How do you solve Time Needed to Rearrange a Binary String on LeetCode?

Time Needed to Rearrange a Binary String (LeetCode #2380) 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: The number of seconds is determined by the distance between the leftmost '0' and the rightmost '1'.

What is the time complexity of Time Needed to Rearrange a Binary String?

The optimal solution for Time Needed to Rearrange a Binary String runs in O(n) time and uses O(1) space. This complexity is linear because we only traverse the string twice — once to find the leftmost '0' and once to find the rightmost '1'.

What is the brute force solution for Time Needed to Rearrange a Binary String?

The brute force approach for Time Needed to Rearrange a Binary String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach simulates the process of replacing '01' with '10' repeatedly until no more replacements can be made. This straightforward method allows us to visualize the changes step by step. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Time Needed to Rearrange a Binary String?

Time Needed to Rearrange a Binary String uses the following concepts: String, Dynamic Programming, Simulation. The recommended patterns to study are: Two Pointers, Sliding Window.

What are the interview tips for Time Needed to Rearrange a Binary String?

Clarify the problem statement and constraints before diving into coding. Think about edge cases and how they might affect your solution. Always consider if there's a more efficient way to solve the problem after the brute force approach.

What are common mistakes when solving Time Needed to Rearrange a Binary String?

Failing to account for cases where there are no '0's or '1's. Overcomplicating the problem by trying to simulate the entire process instead of finding the key positions.

#2380

Time Needed to Rearrange a Binary String

Medium

StringDynamic ProgrammingSimulationTwo PointersSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal approach leverages the observation that the number of seconds needed is determined by the maximum distance between the leftmost '0' and the rightmost '1' in the string. This allows us to calculate the result in linear time.

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Algorithm

3 steps

1Step 1: Initialize variables to track the positions of the leftmost '0' and rightmost '1'.
2Step 2: Traverse the string to find these positions.
3Step 3: The number of seconds needed is the maximum of the distances between these positions.

solution.py6 lines

1def seconds_to_rearrange_optimal(s):
2    leftmost_zero = s.find('0')
3    rightmost_one = s.rfind('1')
4    if leftmost_zero == -1 or rightmost_one == -1 or leftmost_zero > rightmost_one:
5        return 0
6    return rightmost_one - leftmost_zero

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Complexity note: This complexity is linear because we only traverse the string twice — once to find the leftmost '0' and once to find the rightmost '1'.

1The number of seconds is determined by the distance between the leftmost '0' and the rightmost '1'.
2If there are no '0's or '1's, or if '0's are to the right of '1's, the result is 0.

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