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

Author: jeremymanning

problems/3578/gpt5-mini.md

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+# [Problem 3578: Count Partitions With Max-Min Difference at Most K](https://leetcode.com/problems/count-partitions-with-max-min-difference-at-most-k/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+I need to count ways to cut the array into contiguous segments where each segment's max - min <= k. That suggests dynamic programming: dp[i] = number of ways to partition prefix of length i. For dp[i] we sum dp[j] for all j where segment j..i-1 is valid. Naively checking all j for each i is O(n^2) and could be too slow for n up to 5e4.
+
+Observation: for a fixed right end i-1, as j moves left the range max-min is nondecreasing (adding elements can't decrease max-min). So for each right i-1 there is a smallest j* such that any j >= j* yields a valid segment [j..i-1]. So dp[i] = sum_{j=j*..i-1} dp[j]. If we maintain prefix sums of dp, we can compute this sum in O(1) once j* is known. So the remaining challenge is to find j* quickly for each i.
+
+We can maintain a sliding window with two monotonic deques (one for max, one for min). Move the right pointer forward; while current window [L..i] violates (max-min > k) increment L and pop expired indices from deques. L will be nondecreasing across i, giving amortized O(n). So overall O(n) time.
+
+## Refining the problem, round 2 thoughts
+Edge cases:
+- dp indices and prefix sums: be careful with off-by-one. I'll use dp[0]=1 (empty prefix), dp[t] = number ways for first t elements (t from 1..n). When processing element at index i (0-based), we compute dp[i+1].
+- When L==0 need to subtract zero from prefix sum.
+- Maintain deques storing indices; when moving L forward pop indices equal to L from both deques.
+- Use modulo 10**9+7 everywhere and ensure differences are normalized positive.
+
+Complexity:
+- Time: O(n) amortized since each index enters/pops deques at most once.
+- Space: O(n) for dp/prefix arrays and O(n) worst-case for deques but deques are at most n aggregated.
+
+## Attempted solution(s)
+```python
+from collections import deque
+
+class Solution:
+    def countPartitions(self, nums: list[int], k: int) -> int:
+        MOD = 10**9 + 7
+        n = len(nums)
+        # dp[t] = number of ways to partition first t elements (t in [0..n])
+        dp = [0] * (n + 1)
+        pref = [0] * (n + 1)  # pref[t] = sum_{u=0..t} dp[u]
+        dp[0] = 1
+        pref[0] = 1
+
+        L = 0
+        maxdq = deque()  # indices with nums[...] in decreasing order
+        mindq = deque()  # indices with nums[...] in increasing order
+
+        for i in range(n):
+            x = nums[i]
+            # maintain monotonic deques for max and min
+            while maxdq and nums[maxdq[-1]] < x:
+                maxdq.pop()
+            maxdq.append(i)
+            while mindq and nums[mindq[-1]] > x:
+                mindq.pop()
+            mindq.append(i)
+
+            # shrink from left until window [L..i] satisfies max-min <= k
+            while maxdq and mindq and nums[maxdq[0]] - nums[mindq[0]] > k:
+                # if the leftmost index equals L, pop it from the corresponding deque
+                if maxdq and maxdq[0] == L:
+                    maxdq.popleft()
+                if mindq and mindq[0] == L:
+                    mindq.popleft()
+                L += 1
+
+            # Now valid windows are those starting at any j in [L..i]
+            left_pref = pref[L - 1] if L > 0 else 0
+            dp[i + 1] = (pref[i] - left_pref) % MOD
+            pref[i + 1] = (pref[i] + dp[i + 1]) % MOD
+
+        return dp[n]
+```
+- Notes about the approach:
+  - We use dp with prefix sums to get dp[i+1] = sum_{j=L..i} dp[j] = pref[i] - pref[L-1].
+  - Two monotonic deques maintain current window's maximum and minimum in O(1) time per update; incrementing L until the window is valid moves L monotonically, giving amortized O(n) total movement.
+  - Time complexity: O(n). Space complexity: O(n).
+  - All arithmetic is done modulo 10^9 + 7.