Given integer N and values P and Q, The duty is to calculate the minimal variety of powers of P and Q required to generate N.

Notice: The 0^th energy of the values can also be thought-about.

Examples:

Enter: N = 15, P = 2, Q = 3
Output: 3
Clarification: We will make 15 through the use of (8, 4, 3) or (9, 3, 3). Each take 3 numbers.

Enter: N = 19, P = 4, Q = 3
Output: 2
Clarification: Within the second case, we are able to make 19 through the use of (16, 3) which is 2 numbers.

Method: Recursion (Memoization)

The Primary thought is to make use of memoization strategy for this drawback, merely we’ll examine methods to succeed in or to generate N by contemplating each  P and Q powers by making recursive calls.

Pseudo Code: 

To examine the powers being utilized in recursive relation.

‘lengthy lengthy int a=1; 
ans = 1e9;   // to retailer potential reply
  
if(energy = 1){ 
    return n;
}                
whereas(n-a >= 0)
{
    ans = min(ans, dp[n-a]);
    a = a*energy;
}
  
return ans+1;

Observe the steps talked about beneath to implement the thought:

• Initialize a dp[] array of dimension N+1 and initialize it with 1e9.
• Set, the bottom instances, dp[0] = 0 and dp[1] = 1.
• Traverse via 2 to N and discover the methods with powers.
  •  Way1 by contemplating energy of P.
  •  Way2 by contemplating energy of Q.
• Contemplate dp[i] = min(way1, way2).
• After traversing return dp[N].  

Under is the implementation of the above strategy.

Time Complexity: O(N * logN)
Auxiliary House:  O(N * logN)

Associated Articles: