Problem Description

The given LeetCode problem involves operations on a binary string, which is a string composed solely of '0's and '1's. The objective is to make the string "alternating", which means no two consecutive characters in the string are the same. There are two types of operations that can be used to achieve this:

• Type-1: Remove the first character of the string and append it to the end. This is effectively a cyclic permutation of the string.
• Type-2: Flip a character in the string from '0' to '1' or from '1' to '0'.

The challenge is to determine the minimum number of Type-2 operations (flips) required to make the string alternating. We don't need to worry about the number of Type-1 operations since the question only asks for the count of Type-2 operations.

Intuition

The intuition behind the solution comes from observing patterns within an alternating string. Consider the two alternating patterns for a binary string: "010101..." and "101010...". Any valid alternating binary string can be transformed into either of these two patterns with Type-2 operations.

The key insight is to simulate the operations to move towards one of these patterns. Here's how the solution approach works:

1. Calculate the number of flips required to convert the current string into the "010101..." pattern. This is done by comparing each character in the binary string to the expected character in the "01" pattern. Keep count of mismatches.

2. Calculate the number of flips required for the "101010..." pattern. This is simply the length of the string minus the number of flips for "010101..." because every place that matches one pattern mismatches the other and vice versa.

3. Consider the effect of Type-1 operations. Since Type-1 operations are just cyclic permutations, simulate this by moving through the string and adjusting the mismatch count as if the first character is moved to the end, one by one.

4. At each step, recalculate the number of flips needed, taking into account the updated first character. Repeat this for the entire string length and track the minimum number of flips found.

5. The outcome is the minimum of these recalculated flip counts.

This algorithm works because it efficiently combines the permissibility of Type-1 operations to rearrange the string (without actually doing it) with the direct calculation of Type-2 operations required to achieve the pattern.

Learn more about Greedy, Dynamic Programming and Sliding Window patterns.

Solution Approach

The approach to solving this problem is as follows:

1. First, we need initial counts of mismatches for both alternating patterns for the input string s. This can be achieved by iterating over the string and checking each character if it matches the expected character in the pattern "010101...".

2. To implement this, we compare each character in s with the expected character in the pattern. For example, c should match target[i & 1], where & is the bitwise AND operator, used here to alternate between 0 and 1 as we move along the string.

3. We keep a running count cnt of how many characters do not match the expected "01" pattern. Since for an alternating string of length n, the other pattern is just the inverse, the count for the other pattern is n - cnt.

4. We then initialize the answer ans to be the minimum of cnt and n - cnt because we have the choice to convert to either pattern.

5. Next, we simulate Type-1 operations without actually permuting the string. We iterate over the string once again and, for each character, update the mismatch count cnt for the two patterns by considering that the first character is moved to the end. Specifically, we will reduce cnt by one if the first character differs from the expected character in the current position and adjust cnt considering it's now at the end of the string. This effectively simulates a cyclic shift to the left by one position.

6. After each simulated cyclic shift, we update ans with the minimum of the current ans, current cnt, and n - cnt, which keeps track of the minimum number of flips required after each simulated Type-1 operation.

7. After the loop, ans contains the minimum number of flips required to make the string alternating.

In terms of algorithms and data structures, this solution employs a greedy counting approach, where we greedily count mismatches against an expected pattern and use properties of bitwise operations to efficiently alternate between '0' and '1'. The solution also leverages in-place updates to minimize space complexity; no additional data structures are needed beyond simple variables for keeping track of counts and the final answer.

Example Walkthrough

Let's illustrate the solution approach with a small example. Suppose the input binary string is s = "0011". We are to find out the minimum number of flips (Type-2 operations) to make the string alternating ("0101" or "1010").

1. We start by calculating the number of flips needed for the pattern starting with 0 ("0101"). As we scan s, we notice:

   • s[0] = '0', no flip needed (matches "0101").
   • s[1] = '0', flip needed (does not match "0101", should be '1').
   • s[2] = '1', no flip needed (matches "0101").
   • s[3] = '1', flip needed (does not match "0101", should be '0').

   Therefore, we have 2 mismatches for the "0101" pattern, which means cnt = 2.

2. For the pattern starting with 1 ("1010"), we find that we also have 2 mismatches, because it is just the inverse of the other pattern, so n - cnt = 2.

3. We initialize our answer ans = min(cnt, n - cnt), which in this case is min(2, 2) = 2.

4. Now, we simulate the Type-1 operations. We cyclically move the first character to the end and update the flip count. After the first shift, our string becomes "0110":

   • s[1] was '0', and now it's in the second position where it should be '1', so cnt increases to 3.
   • s[0] was '0', and now it's in the last position where it should be '0', so no change in cnt.

5. We then update ans with min(ans, cnt, n - cnt). Since cnt increased, we have min(2, 3, 2), so ans remains 2.

6. We continue the simulation for two more shifts:

   • After second shift to "1100", cnt would decrease since the first '1' at the third position matches, and the last '0' still matches. So, cnt becomes 2.
   • After third shift to "1001", cnt would remain the same since the first '1' at the fourth position still matches, and the last '1' doesn't change the pattern.

7. At every step, we keep checking and updating the minimum flips needed, which in our example remains 2.

The final answer is 2 flips required to make the string alternating, which is the minimum number of flips found during the simulation.

Solution Implementation

Time and Space Complexity

The time complexity of the provided code is O(n). This is because there is a single loop traversing the string of length n. Inside the loop, only constant-time operations are performed, such as updating the cnt variable and comparing values. There are no nested loops or recursive calls that would increase the complexity.

The space complexity of the provided code is O(1). The solution is using only a fixed number of variables (n, cnt, ans, target, and a couple of iterators in the loop) which does not grow with the size of the input. No additional data structures that scale with the input size are used.

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