Table of Contents
Ones and Zeroes – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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 <= 6001 <= strs[i].length <= 100strs[i]consists only of digits'0'and'1'.1 <= m, n <= 100
C++ Ones and Zeroes LeetCode Solution
int findMaxForm(vector<string>& strs, int m, int n) {
vector<vector<int>> memo(m+1, vector<int>(n+1, 0));
int numZeroes, numOnes;
for (auto &s : strs) {
numZeroes = numOnes = 0;
// count number of zeroes and ones in current string
for (auto c : s) {
if (c == '0')
numZeroes++;
else if (c == '1')
numOnes++;
}
// memo[i][j] = 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
// Why? If a cell in the memo is updated(because s is selected),
// we should be adding 1 to memo[i][j] from the previous iteration (when we were not considering s)
// If we go from top left to bottom right, we would be using results from this iteration => overcounting
for (int i = m; i >= numZeroes; i--) {
for (int j = n; j >= numOnes; j--) {
memo[i][j] = max(memo[i][j], memo[i - numZeroes][j - numOnes] + 1);
}
}
}
return memo[m][n];
}
Java Ones and Zeroes LeetCode Solution
public int findMaxForm(String[] strs, int m, int n) {
int[][] dp = new int[m+1][n+1];
for (String s : strs) {
int[] count = count(s);
for (int i=m;i>=count[0];i--)
for (int j=n;j>=count[1];j--)
dp[i][j] = Math.max(1 + dp[i-count[0]][j-count[1]], dp[i][j]);
}
return dp[m][n];
}
public int[] count(String str) {
int[] res = new int[2];
for (int i=0;i<str.length();i++)
res[str.charAt(i) - '0']++;
return res;
}
Python 3 Ones and Zeroes LeetCode Solution
class Solution:
def findMaxForm(self, S: List[str], M: int, N: int) -> int:
dp = [[0 for _ in range(N+1)] for _ in range(M+1)]
for str in S:
zeros = str.count("0")
ones = len(str) - zeros
for i in range(M, zeros - 1, -1):
for j in range(N, ones - 1, -1):
dp[i][j] = max(dp[i][j], dp[i-zeros][j-ones] + 1)
return dp[M][N]
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