LeetCode in Kotlin
1888. Minimum Number of Flips to Make the Binary String Alternating
Medium
You are given a binary string s. You are allowed to perform two types of operations on the string in any sequence:
- Type-1: Remove the character at the start of the string
sand append it to the end of the string. - Type-2: Pick any character in
sand flip its value, i.e., if its value is'0'it becomes'1'and vice-versa.
Return the minimum number of type-2 operations you need to perform such that s becomes alternating.
The string is called alternating if no two adjacent characters are equal.
- For example, the strings
"010"and"1010"are alternating, while the string"0100"is not.
Example 1:
Input: s = “111000”
Output: 2
Explanation: Use the first operation two times to make s = “100011”.
Then, use the second operation on the third and sixth elements to make s = “101010”.
Example 2:
Input: s = “010”
Output: 0
Explanation: The string is already alternating.
Example 3:
Input: s = “1110”
Output: 1
Explanation: Use the second operation on the second element to make s = “1010”.
Constraints:
1 <= s.length <= 105s[i]is either'0'or'1'.
Solution
class Solution {
fun minFlips(s: String): Int {
val n = s.length
val localStr = s + s
val t = localStr.toCharArray()
val a = CharArray(n + n)
val b = CharArray(n + n)
for (i in 0 until n + n) {
if (i % 2 == 0) {
a[i] = '1'
b[i] = '0'
} else {
a[i] = '0'
b[i] = '1'
}
}
var f = 0
var sec = 0
var ans = Int.MAX_VALUE
for (i in 0 until n + n) {
if (a[i] != t[i]) {
f++
}
if (b[i] != t[i]) {
sec++
}
if (i >= n) {
if (a[i - n] != t[i - n]) {
f--
}
if (b[i - n] != t[i - n]) {
sec--
}
}
if (i >= n - 1) {
ans = Math.min(ans, Math.min(f, sec))
}
}
return ans
}
}