# 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 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// CPP implementation to Check whether a given ` `// number N can be converted to the form K ` `// power K by the given operation ` `#include ` `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, becasue 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 kPowKform = ``new` `HashSet(); ` `  `  `// 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, becasue 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, becasue 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     `

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 memoize the recursive approach using a cache or memorization table.

Below is the implementation of the above approach:

 `// CPP implementation to Check whether a given ` `// number N can be converted to the form K ` `// power K by the given operation ` `#include ` `using` `namespace` `std; ` ` `  `unordered_set<``int``> kPowKform; ` `int` `dp; ` ` `  `// 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 numebr 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; ` `} ` ` `  `// Fcuntion 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, becasue 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; ` `} `

Output:

```Yes
```

Time Complexity: O(D * N), where D is the number of digits in N. My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : mohit kumar 29, Rajput-Ji

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