How to Solve Separate Black and White Balls

This problem requires minimizing the number of swaps to group black balls to the right in a binary string. By scanning the string with two pointers, you track the number of swaps needed by checking each black ball and swapping it with any white balls to its right. The approach efficiently solves the problem in O(n) time complexity.

Problem Statement

You are given a binary string s of length n, where each character represents a ball on a table: '1' for black and '0' for white. The task is to group all black balls (1s) to the right side with the fewest adjacent swaps.

In each step, you can choose two adjacent balls and swap them. The goal is to determine the minimum number of swaps required to gather all black balls to the right side of the string.

Examples

Example 1

Input: s = "101"

Output: 1

We can group all the black balls to the right in the following way:

• Swap s[0] and s[1], s = "011". Initially, 1s are not grouped together, requiring at least 1 step to group them to the right.

Example 2

Input: s = "100"

Output: 2

We can group all the black balls to the right in the following way:

• Swap s[0] and s[1], s = "010".
• Swap s[1] and s[2], s = "001". It can be proven that the minimum number of steps needed is 2.

Example 3

Input: s = "0111"

Output: 0

All the black balls are already grouped to the right.

Constraints

• 1 <= n == s.length <= 105
• s[i] is either '0' or '1'.

Solution Approach

Two-Pointer Scanning

Use two pointers to scan the string: one pointer starts at the leftmost position, and the other tracks the position of the last encountered white ball. Each time a black ball (1) is encountered, it is swapped with the white ball at the tracked position.

Tracking the Number of Swaps

Count the number of swaps by calculating how far each black ball is from its correct position, then sum these distances. This gives the total number of swaps needed to group all black balls to the right.

Greedy Approach for Minimizing Swaps

A greedy approach ensures that you minimize the number of swaps by moving each black ball only when necessary, and by swapping each black ball to the furthest right possible position.

Complexity Analysis

The time complexity is O(n), where n is the length of the string, because we iterate over the string once. The space complexity is O(1), as we only use a few extra variables for the two pointers and swap counting.

What Interviewers Usually Probe

• Can the candidate demonstrate efficient two-pointer usage for optimizing adjacent swap operations?
• How well does the candidate recognize and utilize the greedy nature of the problem?
• Does the candidate consider edge cases, such as when all black balls are already grouped?

Common Pitfalls or Variants

Common pitfalls

• Failing to track the correct number of swaps or not considering the distance each black ball needs to move.
• Overcomplicating the solution by using nested loops or unnecessary operations.
• Not understanding the two-pointer method and relying on less efficient approaches.

Follow-up variants

• What if the string was larger, close to the upper limit of 100,000? How would the solution scale?
• What if the balls were not just two colors but had more varieties? How would the solution need to adapt?
• What if the problem asked to group black balls to the left instead of the right?

How GhostInterview Helps

• GhostInterview provides step-by-step explanations of how to effectively apply two-pointer scanning with invariant tracking to solve the problem.
• GhostInterview aids in identifying common pitfalls and helps practice spotting inefficiencies in brute force solutions.
• GhostInterview helps refine understanding of greedy algorithms by focusing on minimizing swap steps while using optimal strategies.

Topic Pages

• Two Pointers
• String
• Greedy

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FAQ

What is the optimal solution for the Separate Black and White Balls problem?

The optimal solution uses a two-pointer approach to minimize the number of adjacent swaps needed to group all black balls to the right of the string.

Why is the two-pointer approach used for this problem?

The two-pointer approach efficiently tracks and swaps balls while maintaining a minimal swap count, ensuring a time complexity of O(n).

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

The time complexity is O(n), where n is the length of the string, as the algorithm performs a single scan of the string.

How does GhostInterview help with the Separate Black and White Balls problem?

GhostInterview assists in mastering the two-pointer technique, helping identify pitfalls and practice optimizing the solution with minimal swaps.

What is the key observation for solving the Separate Black and White Balls problem?

The key observation is that every 1 in the string should be swapped with every 0 on its right side, which can be efficiently tracked using two pointers.