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

This implementation is for @lars.

from __future__ import (print_function)
import collections
import sys

try:
    xrange
except NameError:  # python3
    xrange = range


def max_product(s, n):
    """Return the maximum product of digits from the string s using m
    multiplication operators.

    """
    # Guard condition.
    if len(s) <= n:
        return None

    # A type for our partial solutions.
    partial_solution = collections.namedtuple("product",
                                              ["value", "expression"])

    # Initialize the best_answers dictionary with the leading terms
    best_answers = {}
    for i in xrange(len(s)):
        term = s[0: i+1]
        best_answers[i+1] = partial_solution(int(term), term)

    # We then replace best_answers n times.
    for prev_product_count in [x for x in xrange(n)]:
        product_count = prev_product_count + 1
        old_best_answers = best_answers
        best_answers = {}
        # For each position, find the best answer with the last * there.
        for position in xrange(product_count+1, len(s)+1):
            candidates = []
            for old_position in xrange(product_count, position):
                prior_product = old_best_answers[old_position]
                term = s[old_position:position]
                value = prior_product.value * int(term)
                expression = prior_product.expression + "*" + term
                candidates.append(partial_solution(value, expression))
            # max will choose the biggest value, breaking ties by the expression
            best_answers[position] = max(candidates)

    # We want the answer with the next * going at the end of the string.
    return best_answers[len(s)]

print(max_product(sys.argv[1], int(sys.argv[2])))

Here is a sample run:

$ python mult.py 99287 2
product(value=72036, expression='9*92*87')

Hopefully the logic is clear from the implementation.