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This approach involves traversing the string and counting the balance between opening and closing brackets. You increase a count when you find an opening bracket and decrease the count when you find a closing bracket. If at any point the count goes negative, a swap is needed to balance out the string, and the swap count is incremented. The final swap count will be the required number of swaps to make the string balanced.
Time Complexity: O(n), where n is the length of string s.
Space Complexity: O(1), as no extra space is used aside from a few variables.
1using System;
2
3public class MinSwaps {
4 public static int MinSwapsToBalance(string s) {
5 int balance = 0, swaps = 0;
6
7 foreach (char ch in s) {
8 balance += (ch == '[') ? 1 : -1;
9
10 if (balance < 0) {
11 swaps++;
12 balance += 2;
13 }
14 }
15 return swaps;
16 }
17
18 public static void Main(string[] args) {
19 string s = "][][";
20 Console.WriteLine("Minimum swaps: " + MinSwapsToBalance(s));
21 }
22}The C# solution employs the same logic. By iterating through the string and keeping track of the balance, it identifies the swap points for minimum swaps needed.
This approach uses two pointers to traverse the string efficiently. One pointer starts from the beginning of the string, and the other starts at the end. Using these pointers, swaps are performed when an excessive number of closing brackets on one side can be paired with an opening bracket on the other side.
Time Complexity: O(n), where n is the length of string s, due to traversing the string once.
Space Complexity: O(1), as we're only using constants amount of space for pointers and variables.
In the Java solution, two indices track opening and closing brackets starting from different ends, swap when mismatch occurs, and converge towards the center of the string, ensuring efficient operation.