Description
You are given an array of binary strings strs
and two integers m
and n
.
Return the size of the largest subset of strs
such that there are at most m
0
’s and n
1
’s in the subset.
A set x
is a subset of a set y
if all elements of x
are also elements of y
.
Example 1:
Input: strs = [“10”,“0001”,“111001”,“1”,“0”], m = 5, n = 3 Output: 4 Explanation: The largest subset with at most 5 0’s and 3 1’s is {“10”, “0001”, “1”, “0”}, so the answer is 4. Other valid but smaller subsets include {“0001”, “1”} and {“10”, “1”, “0”}. {“111001”} is an invalid subset because it contains 4 1’s, greater than the maximum of 3.
Example 2:
Input: strs = [“10”,“0”,“1”], m = 1, n = 1 Output: 2 Explanation: The largest subset is {“0”, “1”}, so the answer is 2.
Constraints:
1 <= strs.length <= 600
1 <= strs[i].length <= 100
strs[i]
consists only of digits'0'
and'1'
.1 <= m, n <= 100
Code
和 Partition Equal Subset Sum 一樣是 knapsack 類型的題目。不過這題是一個 3D 的 DP。
Time Complexity: , Space Complexity:
class Solution {
public:
int findMaxForm(vector<string>& strs, int m, int n) {
int k = strs.size();
int dp[k + 1][m + 1][n + 1];
memset(dp, 0, sizeof(dp));
for(int i = 1; i < k + 1; i++) {
auto count = convert(strs[i - 1]);
auto zeros = count[0];
auto ones = count[1];
for(int j = 0; j < m + 1; j++) {
for(int l = 0; l < n + 1; l++) {
int res = dp[i - 1][j][l];
if((j >= zeros) && (l >= ones)) {
res = max(res, dp[i - 1][j - zeros][l - ones] + 1);
}
dp[i][j][l] = res;
}
}
}
return dp[k][m][n];
}
vector<int> convert(string s) {
int zero = 0, one = 0;
for(auto c: s) {
if(c == '0') {
zero++;
} else if (c == '1') {
one++;
}
}
return {zero, one};
}
};