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

Author: jeremymanning

problems/1458/gpt5-mini.md

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+# [Problem 1458: Max Dot Product of Two Subsequences](https://leetcode.com/problems/max-dot-product-of-two-subsequences/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+We need the maximum dot product between two non-empty subsequences of equal length. Sounds like a dynamic programming problem over prefixes of both arrays. A naive combinatorial approach (try all subsequence pairs) is exponential. We want an optimal substructure: consider prefixes of nums1 and nums2 and decide whether to match the current elements or skip one of them.
+
+The dot product accumulates pairwise multiplications in order. For indices i and j (pointing to last elements of chosen subsequences), either we use nums1[i] with nums2[j] (and add their product to the best result for previous prefixes) or we skip one side (move left in one array). Need to ensure subsequences are non-empty: we must allow starting the pairing at some point (choose a single pair as baseline). Negative values complicate things: the best may be a single negative*positive giving a negative, or multiple pairs combining to a positive sum.
+
+So dp on prefixes seems right. Need to carefully initialize to allow starting new pair (single product) and not force choosing pairs.
+
+## Refining the problem, round 2 thoughts
+Define dp[i][j] = maximum dot product you can get using some non-empty subsequences taken from nums1[:i] and nums2[:j] (i and j are lengths, 1-based). For each (i,j):
+
+- We can skip nums1[i-1]: dp[i-1][j]
+- We can skip nums2[j-1]: dp[i][j-1]
+- We can take the pair nums1[i-1]*nums2[j-1] and add it to dp[i-1][j-1] (if dp[i-1][j-1] exists)
+- Or we can start a new subsequence pair with just this product (choose the single pair), because dp[i-1][j-1] might be undefined or very negative; so include product itself explicitly.
+
+Thus:
+dp[i][j] = max(dp[i-1][j], dp[i][j-1], dp[i-1][j-1] + prod, prod)
+
+Initialize dp with negative infinity for zero prefixes so that skipping until a first match is allowed and single product is always an option. Finally dp[n][m] is the answer.
+
+Complexity: O(n*m) time, O(n*m) space (or O(m) space if rolling rows used). n,m <= 500 so both fine.
+
+Edge cases:
+- All numbers negative or mixed signs — including the single product option ensures we can return the best single-pair negative if needed.
+- Must ensure subsequences are non-empty; dp definition and initialization ensure we only return values derived from at least one pair.
+
+## Attempted solution(s)
+```python
+from typing import List
+
+class Solution:
+    def maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int:
+        n, m = len(nums1), len(nums2)
+        NEG_INF = -10**18  # sufficiently small sentinel
+
+        # dp[i][j]: max dot product using non-empty subsequences from nums1[:i] and nums2[:j]
+        dp = [[NEG_INF] * (m + 1) for _ in range(n + 1)]
+
+        for i in range(1, n + 1):
+            for j in range(1, m + 1):
+                prod = nums1[i-1] * nums2[j-1]
+                # Option 1,2: skip an element from one side
+                skip1 = dp[i-1][j]
+                skip2 = dp[i][j-1]
+                # Option 3: extend a previous pairing
+                extend = dp[i-1][j-1] + prod if dp[i-1][j-1] != NEG_INF else NEG_INF
+                # Option 4: start a new subsequence with this single pair
+                start_new = prod
+
+                dp[i][j] = max(skip1, skip2, extend, start_new)
+
+        return dp[n][m]
+```
+- Notes about the solution:
+  - Approach: 2D dynamic programming over prefixes. For each pair of prefix endpoints, consider skipping elements or pairing the endpoints (either as extension of a prior pairing or as a new single-pair subsequence).
+  - Correctness: dp always represents the best non-empty subsequence dot product from those prefixes. Including the product itself handles starting subsequences and guarantees non-empty result. Skipping options allow choosing the best placement.
+  - Time complexity: O(n * m), where n = len(nums1) and m = len(nums2).
+  - Space complexity: O(n * m) for the dp table. This can be reduced to O(m) by keeping only previous and current rows because dp[i][j] depends only on dp[i-1][*] and dp[i][j-1]/dp[i-1][j-1].
+  - Implementation detail: Use a very small sentinel (NEG_INF) to represent impossible/unused states for empty prefixes. This prevents invalid accumulation when extending.