Auto-solve daily LeetCode problem using GPT-5-mini

Author: jeremymanning

problems/1888/gpt5-mini.md

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+# [Problem 1888: Minimum Number of Flips to Make the Binary String Alternating](https://leetcode.com/problems/minimum-number-of-flips-to-make-the-binary-string-alternating/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+We can perform arbitrary rotations (taking first char to end) and flip arbitrary characters. Rotations allow us to consider every cyclic shift of s; for each rotation we want the minimum flips required to make it alternating. A brute force that rotates and counts flips for each rotation would be O(n^2) if done naively.
+
+Observations:
+- An alternating string has only two possible patterns for a given length: "0101..." (start with '0') or "1010..." (start with '1').
+- Considering rotations is equivalent to checking all length-n substrings of the doubled string s+s.
+- So if we build s2 = s + s and slide a window of length n across s2, we can count mismatches to the two alternating templates for each window in O(1) amortized time using a sliding-window mismatch count.
+
+This suggests an O(n) time, O(1) extra space solution.
+
+## Refining the problem, round 2 thoughts
+Edge cases:
+- n = 1: string is already alternating, answer is 0.
+- The sliding-window approach must carefully add the contribution of the incoming character and remove the outgoing one (when window exceeds size n).
+- For each index i in s2 we can determine the expected char for both templates by parity: for template0 (start with '0') expected is '0' if i%2==0 else '1'; for template1 it's the opposite.
+- Maintain two counters: mismatches to template0 and mismatches to template1 in the current window. When the window size reaches n, update the answer with the minimum of the two counters.
+- Complexity: O(n) time (we iterate over 2n chars once, with O(1) updates each step) and O(1) extra space (only counters and indices).
+
+Alternative solutions:
+- For even n there's a small optimization: rotation parity doesn't change counts in some way, but the sliding-window approach is simple and uniform for both even and odd n.
+
+## Attempted solution(s)
+```python
+class Solution:
+    def minFlips(self, s: str) -> int:
+        n = len(s)
+        # If length 1, already alternating
+        if n == 1:
+            return 0
+
+        s2 = s + s
+        # mismatch counts for templates starting with '0' and starting with '1'
+        mismatch0 = 0  # mismatches to pattern "0101..."
+        mismatch1 = 0  # mismatches to pattern "1010..."
+        ans = float('inf')
+
+        for i, ch in enumerate(s2):
+            # expected characters for index i for both patterns
+            if i % 2 == 0:
+                expected0 = '0'
+                expected1 = '1'
+            else:
+                expected0 = '1'
+                expected1 = '0'
+
+            if ch != expected0:
+                mismatch0 += 1
+            if ch != expected1:
+                mismatch1 += 1
+
+            # once we have more than n characters in window, remove the leftmost
+            if i >= n:
+                left = s2[i - n]
+                j = i - n
+                # expected chars at position j
+                if j % 2 == 0:
+                    left_expected0 = '0'
+                    left_expected1 = '1'
+                else:
+                    left_expected0 = '1'
+                    left_expected1 = '0'
+
+                if left != left_expected0:
+                    mismatch0 -= 1
+                if left != left_expected1:
+                    mismatch1 -= 1
+
+            # when window size is exactly n, update answer
+            if i >= n - 1:
+                ans = min(ans, mismatch0, mismatch1)
+
+        return ans if ans != float('inf') else 0
+```
+- Approach: Duplicate the string to consider all rotations as length-n windows of s+s. Maintain mismatch counts to the two alternating templates while sliding a window of size n. For each window compute the minimal flips needed (the fewer mismatches between the two templates).
+- Time complexity: O(n) — we scan 2n characters and perform O(1) work per character.
+- Space complexity: O(1) extra space (only counters and indices, plus the doubled string s2 which is O(n) but can be considered input-expanded; if strict extra-space counting excludes s2, then O(1)).
+- Important detail: computing expected characters by parity avoids building full pattern strings and keeps checks constant-time.