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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.
1function minSwaps(s) {
2 let balance = 0, swaps = 0;
3
4 for (const ch of s) {
5 balance += (ch === '[') ? 1 : -1;
6
7 if (balance < 0) {
8 swaps++;
9 balance += 2;
10 }
11 }
12 return swaps;
13}
14
15const s = '][[';
16console.log('Minimum swaps:', minSwaps(s));In this JavaScript solution, we iterate through the string characters, adjusting the balance counter and counting swaps when needed to balance the brackets.
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.
Two pointers are initialized at the start and end of the string. The inner mismatched brackets are swapped until the string becomes balanced. The solution ensures that excessive closing brackets found earlier are balanced by swapping them with opening brackets found later in the string.