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

Author: jeremymanning

problems/1590/gpt5-mini.md

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+# [Problem 1590: Make Sum Divisible by P](https://leetcode.com/problems/make-sum-divisible-by-p/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+We need to remove the smallest contiguous subarray so that the remaining sum is divisible by p. Let total = sum(nums). If total % p == 0, return 0 immediately. Otherwise we need to remove a subarray whose sum % p equals total % p (call this need). So the problem becomes: find the shortest subarray with sum % p == need. Brute force O(n^2) checking all subarrays is too slow for n up to 1e5.
+
+Prefix sums modulo p can help: sum(i..j) % p = (pref[j] - pref[i-1]) % p. So for each prefix remainder pref_i, we want to find a previous prefix remainder pref_j such that (pref_i - pref_j) % p == need, equivalently pref_j == (pref_i - need) % p. We can keep a map from remainder -> latest index and update as we scan. Initialize map{0: -1} to represent empty prefix. Track minimal length. If minimal length equals the whole array, not allowed (must not remove whole array), return -1.
+
+## Refining the problem, round 2 thoughts
+- Edge cases: need == 0 (already divisible) -> return 0. If no valid subarray found -> return -1.
+- Must ensure we don't allow removing whole array: if best length == n -> return -1.
+- Implementation detail: use modulo arithmetic carefully: (pref - need) % p gives target remainder in Python (non-negative).
+- Complexity: single pass O(n) time, O(n) extra space for the map (at most n distinct remainders stored). Works for large nums since we only use remainders.
+- Alternative: sliding window doesn't apply directly because elements are arbitrary and we want modulo condition, not fixed target sum.
+
+## Attempted solution(s)
+```python
+class Solution:
+    def minSubarray(self, nums: list[int], p: int) -> int:
+        total = sum(nums)
+        need = total % p
+        if need == 0:
+            return 0
+
+        pref = 0
+        best = len(nums) + 1
+        last_index = {0: -1}  # remainder -> latest index
+
+        for i, x in enumerate(nums):
+            pref = (pref + x) % p
+            # we need previous remainder such that (pref - prev) % p == need
+            target = (pref - need) % p
+            if target in last_index:
+                length = i - last_index[target]
+                if length < best:
+                    best = length
+            # update with current prefix remainder (store latest index)
+            last_index[pref] = i
+
+        if best <= 0 or best > len(nums) - 1:
+            # best == len(nums) means removing whole array (not allowed)
+            return -1
+        return best
+```
+- Notes about solution:
+  - Approach: Use prefix-sum remainders and a hashmap from remainder to latest index to find shortest subarray whose sum % p == need in one pass.
+  - Correctness: For each index i, if a previous index j has remainder equal to (pref_i - need) % p, then the subarray (j+1..i) has sum % p == need. We minimize i-j.
+  - Time complexity: O(n) single pass.
+  - Space complexity: O(n) for the hashmap (in worst case storing one entry per index).
+  - Implementation details: Initialize map with 0 -> -1 to allow subarrays starting at index 0. Ensure we do not return length equal to n (can't remove entire array).