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
and1s
will both not exceed100
- 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