Separate Black and White Balls

Separate Black and White Balls - Problem

There are n balls on a table, each ball has a color black or white. You are given a 0-indexed binary string s of length n, where 1 and 0 represent black and white balls, respectively.

In each step, you can choose two adjacent balls and swap them. Return the minimum number of steps to group all the black balls to the right and all the white balls to the left.

Input & Output

Example 1 — Basic Mixed Sequence

$ Input: s = "101"

› Output: 1

💡 Note: We can group all black balls to the right by swapping s[0] and s[1]: "101" → "011". This takes 1 step.

Example 2 — Multiple Swaps Needed

$ Input: s = "0110"

› Output: 2

💡 Note: We can group all black balls to the right in 2 steps: "0110" → "1010" → "1100". The white ball at index 0 needs 2 swaps to move past both black balls.

Example 3 — Already Sorted

$ Input: s = "0011"

› Output: 0

💡 Note: All white balls are already on the left and black balls on the right. No swaps needed.

Constraints

1 ≤ s.length ≤ 10⁵
s[i] is either '0' or '1'

Visualization

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Asked in

M Meta 25 G Google 20 μ Microsoft 15

The key insight is that each white ball needs to move past all black balls to its right. The optimal greedy approach scans right to left, counting black balls and adding this count to the total for each white ball encountered. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force Simulation

⏱️ Time: O(n³) Space: O(n)

Keep swapping adjacent balls until all white balls are on the left and all black balls are on the right. Continue until no more swaps are needed.

Greedy Counting

⏱️ Time: O(n) Space: O(1)

Scan from right to left, counting black balls. For each white ball encountered, it needs to move past all the black balls we've seen so far.

Brute Force Simulation — Algorithm Steps

Step 1: Scan for any white ball that has black balls to its right
Step 2: Swap the white ball with its right neighbor if it's black
Step 3: Repeat until all balls are in correct positions

Visualization

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Step-by-Step Walkthrough

Initial State

Mixed black and white balls

Scan & Swap

Find white balls followed by black balls and swap them

Repeat

Continue until all balls are sorted

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(char* s) {
    int n = strlen(s);
    int steps = 0;
    int blackCount = 0;
    
    // Count how many black balls each white ball needs to pass
    for (int i = 0; i < n; i++) {
        if (s[i] == '1') {
            blackCount++;
        } else { // s[i] == '0'
            steps += blackCount;
        }
    }
    
    return steps;
}

int main() {
    char s[100001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    int result = solution(s);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

In worst case, need O(n²) swaps and each scan takes O(n) time

⚠ Quadratic Growth

Space Complexity

O(n)

Need to store the string as mutable array for swapping

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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Ln 1, Col 1

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Simulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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