Ones and Zeroes
Given an array, strs
, with strings consisting of only 0s
and 1s
. Also two integers m
and n
.
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.
Example 1:
Input: strs = ["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: strs = ["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".
Constraints:
1 <= strs.length <= 600
1 <= strs[i].length <= 100
strs[i]
consists only of digits '0' and '1'.1 <= m, n <= 100
/*
The problem can be interpreted as: What's the max number of str can we pick from strs with limitation of m "0"s and n "1"s.
Thus we can define dp[i][j] stands for max number of str can we pick from strs with limitation of i "0"s and j "1"s.
For each str, assume it has a "0"s and b "1"s, we update the dp array iteratively and set dp[i][j] = Math.max(dp[i][j], dp[i - a][j - b] + 1). So at the end, dp[m][n] is the answer.
*/
class Solution {
public int findMaxForm(String[] strs, int m, int n) {
int[][] list = new int[strs.length][];
int index = 0;
for (String s : strs) {
int zeros = 0, ones = 0;
for (char c : s.toCharArray())
if (c == '0')
zeros++;
else
ones++;
list[index++] = new int[] { zeros, ones };
}
int[][] dp = new int[m + 1][n + 1];
for (int[] x : list) {
for (int i = m; i - x[0] >= 0; i--)
for (int j = n; j - x[1] >= 0; j--)
dp[i][j] = Math.max(1 + dp[i - x[0]][j - x[1]], dp[i][j]);
}
return dp[m][n];
}
}
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