LeetCode in Kotlin
1869. Longer Contiguous Segments of Ones than Zeros
Easy
Given a binary string s, return true if the longest contiguous segment of 1‘s is strictly longer than the longest contiguous segment of 0‘s in s, or return false otherwise.
- For example, in
s = "110100010"the longest continuous segment of1s has length2, and the longest continuous segment of0s has length3.
Note that if there are no 0’s, then the longest continuous segment of 0’s is considered to have a length 0. The same applies if there is no 1’s.
Example 1:
Input: s = “1101”
Output: true
Explanation:
The longest contiguous segment of 1s has length 2: “1101”
The longest contiguous segment of 0s has length 1: “1101”
The segment of 1s is longer, so return true.
Example 2:
Input: s = “111000”
Output: false
Explanation:
The longest contiguous segment of 1s has length 3: “111000”
The longest contiguous segment of 0s has length 3: “111000”
The segment of 1s is not longer, so return false.
Example 3:
Input: s = “110100010”
Output: false
Explanation:
The longest contiguous segment of 1s has length 2: “110100010”
The longest contiguous segment of 0s has length 3: “110100010”
The segment of 1s is not longer, so return false.
Constraints:
1 <= s.length <= 100s[i]is either'0'or'1'.
Solution
class Solution {
fun checkZeroOnes(s: String): Boolean {
var zeroes = 0
var ones = 0
var i = 0
while (i < s.length) {
var start = i
while (i < s.length && s[i] == '0') {
i++
}
if (i > start) {
zeroes = Math.max(zeroes, i - start)
}
start = i
while (i < s.length && s[i] == '1') {
i++
}
if (i > start) {
ones = Math.max(ones, i - start)
}
}
return ones > zeroes
}
}