Time Needed to Rearrange a Binary String

Time Needed to Rearrange a Binary String - Problem

You are given a binary string s. In one second, all occurrences of "01" are simultaneously replaced with "10".

This process repeats until no occurrences of "01" exist in the string.

Return the number of seconds needed to complete this process.

Input & Output

Example 1 — Basic Case

$ Input: s = "0110"

› Output: 2

💡 Note: Second 1: "0110" → "1001" (replace 01 with 10). Second 2: "1001" → "1010" → "1100". No more "01" patterns exist, so 2 seconds total.

Example 2 — Single Swap

$ Input: s = "01"

› Output: 1

💡 Note: Second 1: "01" → "10". No more "01" patterns, so 1 second total.

Example 3 — No Swaps Needed

$ Input: s = "1100"

› Output: 0

💡 Note: Already sorted with all 1's before 0's. No "01" patterns exist, so 0 seconds needed.

Constraints

1 ≤ s.length ≤ 1000
s[i] is either '0' or '1'

Visualization

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Asked in

G Google 15 f Facebook 12

The key insight is that each '1' will eventually move past all '0's to its right. Instead of simulating, count zeros to the right of each '1' - the maximum count is the answer. Best approach is the Count Zeros method. Time: O(n), Space: O(1)

Common Approaches

✓ Count Zeros Approach

⏱️ Time: O(n) Space: O(1)

Instead of simulating, observe that each 1 will eventually move past all 0's to its right. Count zeros to the right of each 1 - the maximum count determines the answer.

Simulation Approach

⏱️ Time: O(n²) Space: O(n)

Simulate the exact process described in the problem. In each second, find all occurrences of "01" and replace them with "10" simultaneously. Continue until no "01" patterns remain.

Count Zeros Approach — Algorithm Steps

Step 1: Traverse string from right to left, counting zeros encountered
Step 2: For each 1, record how many zeros it needs to pass
Step 3: Return maximum count among all 1's

Visualization

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Step-by-Step Walkthrough

Scan Right to Left

Count zeros encountered while moving left

Track Maximum

For each 1, record current zero count

Return Max

Maximum zero count is the answer

Code -

solution.c — C

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

int solution(char* s) {
    int result = 0;
    int onesSeen = 0;
    int len = strlen(s);
    
    for (int i = 0; i < len; i++) {
        if (s[i] == '1') {
            onesSeen++;
        } else {
            result += onesSeen;
        }
    }
    
    return result;
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    printf("%d\n", solution(s));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the string from right to left

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to track counts

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Count Zeros Approach — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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