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 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:

1. The given numbers of 0s and 1s will both not exceed 100
2. 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