LeetCode in Kotlin
3234. Count the Number of Substrings With Dominant Ones
Medium
You are given a binary string s.
Return the number of substrings with dominant ones.
A string has dominant ones if the number of ones in the string is greater than or equal to the square of the number of zeros in the string.
Example 1:
Input: s = “00011”
Output: 5
Explanation:
The substrings with dominant ones are shown in the table below.
| i | j | s[i..j] | Number of Zeros | Number of Ones |
|---|---|---|---|---|
| 3 | 3 | 1 | 0 | 1 |
| 4 | 4 | 1 | 0 | 1 |
| 2 | 3 | 01 | 1 | 1 |
| 3 | 4 | 11 | 0 | 2 |
| 2 | 4 | 011 | 1 | 2 |
Example 2:
Input: s = “101101”
Output: 16
Explanation:
The substrings with non-dominant ones are shown in the table below.
Since there are 21 substrings total and 5 of them have non-dominant ones, it follows that there are 16 substrings with dominant ones.
| i | j | s[i..j] | Number of Zeros | Number of Ones |
|---|---|---|---|---|
| 1 | 1 | 0 | 1 | 0 |
| 4 | 4 | 0 | 1 | 0 |
| 1 | 4 | 0110 | 2 | 2 |
| 0 | 4 | 10110 | 2 | 3 |
| 1 | 5 | 01101 | 2 | 3 |
Constraints:
1 <= s.length <= 4 * 104sconsists only of characters'0'and'1'.
Solution
import kotlin.math.min
class Solution {
fun numberOfSubstrings(s: String): Int {
val zero: MutableList<Int> = ArrayList()
zero.add(-1)
var result = 0
for (i in s.indices) {
if (s[i] == '0') {
zero.add(i)
} else {
result += i - zero[zero.size - 1]
}
for (j in 1 until zero.size) {
val len = j * (j + 1)
if (len > i + 1) {
break
}
val prev = zero[zero.size - j - 1]
val from = min((i - len + 1), zero[zero.size - j])
if (from > prev) {
result += from - prev
}
}
}
return result
}
}