What is the time complexity of Minimum Number of Swaps to Make the String Balanced?

The optimal solution for Minimum Number of Swaps to Make the String Balanced runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only traverse the string once, and the space complexity is O(1) since we only use a few variables for counting.

What is the brute force solution for Minimum Number of Swaps to Make the String Balanced?

The brute force approach for Minimum Number of Swaps to Make the String Balanced is Brute Force, with O(n²) time complexity and O(n!) space complexity. The brute force approach attempts to generate all possible configurations of the string by swapping brackets and checks if the resulting string is balanced. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Swaps to Make the String Balanced?

Minimum Number of Swaps to Make the String Balanced uses the following concepts: Two Pointers, String, Stack, Greedy. The recommended patterns to study are: Greedy, Two Pointers.

What are the interview tips for Minimum Number of Swaps to Make the String Balanced?

Explain your thought process clearly as you work through the problem. Be prepared to discuss the time and space complexity of your solution. Practice writing clean and efficient code, as this is often evaluated in interviews.

What are common mistakes when solving Minimum Number of Swaps to Make the String Balanced?

Not resetting the balance after counting a swap. Confusing the number of swaps needed with the number of misplaced brackets.

#1963

Minimum Number of Swaps to Make the String Balanced

Medium

Two PointersStringStackGreedyGreedyTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n!)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a greedy approach to count the number of misplaced closing brackets as we traverse the string. This allows us to calculate the minimum swaps needed without generating permutations.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to track the balance of brackets.
2Step 2: Traverse the string and update the balance for each character: +1 for '[' and -1 for ']'.
3Step 3: Whenever the balance goes negative, it indicates a misplaced closing bracket, so increment the swap count.
4Step 4: The number of swaps needed is equal to the number of times the balance went negative divided by 2.

solution.py15 lines

1# Full working Python code
2
3def min_swaps_optimal(s):
4    balance = 0
5    swaps = 0
6    for char in s:
7        if char == '[':
8            balance += 1
9        else:
10            balance -= 1
11        if balance < 0:
12            swaps += 1
13            balance = 0  # Reset balance after counting a swap
14    return swaps
15

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once, and the space complexity is O(1) since we only use a few variables for counting.

1The balance of brackets can be tracked to determine how many swaps are needed.
2Each time the balance goes negative, it indicates a misplaced closing bracket.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.