Welcome to Subscribe On Youtube
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; };