LeetCode-in-Java
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
public class Solution {
public int minFlips(String s) {
int n = s.length();
String localStr = s + s;
char[] t = localStr.toCharArray();
char[] a = new char[n + n];
char[] b = new char[n + n];
for (int i = 0; i < n + n; i++) {
if (i % 2 == 0) {
a[i] = '1';
b[i] = '0';
} else {
a[i] = '0';
b[i] = '1';
}
}
int f = 0;
int sec = 0;
int ans = Integer.MAX_VALUE;
for (int i = 0; i < n + n; i++) {
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;
}
}