LeetCode in Kotlin
474. Ones and Zeroes
Medium
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
Solution
class Solution {
/*
* 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] as it 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.
*/
fun findMaxForm(strs: Array<String>, m: Int, n: Int): Int {
val dp = Array(m + 1) { IntArray(n + 1) }
for (str in strs) {
val count = count(str)
for (i in m downTo count[0]) {
for (j in n downTo count[1]) {
dp[i][j] = Math.max(dp[i][j], dp[i - count[0]][j - count[1]] + 1)
}
}
}
return dp[m][n]
}
private fun count(str: String): IntArray {
val res = IntArray(2)
for (i in 0 until str.length) {
res[str[i].code - '0'.code]++
}
return res
}
}