ones-and-zeroes.py

# Time:  O(s * m * n), s is the size of the array.
# Space: O(m * n)

# In the computer world, use restricted resource you have to
# generate maximum benefit is what we always want to pursue.
#
# For now, suppose you are a dominator of m 0s and n 1s respectively.
# On the other hand, there is an array with strings consisting of only 0s and 1s.
#
# Now your task is to find the maximum number of strings that you can form
# with given m 0s and n 1s. Each 0 and 1 can be used at most once.
#
# Note:
# The given numbers of 0s and 1s will both not exceed 100
# The size of given string array won't exceed 600.
# Example 1:
# Input: Array = {"10", "0001", "111001", "1", "0"}, m = 5, n = 3
# Output: 4
#
# Explanation: This are totally 4 strings can be formed 
# by the using of 5 0s and 3 1s, which are “10,”0001”,”1”,”0”
# Example 2:
# Input: Array = {"10", "0", "1"}, m = 1, n = 1
# Output: 2
#
# Explanation: You could form "10", but then you'd have nothing left. Better form "0" and "1".

class Solution(object):
    def findMaxForm(self, strs, m, n):
        """
        :type strs: List[str]
        :type m: int
        :type n: int
        :rtype: int
        """
        dp = [[0 for _ in xrange(n+1)] for _ in xrange(m+1)]
        for s in strs:
            zero_count, one_count = 0, 0
            for c in s:
                if c == '0':
                    zero_count += 1
                elif c == '1':
                    one_count += 1

            for i in reversed(xrange(zero_count, m+1)): 
            	for j in reversed(xrange(one_count, n+1)): 
                    dp[i][j] = max(dp[i][j], dp[i-zero_count][j-one_count]+1)
        return dp[m][n]