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 KK 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 KK 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(DN), 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> |
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> |
Yes
Time Complexity: O(D * N), where D is the number of digits in N.
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