1864. Minimum Number of Swaps to Make the Binary String Alternating

Problem Description

In this problem, we are given a binary string s which only contains '0's and '1's. Our goal is to find the minimum number of character swaps required to convert the string into an alternating binary string, where no two adjacent characters are the same. For example, "010101" and "101010" are alternating strings, but "110" is not as there are two '1's adjacent to each other.

If making the string alternating is not possible, we should return -1.

A character swap means choosing any two characters in the string (not necessarily adjacent) and exchanging their positions. The challenge is to do the minimum number of such swaps to achieve an alternating pattern.

Intuition

To arrive at the solution, we first need to understand that there are only two possible alternating patterns for any string: "010101..." or "101010...". Also, for a binary string that can be made alternating, the difference between the counts of '0's and '1's cannot be more than one. If the difference is more than one, it is impossible to form an alternating string because there would be extra characters of one type that we cannot place without creating adjacent duplicates.

With this understanding, the next insight is to count the number of misplaced '0's and '1's in both possible alternating patterns. We need to track the following:

• s0n0: Number of '0's that are in the wrong place if we are trying to form the "010101..." pattern.
• s0n1: Number of '1's that are in the wrong place if we are trying to form the "010101..." pattern.
• s1n0: Number of '0's that are in the wrong place if we are trying to form the "101010..." pattern.
• s1n1: Number of '1's that are in the wrong place if we are trying to form the "101010..." pattern.

For a valid alternating string:

• The counts of '0's and '1's should be equal, or there should be exactly one more '0' or exactly one more '1'.
• The numbers of misplaced '0's should equal the numbers of misplaced '1's for a given pattern. That is, s0n0 should equal s0n1 and s1n0 should equal s1n1. If they are not equal, it means we can't swap a '0' with a '1' to correct the string since there's an unequal amount to swap.

After counting, we have the following cases:

• If both the counts of s0n0 and s0n1 aren't equal, and the counts of s1n0 and s1n1 aren't equal either, we return -1 because it's impossible to make the string alternating.
• If s0n0 and s0n1 aren't equal, which means we can't form "010101..." pattern, we should check if we can form "101010..." pattern and return s1n0 (or s1n1 since they are equal).
• If s1n0 and s1n1 aren't equal, meaning we can't form "101010..." pattern, we should return s0n0 (or s0n1 since they are equal).
• If we can form both "010101..." and "101010..." patterns, we return the minimum of s0n0 and s1n0.

Therefore, by counting the number of misplaced characters and comparing them, we can determine the minimum number of swaps required. If any configuration allows for alternating characters, that minimum count is the answer. If neither does, then it is impossible to form an alternating string, and we return -1.

Learn more about Greedy patterns.

Solution Approach

The solution provided follows a simple but effective approach without the need for any complex algorithms or data structures. Here's the breakdown:

1. Initialize four counters: s0n0, s0n1, s1n0, s1n1 to zero. These will count the number of misplaced '0's and '1's for both potential alternating patterns "010101..." (s0n0 and s0n1) and "101010..." (s1n0 and s1n1).

2. Iterate through the string s using a for-loop and the range function, checking each character:

   • If we're at an even index (using i & 1 to check if i is odd or even), we compare the character with '0'.
     • If it's not '0', then it's in the wrong place for the "010101..." pattern, so we increment s0n0.
     • If it is a '0', then it's in the wrong place for the "101010..." pattern, so we increment s1n1.
   • If we're at an odd index, we do the reverse:
     • If the character is not '0', it's wrong for "101010..." and we increment s1n0.
     • If it is a '0', it's wrong for "010101..." and we increment s0n1.

3. After the loop, we have the counts of misplaced '0's and '1's for both patterns. We then examine these counts:

   • If s0n0 does not equal s0n1, and s1n0 does not equal s1n1, we can't form an alternating string. We return -1.
   • If s0n0 does not equal s0n1, we know that the "010101..." pattern is not possible, but the "101010..." pattern is, so we return s1n0 (or s1n1 as they are the same).
   • Conversely, if s1n0 does not equal s1n1, we can't form "101010...", but we know "010101..." is possible, so we return s0n0 (or s0n1 since they are equal).
   • If both sets of counts are equal, meaning either pattern could be formed, we return the minimum of s0n0 and s1n0.

This approach ensures we find the minimum swaps needed for either pattern if it's possible to build an alternating string out of s. It efficiently utilizes bitwise operations and simple if-else constructs without the need for additional space, hence operating in O(n) time complexity and O(1) space complexity, where n is the length of string s.

Example Walkthrough

Let's use a small example to illustrate the solution approach. Consider the binary string s = "1001". We want to determine the minimum number of swaps required to make this string into an alternating binary string, either "1010" or "0101".

1. Initialize our counters: s0n0 = 0, s0n1 = 0, s1n0 = 0, s1n1 = 0.

2. We begin iterating through the string:

   • For the first character (index 0, even), we have '1'.

     • It's not '0', so it's misplaced for the "010101..." pattern. Increment s0n0 to 1.
     • Since it's '1', it's correctly placed for the "101010..." pattern, so s1n1 remains 0.

   • For the second character (index 1, odd), we have '0'

     • It's not '1', so it's misplaced for the "101010..." pattern. Increment s1n0 to 1.
     • Since it's '0', it's correctly placed for the "010101..." pattern, so s0n1 remains 0.

   • For the third character (index 2, even), we have '0'.

     • It's correctly placed for the "010101..." pattern, so s0n0 remains 1.
     • Since it's not '1', it's misplaced for the "101010..." pattern. Increment s1n1 to 1.

   • For the fourth character (index 3, odd), we have '1'.

     • It's not '0', so it's misplaced for the "010101..." pattern. Increment s0n1 to 1.
     • Since it's '1', it's correctly placed for the "101010..." pattern, so s1n0 remains 1.

3. After iterating, our counts are as follows: s0n0 = 1, s0n1 = 1, s1n0 = 1, s1n1 = 1.

4. We examine our counts:

   • Since s0n0 equals s0n1, and s1n0 equals s1n1, either pattern "010101..." or "101010..." can be formed.

   • We return the minimum of s0n0 and s1n0, which is min(1, 1) = 1.

Thus, it only takes a single swap to turn the string "1001" into an alternating binary string. We can swap the second and third characters to obtain "1010" or swap the first and second characters to obtain "0101".

Solution Implementation

Time and Space Complexity

Time Complexity

The given code iterates through the string s once, which means that the loop runs for n iterations, where n is the length of the string s. Within each iteration of the loop, the code performs a constant number of operations, such as comparison, bitwise AND, and increment operations, all of which take constant O(1) time. Therefore, the time complexity of the entire function is directly proportional to the length of the string, which gives us a time complexity of O(n).

Space Complexity

Regarding space complexity, the code allocates a constant amount of extra space. It uses four integer variables s0n0, s0n1, s1n0, and s1n1 to keep track of counts, no matter how large the input string s is. Since these variables do not depend on the size of the input, and no other significant space is used (no dynamic data structures or arrays are allocated), the space complexity is constant, O(1).

Learn more about how to find time and space complexity quickly using problem constraints.