leetcode - 1963. Minimum Number of Swaps to Make the String Balanced

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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 <= 10^6
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.

Solution

Solved after hints…

Intuitive

When there’s an imbalanced ], use the rightest [ to swap.

Time complexity: $o (n)$
Space complexity: $o (1)$

Greedy

Keep track of the maximum imbalance, to solve this imbalance, we need to do (im+1)//2 swaps.

Time complexity: $o (n)$
Space complexity: $o (1)$

Code

Intuitive

class Solution:def minSwaps(self, s: str) -> int:open_to_swap = len(s) - 1while open_to_swap >= 0 and s[open_to_swap] == ']':open_to_swap -= 1cnter = 0res = 0i = 0while i < open_to_swap:if s[i] == '[':cnter += 1else:cnter -= 1if cnter < 0:res += 1open_to_swap -= 1while open_to_swap >= i and s[open_to_swap] == ']':open_to_swap -= 1cnter += 2i += 1return res

Greedy

class Solution:def minSwaps(self, s: str) -> int:cnter = 0max_imbalance = 0for each_c in s:if each_c == ']':cnter -= 1else:cnter += 1max_imbalance = min(max_imbalance, cnter)return (-max_imbalance + 1) // 2