LeetCode in Kotlin
3955. Valid Binary Strings With Cost Limit
Medium
You are given two integers n and k.
The cost of a binary string s is defined as the sum of all indices i (0-based) such that s[i] == '1'.
A binary string is considered valid if:
- It does not contain two consecutive
'1'characters. - Its cost is less than or equal to
k.
Return a list of all valid binary strings of length n in any order.
Example 1:
Input: n = 3, k = 1
Output: [“000”,”010”,”100”]
Explanation:
The binary strings of length 3 without consecutive '1' characters are:
"000":cost = 0- “
100":cost = 0 "010":cost = 1"001":cost = 2"101":cost = 0 + 2 = 2
Among these, the strings with cost less than or equal to k = 1 are "000", "010" and "100".
Thus, the valid strings are ["000", "010", "100"].
Example 2:
Input: n = 1, k = 0
Output: [“0”,”1”]
Explanation:
The valid binary strings of length 1 are "0" and "1".
Thus the answer is ["0", "1"].
Constraints:
1 <= n <= 120 <= k <= n * (n - 1) / 2
Solution
class Solution {
private lateinit var result: MutableList<String>
private var n = 0
private var k = 0
fun generateValidStrings(n: Int, k: Int): MutableList<String> {
this.result = ArrayList<String>()
this.n = n
this.k = k
val path = CharArray(n)
dfs(0, 0, path)
return result
}
private fun dfs(index: Int, cost: Int, path: CharArray) {
if (index == n) {
result.add(String(path))
return
}
path[index] = '0'
dfs(index + 1, cost, path)
val canPlaceOne = (index == 0 || path[index - 1] != '1') && (cost + index <= k)
if (canPlaceOne) {
path[index] = '1'
dfs(index + 1, cost + index, path)
}
}
}