LeetCode in Kotlin
1573. Number of Ways to Split a String
Medium
Given a binary string s, you can split s into 3 non-empty strings s1, s2, and s3 where s1 + s2 + s3 = s.
Return the number of ways s can be split such that the number of ones is the same in s1, s2, and s3. Since the answer may be too large, return it modulo 109 + 7.
Example 1:
Input: s = “10101”
Output: 4
Explanation: There are four ways to split s in 3 parts where each part contain the same number of letters ‘1’.
“1|010|1”
“1|01|01”
“10|10|1”
“10|1|01”
Example 2:
Input: s = “1001”
Output: 0
Example 3:
Input: s = “0000”
Output: 3
Explanation: There are three ways to split s in 3 parts.
“0|0|00”
“0|00|0”
“00|0|0”
Constraints:
3 <= s.length <= 105s[i]is either'0'or'1'.
Solution
class Solution {
fun numWays(s: String): Int {
var totalOnesCount: Long = 0
val mod: Long = 1000000007
var waysOfFirstString: Long = 0
var waysOfSecondString: Long = 0
var onesCount: Long = 0
val n = s.length.toLong()
for (i in 0 until s.length) {
if (s[i] == '1') {
totalOnesCount += 1
}
}
if (totalOnesCount % 3 != 0L) {
return 0
}
val onesFirstPart = totalOnesCount / 3
val onesSecondPart = onesFirstPart * 2
if (totalOnesCount == 0L) {
return ((n - 1) * (n - 2) / 2 % mod).toInt()
}
for (i in 0 until s.length) {
if (s[i] == '1') {
onesCount += 1
}
if (onesCount == onesFirstPart) {
waysOfFirstString += 1
} else if (onesCount == onesSecondPart) {
waysOfSecondString += 1
} else if (onesCount > onesSecondPart) {
break
}
}
return (waysOfFirstString * waysOfSecondString % mod).toInt()
}
}