474. Ones and Zeroes

最新推荐文章于 2025-09-06 18:22:26 发布

转载最新推荐文章于 2025-09-06 18:22:26 发布 · 125 阅读

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原文链接：http://www.cnblogs.com/ruruozhenhao/p/10486251.html

文章标签：

#c/c++

本文介绍了一个计算机科学中的动态规划问题，即如何利用有限的0和1字符资源，从一组仅由0和1组成的字符串中形成尽可能多的新字符串。通过详细解析一个C++动态规划解决方案，文章展示了如何在不超过给定数量的0和1的情况下，找到可以形成的最大字符串数目。

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 0sand 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".

Approach #1: DP. [C++]

class Solution {
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        vector<vector<int>> memo(m+1, vector<int>(n+1, 0));
        int countOfZeros, countOfOnes;
        for (auto& s : strs) {
            countOfZeros = 0, countOfOnes = 0;
            for (auto c : s) {
                if (c == '0') countOfZeros++;
                else if (c == '1') countOfOnes++;
            }
            
            for (int i = m; i >= countOfZeros; --i) {
                for (int j = n; j >= countOfOnes; --j) {
                    memo[i][j] = max(memo[i][j], memo[i-countOfZeros][j-countOfOnes] + 1);
                }
            }
        }
        return memo[m][n];
    }
};

Analysis:

memo[i][j] represent the max number of strings that can be formed with i 0's and j 1's.

from the first few strings up to the current string s

Catch: have to go from bottom right to top left

If we go from top left to bottom right, we would be using results from this iteration => overcounting

Reference:

https://leetcode.com/problems/ones-and-zeroes/discuss/95814/c%2B%2B-DP-solution-with-comments

转载于:https://www.cnblogs.com/ruruozhenhao/p/10486251.html