LeetCode in Kotlin
3922. Minimum Flips to Make Binary String Coherent
Medium
You are given a binary string s.
A string is considered coherent if it does not contain "011" or "110" as subsequences.
In one operation, you can flip any character in s ('0' to '1' or '1' to '0').
Return an integer denoting the minimum number of operations required to make s coherent.
Example 1:
Input: s = “1010”
Output: 1
Explanation:
Flip s[0] to get "0010", which contains no "011" or "110" subsequences.
Example 2:
Input: s = “0110”
Output: 1
Explanation:
Flip s[1] to get "0010", removing all forbidden subsequences "011" and "110".
Example 3:
Input: s = “1000”
Output: 0
Explanation:
The string already has no "011" or "110" subsequences, so no flips are needed.
Constraints:
1 <= s.length <= 105s[i]is either'0'or'1'.
Solution
import kotlin.math.min
class Solution {
fun minFlips(s: String): Int {
val n = s.length
var count1 = 0
for (c in s.toCharArray()) {
count1 += c.code - '0'.code
}
val count0 = n - count1
// Option 1: all 0s
var ans = count1
// Option 2: all 1s
ans = min(ans, count0)
// Option 3: exactly one 1 (at any position)
ans = if (count1 >= 1) {
min(ans, count1 - 1)
} else {
min(ans, 1)
}
// Option 4: "10...01" (1 at position 0 and position n-1)
if (n >= 2 && count1 >= 2) {
var cost4 = count1
if (s[0] == '1') {
cost4--
}
if (s[n - 1] == '1') {
cost4--
}
ans = min(ans, cost4)
}
return ans
}
}