Time Needed To Rearrange A Binary String - Leetcode Solution

The problem "Time Needed To Rearrange A Binary String" on LeetCode is to find the minimum time required to rearrange a binary string such that there is no consecutive 0's or 1's in the final arrangement.

Solution

To solve this problem, we need to find a way to arrange the binary string in such a way that there are no consecutive 0's or 1's. One way to do that is to find the number of consecutive 0's and 1's in the string and then arrange them alternatively.

For example, let's say the input binary string is "0010110010". We can count the number of consecutive 0's and 1's in the string as follows:

Groups of 0's: 00, 1, 0, 00 Groups of 1's: 0, 11, 0, 1, 0

Now we can arrange the groups alternatively to get the final string without consecutive 0's or 1's:

Alternate: 00 0 11 1 0 0 1 0 0 0

Thus, we need to find the number of groups of consecutive 0's and 1's in the string and then arrange them alternatively to get the final string.

Code:

Here is the Python code to solve this problem:

class Solution:
    def minFlips(self, s: str) -> int:
        zero_groups = s.split('1')
        one_groups = s.split('0')
        zero_groups = [group for group in zero_groups if group != '']
        one_groups = [group for group in one_groups if group != '']

        if len(zero_groups) < len(one_groups):
            zero_groups, one_groups = one_groups, zero_groups

        if len(one_groups) == 1:
            return 0

        if len(zero_groups) == 0:
            return len(one_groups) - 1

        return len(zero_groups) + len(one_groups) - (len(one_groups) % 2 == 0)

Explanation:

First, we split the binary string into groups of consecutive 0's and 1's. We discard empty groups if any.

Then, we check which one is greater in count, zero groups or one groups. If one groups is greater, we swap them.

Next, we check if there is only one group of 1's, then no flips are needed, we return 0.

If there are no zero groups, we need to flip half the one groups because we can start with either 0 or 1 without affecting the result. Thus, we return the number of one groups minus 1.

Finally, we arrange the groups alternatively by adding the total number of groups minus 1. If the total number of groups is even, we need to subtract 1 for the last group.

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