static void canBeConverted(int n)
  {

    // Initialize the dp table
    boolean[] dp = new boolean[n + 1];
    Arrays.fill(dp, false);

    // Pre-store K power K form of numbers
    // Loop till 8, because 8^8 > 10^7
    Set<Integer> kPowKform = new HashSet<Integer>();
    for (int i = 1; i <= 8; i++) {
      int val = 1;
      for (int j = 1; j <= i; j++) {
        val *= i;
      }
      kPowKform.add(val);
    }

    // Base case
    dp[0] = true;

    // Solve subproblems in a bottom-up manner
    for (int i = 1; i <= n; i++)
    {

      // Check if i is of the form k^k
      if (kPowKform.contains(i)) {
        dp[i] = true;
        continue;
      }

      int x = i;
      boolean found = false;

      // Iterate through each digit of i
      while (x > 0) {
        int d = x % 10;
        if (d != 0)
        {

          // Check if it is possible to obtain number of given form
          if (i - d * d >= 0 && dp[i - d * d]) {
            found = true;
            break;
          }
        }

        // Reduce the number each time
        x /= 10;
      }

      // Store the answer to this subproblem
      dp[i] = found;

    }

    // Print the final solution
    if (dp[n]) {
      System.out.println("Yes");
    } else {
      System.out.println("No");
    }
  }

  // Driver code
  public static void main(String[] args) {
    int n = 13;
    canBeConverted(n);
  }
}

Python

def canBeConverted(n):
    # Initialize the dp table
    dp = [0] * (n+1)

    # Pre-store K power K form of numbers
    # Loop till 8, because 8^8 > 10^7
    kPowKform = set()
    for i in range(1, 9):
        val = 1
        for j in range(1, i+1):
            val *= i
        kPowKform.add(val)

    # Base case
    dp[0] = 0

    # Solve subproblems in a bottom-up manner
    for i in range(1, n+1):
        # Check if i is of the form k^k
        if i in kPowKform:
            dp[i] = 1
            continue

        x = i
        found = False

        # Iterate through each digit of i
        while x > 0:
            d = x % 10
            if d != 0:
                # Check if it is possible to obtain number of given form
                if i - d*d >= 0 and dp[i - d*d] == 1:
                    found = True
                    break


            # Reduce the number each time
            x //= 10

        # Store the answer to this subproblem
        dp[i] = found

    # Print the final solution
    if dp[n] == 1:
        print("Yes")
    else:
        print("No")

# Driver code
n = 13
canBeConverted(n)

C#

// C# code addition
using System;
using System.Collections.Generic;

class MainClass
{

  // Function to check if a number is convertible
  public static void canBeConverted(int n)
  {

    // Initialize the dp table
    bool[] dp = new bool[n + 1];
    Array.Fill(dp, false);

    // Pre-store K power K form of numbers
    // Loop till 8, because 8^8 > 10^7
    HashSet < int > kPowKform = new HashSet < int > ();
    for (int i = 1; i <= 8; i++) {
      int val = 1;
      for (int j = 1; j <= i; j++) {
        val *= i;
      }
      kPowKform.Add(val);
    }

    // Base case
    dp[0] = true;

    // Solve subproblems in a bottom-up manner
    for (int i = 1; i <= n; i++) {

      // Check if i is of the form k^k
      if (kPowKform.Contains(i)) {
        dp[i] = true;
        continue;
      }

      int x = i;
      bool found = false;

      // Iterate through each digit of i
      while (x > 0) {
        int d = x % 10;
        if (d != 0) {

          // Check if it is possible to obtain number of given form
          if (i - d * d >= 0 && dp[i - d * d]) {
            found = true;
            break;
          }
        }

        // Reduce the number each time
        x /= 10;
      }

      // Store the answer to this subproblem
      dp[i] = found;

    }

    // Print the final solution
    if (dp[n]) {
      Console.WriteLine("Yes");
    } else {
      Console.WriteLine("No");
    }
  }

  // Driver code
  public static void Main(string[] args) {
    int n = 13;
    canBeConverted(n);
  }
}

// The code is contributed by Nidhi goel.

Javascript

// Function to check if a number is convertible
function canBeConverted(n) {
    // Initialize the dp table
    let dp = new Array(n + 1).fill(0);

    // Pre-store K power K form of numbers
    // Loop till 8, because 8^8 > 10^7
    let kPowKform = new Set();
    for (let i = 1; i <= 8; i++) {
        let val = 1;
        for (let j = 1; j <= i; j++)
            val *= i;
        kPowKform.add(val);
    }

    // Base case
    dp[0] = 0;

    // Solve subproblems in a bottom-up manner
    for (let i = 1; i <= n; i++) {
        // Check if i is of the form k^k
        if (kPowKform.has(i)) {
            dp[i] = 1;
            continue;
        }

        let x = i;
        let found = false;

        // Iterate through each digit of i
        while (x > 0) {
            let d = x % 10;
            if (d != 0) {
                // Check if it is possible to obtain number of given form
                if (dp[i - d * d] == 1) {
                    found = true;
                    break;
                }
            }

            // Reduce the number each time
            x = Math.floor(x / 10);
        }

        // Store the answer to this subproblem
        dp[i] = found;
    }

    // Print the final solution
    if (dp[n] == 1)
        console.log("Yes");
    else
        console.log("No");
}

// Driver code
let n = 13;
canBeConverted(n);

Output

Yes

Time Complexity: O(NlogN)
Auxiliary Space: O(N)