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1963. Minimum Number of Swaps to Make the String Balanced

Description

You are given a 0-indexed string s of even length n. The string consists of exactly n / 2 opening brackets '[' and n / 2 closing brackets ']'.

A string is called balanced if and only if:

  • It is the empty string, or
  • It can be written as AB, where both A and B are balanced strings, or
  • It can be written as [C], where C is a balanced string.

You may swap the brackets at any two indices any number of times.

Return the minimum number of swaps to make s balanced.

 

Example 1:

Input: s = "][]["
Output: 1
Explanation: You can make the string balanced by swapping index 0 with index 3.
The resulting string is "[[]]".

Example 2:

Input: s = "]]][[["
Output: 2
Explanation: You can do the following to make the string balanced:
- Swap index 0 with index 4. s = "[]][][".
- Swap index 1 with index 5. s = "[[][]]".
The resulting string is "[[][]]".

Example 3:

Input: s = "[]"
Output: 0
Explanation: The string is already balanced.

 

Constraints:

  • n == s.length
  • 2 <= n <= 106
  • n is even.
  • s[i] is either '[' or ']'.
  • The number of opening brackets '[' equals n / 2, and the number of closing brackets ']' equals n / 2.

Solutions

Solution 1: Greedy

We use a variable \(x\) to record the current number of unmatched left brackets. We traverse the string \(s\), for each character \(c\):

  • If \(c\) is a left bracket, then we increment \(x\) by one;
  • If \(c\) is a right bracket, then we need to check whether \(x\) is greater than zero. If it is, we match the current right bracket with the nearest unmatched left bracket on the left, i.e., decrement \(x\) by one.

After the traversal, we will definitely get a string of the form "]]]...[[[...". We then greedily swap the brackets at both ends each time, which can eliminate \(2\) unmatched left brackets at a time. Therefore, the total number of swaps needed is \(\left\lfloor \frac{x + 1}{2} \right\rfloor\).

The time complexity is \(O(n)\), where \(n\) is the length of the string \(s\). The space complexity is \(O(1)\).

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class Solution:
    def minSwaps(self, s: str) -> int:
        x = 0
        for c in s:
            if c == "[":
                x += 1
            elif x:
                x -= 1
        return (x + 1) >> 1

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class Solution {
    public int minSwaps(String s) {
        int x = 0;
        for (int i = 0; i < s.length(); ++i) {
            char c = s.charAt(i);
            if (c == '[') {
                ++x;
            } else if (x > 0) {
                --x;
            }
        }
        return (x + 1) / 2;
    }
}

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class Solution {
public:
    int minSwaps(string s) {
        int x = 0;
        for (char& c : s) {
            if (c == '[') {
                ++x;
            } else if (x) {
                --x;
            }
        }
        return (x + 1) / 2;
    }
};

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func minSwaps(s string) int {
    x := 0
    for _, c := range s {
        if c == '[' {
            x++
        } else if x > 0 {
            x--
        }
    }
    return (x + 1) / 2
}

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function minSwaps(s: string): number {
    let x = 0;
    for (const c of s) {
        if (c === '[') {
            ++x;
        } else if (x) {
            --x;
        }
    }
    return (x + 1) >> 1;
}

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