# 1864. Minimum Number of Swaps to Make the Binary String Alternating

## Description

Given a binary string s, return the minimum number of character swaps to make it alternating, or -1 if it is impossible.

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.

Any two characters may be swapped, even if they are not adjacent.

Example 1:

Input: s = "111000"
Output: 1
Explanation: Swap positions 1 and 4: "111000" -> "101010"
The string is now alternating.


Example 2:

Input: s = "010"
Output: 0
Explanation: The string is already alternating, no swaps are needed.


Example 3:

Input: s = "1110"
Output: -1


Constraints:

• 1 <= s.length <= 1000
• s[i] is either '0' or '1'.

## Solutions

• class Solution {
public int minSwaps(String s) {
int s0n0 = 0, s0n1 = 0;
int s1n0 = 0, s1n1 = 0;
for (int i = 0; i < s.length(); ++i) {
if ((i & 1) == 0) {
if (s.charAt(i) != '0') {
s0n0 += 1;
} else {
s1n1 += 1;
}
} else {
if (s.charAt(i) != '0') {
s1n0 += 1;
} else {
s0n1 += 1;
}
}
}
if (s0n0 != s0n1 && s1n0 != s1n1) {
return -1;
}
if (s0n0 != s0n1) {
return s1n0;
}
if (s1n0 != s1n1) {
return s0n0;
}
return Math.min(s0n0, s1n0);
}
}

• class Solution:
def minSwaps(self, s: str) -> int:
s0n0 = s0n1 = s1n0 = s1n1 = 0
for i in range(len(s)):
if (i & 1) == 0:
if s[i] != '0':
s0n0 += 1
else:
s1n1 += 1
else:
if s[i] != '0':
s1n0 += 1
else:
s0n1 += 1
if s0n0 != s0n1 and s1n0 != s1n1:
return -1
if s0n0 != s0n1:
return s1n0
if s1n0 != s1n1:
return s0n0
return min(s0n0, s1n0)


• /**
* @param {string} s
* @return {number}
*/
var minSwaps = function (s) {
let n = s.length;
let n1 = [...s].reduce((a, c) => parseInt(c) + a, 0);
let n0 = n - n1;
let count = Infinity;
let half = n / 2;
// 101、1010
if (n1 == Math.ceil(half) && n0 == Math.floor(half)) {
let cur = 0;
for (let i = 0; i < n; i++) {
if (i % 2 == 0 && s.charAt(i) != '1') cur++;
}
count = Math.min(count, cur);
}
// 010、0101
if (n0 == Math.ceil(half) && n1 == Math.floor(half)) {
let cur = 0;
for (let i = 0; i < n; i++) {
if (i % 2 == 0 && s.charAt(i) != '0') cur++;
}
count = Math.min(count, cur);
}
return count == Infinity ? -1 : count;
};