Week 10

474. Ones and Zeroes

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".

my solution:

#include<vector>
#include<iostream>
#include<algorithm>
using namespace std;
class Solution {
public:
  int findMaxForm(vector<string>& strs, int m, int n) {     
    vector<int> list;
    bool flag = true;
    for (vector<string>::iterator i = strs.begin(); i != strs.end(); i++) {
      int count0 = 0, count1 = 0;
      for (auto p : *i) { if (p == '0') count0++; else count1++; }
      if ((m >= count0 && n >= count1)) {
        flag = false;
        vector<string> strs_cp = strs;
        strs_cp.erase(strs_cp.begin() + (i - strs.begin()));
        list.push_back(1 + findMaxForm(strs_cp, m - count0, n - ((*i).length() - count0)));
      }
    }
    if (!flag) return *max_element(list.begin(), list.end()); else return 0;
  }
};

在自己做这道题的过程中我使用的是递归的想法，能做出正确的答案，但是提交到leetcode显示Time limit exceeded。递归的做法虽然能做但是并没有利用到相同的子结构来降低复杂度。于是改出了一个非递归的做法。

#include<vector>
#include<iostream>
#include<algorithm>
using namespace std;
class Solution {
public:
  int findMaxForm(vector<string>& strs, int m, int n) {     
    vector<vector<int>> memory(m + 1, vector<int>(n + 1, 0));
    for (auto &s : strs) {
      int count0 = 0, count1 = 0;
      for (auto p : s) { if (p == '0') count0++; else count1++; }
      for (int i = m; i >= count0; i--) {
        for (int j = n; j >= count1; j--) {
          memory[i][j] = max(memory[i][j], memory[i - count0][j - count1] + 1);
        }
      }
    }
    return memory[m][n];
  }
};

用一个表来记录每组成一个字符串之后的状态。