How do you solve Separate Black and White Balls on LeetCode?

Separate Black and White Balls (LeetCode #2938) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Swapping adjacent elements can be inefficient; counting jumps is often faster.

What is the time complexity of Separate Black and White Balls?

The optimal solution for Separate Black and White Balls runs in O(n) time and uses O(1) space. This approach is linear because we only make a single pass through the string, counting '0's and calculating steps simultaneously.

What is the brute force solution for Separate Black and White Balls?

The brute force approach for Separate Black and White Balls is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves repeatedly swapping adjacent balls until all black balls are grouped to the right. This is simple to understand but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Separate Black and White Balls?

Separate Black and White Balls uses the following concepts: Two Pointers, String, Greedy. The recommended patterns to study are: Two Pointers, Greedy Algorithms.

What are the interview tips for Separate Black and White Balls?

Always consider edge cases, such as strings that are already sorted. Think about how you can reduce the number of operations needed. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Separate Black and White Balls?

Not realizing that multiple swaps can be avoided by counting instead. Overcomplicating the problem by trying to simulate each swap.

#2938

Separate Black and White Balls

Medium

Two PointersStringGreedyTwo PointersGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Instead of swapping, we can count how many '0's each '1' has to jump over to reach its final position. This allows us to calculate the minimum swaps needed in a single pass.

⚙️

Algorithm

3 steps

1Step 1: Initialize a counter for the number of '0's encountered as we traverse the string.
2Step 2: For each '1' encountered, add the current count of '0's to the total steps.
3Step 3: Continue until the end of the string.

solution.py11 lines

1def minSwaps(s):
2    count_zeros = 0
3    steps = 0
4    for char in s:
5        if char == '0':
6            count_zeros += 1
7        else:
8            steps += count_zeros
9    return steps
10
11print(minSwaps('101'))

ℹ

Complexity note: This approach is linear because we only make a single pass through the string, counting '0's and calculating steps simultaneously.

1Swapping adjacent elements can be inefficient; counting jumps is often faster.
2Understanding the problem allows you to identify patterns, like counting instead of swapping.

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