Given a string of numbers and a number of multiplication operators, what is the highest number one can calculate?

Question

This was an interview question I had and I was embarrassingly pretty stumped by it. Wanted to know if anyone could think up an answer to it and provide the big O notation for it

Answer 1

Here's an iterative dynamic programming solution.

As opposed to the recursive version (which should have a similar running time).

The basic idea:

A[position][count] is the highest number that can be obtained ending at position position, using count multiplications.

So:

A[position][count] = max(for i = 0 to position
                           A[i][count-1] * input.substring(i, position))

Do this for each position and each count, then multiply each of these at the required number of multiplications with the entire remaining string.

Complexity:

Given a string |s| with m multiplication operators to be inserted...

O(m|s|2g(s)) where g(s) is the complexity of multiplication.

Java code:

static long solve(String digits, int multiplications)
{
  if (multiplications == 0)
     return Long.parseLong(digits);

  // Preprocessing - set up substring values
  long[][] substrings = new long[digits.length()][digits.length()+1];
  for (int i = 0; i < digits.length(); i++)
  for (int j = i+1; j <= digits.length(); j++)
     substrings[i][j] = Long.parseLong(digits.substring(i, j));

  // Calculate multiplications from the left
  long[][] A = new long[digits.length()][multiplications+1];
  A[0][0] = 1;
  for (int i = 1; i < A.length; i++)
  {
     A[i][0] = substrings[0][i];
     for (int j = 1; j < A[0].length; j++)
     {
        long max = -1;
        for (int i2 = 0; i2 < i; i2++)
        {
           long l = substrings[i2][i];
           long prod = l * A[i2][j-1];
           max = Math.max(max, prod);
        }
        A[i][j] = max;
     }
  }

  // Multiply left with right and find maximum
  long max = -1;
  for (int i = 1; i < A.length; i++)
  {
     max = Math.max(max, substrings[i][A.length] * A[i][multiplications]);
  }
  return max;
}

A very basic test:

System.out.println(solve("99287", 1));
System.out.println(solve("99287", 2));
System.out.println(solve("312", 1));

Prints:

86304
72036
62

Yes, it just prints the maximum. It's not too difficult to have it actually print the sums, if required.