[leetcode]ones and zeros

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本文探讨了一个计算机科学领域的经典问题——如何利用有限数量的0和1形成尽可能多的字符串。通过动态规划方法解决这一问题，并提供了两种不同的实现方案，一种是简单实现的空间复杂度为O(M×N×ArrayLength)，另一种进行了优化后的空间复杂度为O(M×N)。

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Ones and Zeroes Add to List.

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".

This problem is similar to knapsack problem, which is a classical dynamic programming problem. Here is a great slide that you can learn how to solve it.
Use $dp[i][m][n]$ denotes max number of string can form with $m$ 0s and $n$ 1s.
$count0$ denotes 0s’ number in $s_i$ , $count1$ denotes 1s’ number in $s_i$ .

if $count0$ > $m$ OR $count1$ > $n$ : $dp[i][m][n] = dp[i - 1][m][n]$ .
else: $dp[i][m][n] = max(dp[i - 1][m][n], 1 +dp[i-1][m-count0][n-count1])$ .

Simple implementation: $O(M\times N\times ArrayLength)$ space complexity

int dp[601][101][101] = { 0 };
int findMaxForm(vector<string>& strs, int m, int n)
{
    int N = strs.size();
    for (int i = 1; i <= N; ++i)
    {
        int count0 = 0, count1 = 0;
        for (int l = 0; l < strs[i - 1].length(); ++l)
        {
            if (strs[i - 1][l] == '0') ++count0;
            else ++count1;
        }
        for (int n0 = 0; n0 <= m; ++n0)
        {
            for (int n1 = 0; n1 <= n; ++n1)
            {
                if (count0 > n0 || count1 > n1)
                    dp[i][n0][n1] = dp[i - 1][n0][n1];
                else
                    dp[i][n0][n1] = max(dp[i - 1][n0][n1], dp[i - 1][n0 - count0][n1 - count1] + 1);
            }
        }
    }
    return dp[N][m][n];
}

Little optimization: $O(M\times N)$ space

int findMaxForm(vector<string>& strs, int m, int n)
{
    vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
    for (auto s : strs)
    {
        int count0 = 0, count1 = 0;
        for (auto c : s)
            c == '0' ? ++count0 : ++count1;
        for (int n0 = m; n0 >= count0; --n0)
            for (int n1 = n; n1 >= count1; --n1)
                dp[n0][n1] = max(dp[n0][n1], dp[n0 - count0][n1 - count1] + 1);
    }
    return dp[m][n];
}