How do you solve Minimum Operations to Equalize Binary String on LeetCode?

Minimum Operations to Equalize Binary String (LeetCode #3666) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the parity of zeros after flips is crucial.

What is the time complexity of Minimum Operations to Equalize Binary String?

The optimal solution for Minimum Operations to Equalize Binary String runs in O(n) time and uses O(n) space. The BFS explores states efficiently, leading to linear time complexity in terms of the number of zeros.

What is the brute force solution for Minimum Operations to Equalize Binary String?

The brute force approach for Minimum Operations to Equalize Binary String is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try all possible combinations of flipping k indices to see how many operations it takes to convert all '0's to '1's. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Minimum Operations to Equalize Binary String?

Explain your thought process clearly. Discuss edge cases upfront. Practice BFS and state exploration problems.

What are common mistakes when solving Minimum Operations to Equalize Binary String?

Not considering the parity of zeros. Assuming all configurations can be reached.

#3666

Minimum Operations to Equalize Binary String

Hard

MathStringBreadth-First SearchUnion-FindOrdered SetBFSState Space Exploration

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Instead of brute-forcing through combinations, we can model the number of zeros and track their changes through valid operations using BFS.

⚙️

Algorithm

3 steps

1Step 1: Count the number of zeros in the string.
2Step 2: Use BFS to explore how many operations are needed to reach zero zeros, considering the parity of zeros after each operation.
3Step 3: Return the minimum operations or -1 if not possible.

solution.py17 lines

1from collections import deque
2
3def min_operations(s, k):
4    z = s.count('0')
5    n = len(s)
6    queue = deque([(z, 0)])
7    visited = set([z])
8    while queue:
9        curr_z, ops = queue.popleft()
10        if curr_z == 0:
11            return ops
12        for i in range(max(0, k - (n - curr_z)), min(k, curr_z) + 1):
13            next_z = curr_z + k - 2 * i
14            if next_z not in visited:
15                visited.add(next_z)
16                queue.append((next_z, ops + 1))
17    return -1

ℹ

Complexity note: The BFS explores states efficiently, leading to linear time complexity in terms of the number of zeros.

1Understanding the parity of zeros after flips is crucial.
2BFS allows us to explore states efficiently.

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