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

Author: jeremymanning

problems/3321/gpt5-mini.md

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+# [Problem 3321: Find X-Sum of All K-Long Subarrays II](https://leetcode.com/problems/find-x-sum-of-all-k-long-subarrays-ii/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+I need the x-sum of every k-length sliding window. The x-sum keeps the occurrences of the top x most frequent distinct elements (tie-break by larger value) and sums the resulting array (i.e., sum value * frequency for each chosen distinct value). A sliding window suggests maintaining counts incrementally, and deciding which distinct values belong to the top-x set as the window moves.
+
+We must frequently compare elements by (frequency, value) ordering and maintain the top-x set. This is similar to maintaining two multisets/heaps (top and bottom) like the two-heap approach for sliding-window median. Python doesn't have a balanced tree, but we can use two heaps with lazy deletion and a freq map + an "in_top" marker. Maintain:
+- top heap: the chosen top-x distinct values (we must be able to access the worst among them quickly)
+- bottom heap: the remaining distinct values (we must be able to access the best among them quickly)
+Also keep sum_top = sum(freq[val] * val for val in top-set). Rebalance after each add/remove to ensure top contains exactly t = min(x, distinct_count) best elements.
+
+Lazy deletion: when an element's frequency changes we push a new tuple to its current heap and mark its current in_top. When popping, we skip stale entries by checking the current freq and in_top status.
+
+This should give O(n log m) time where m is number of distinct values (<= n).
+
+## Refining the problem, round 2 thoughts
+Edge cases:
+- k == x: x-sum equals full window sum (algorithm still handles this since top will include all distinct or x distinct).
+- frequencies hitting zero: must remove distinct and update t and top_size. If an element is in top and gets removed (freq 0), we must update sum_top and top_size immediately (can't wait for lazy pops).
+- tie-breaker: when freq equal, larger value is considered more frequent. So ordering is by (freq, value) descending. For comparisons, I'll treat pair (freq, value) and compare lexicographically.
+
+Implementation details:
+- top heap should let us access worst among top (min by (freq, value)), so top_heap uses (freq, value, val) as min-heap.
+- bottom heap should let us access best among bottom (max by (freq, value)), so bottom_heap stores (-freq, -value, val).
+- Maintain freq dict and in_top dict (bool), top_size int, D distinct count, sum_top integer.
+- Provide helper functions to get/pop valid entries from heaps and to rebalance.
+
+Complexities:
+- Each add/remove pushes at most one new entry and each rebalance step moves/pops elements; each pop processes at least one stale entry or valid element, so overall number of heap operations is O(n log m). Space O(m).
+
+## Attempted solution(s)
+```python
+import heapq
+from typing import List
+
+class Solution:
+    def xSum(self, nums: List[int], k: int, x: int) -> List[int]:
+        n = len(nums)
+        # Heaps and data structures:
+        # top_heap: min-heap of (freq, value, val) -> worst among top is top_heap[0]
+        # bottom_heap: min-heap used as max-heap storing (-freq, -value, val) -> best among bottom is bottom_heap[0]
+        top_heap = []
+        bottom_heap = []
+        freq = {}        # current freq of value in window
+        in_top = {}      # whether value is considered in top set
+        D = 0            # number of distinct elements with freq > 0
+        top_size = 0     # number of distinct elements currently in top set
+        sum_top = 0      # sum of freq[val] * val over elements in top set
+
+        def push_top(v):
+            # push current tuple for v into top heap.
+            f = freq.get(v, 0)
+            if f <= 0:
+                return
+            heapq.heappush(top_heap, (f, v, v))
+            in_top[v] = True
+
+        def push_bottom(v):
+            # push current tuple for v into bottom heap.
+            f = freq.get(v, 0)
+            if f <= 0:
+                return
+            heapq.heappush(bottom_heap, (-f, -v, v))
+            in_top[v] = False
+
+        def pop_valid_top():
+            # pop until a valid top entry is found and return (f, v)
+            while top_heap:
+                f, val_key, v = heapq.heappop(top_heap)
+                # valid if current freq matches and in_top is True
+                if freq.get(v, 0) == f and in_top.get(v, False) == True:
+                    return f, v
+                # else stale, continue popping
+            return None
+
+        def peek_valid_top():
+            while top_heap:
+                f, val_key, v = top_heap[0]
+                if freq.get(v, 0) == f and in_top.get(v, False) == True:
+                    return f, v
+                heapq.heappop(top_heap)  # drop stale
+            return None
+
+        def pop_valid_bottom():
+            while bottom_heap:
+                nf, nval, v = heapq.heappop(bottom_heap)
+                f = -nf
+                # valid if current freq matches and in_top is False
+                if freq.get(v, 0) == f and in_top.get(v, True) == False:
+                    return f, v
+                # else stale, continue popping
+            return None
+
+        def peek_valid_bottom():
+            while bottom_heap:
+                nf, nval, v = bottom_heap[0]
+                f = -nf
+                if freq.get(v, 0) == f and in_top.get(v, True) == False:
+                    return f, v
+                heapq.heappop(bottom_heap)
+            return None
+
+        def rebalance():
+            nonlocal top_size, sum_top
+            tgt = min(x, D)
+            # 1) Grow top if needed
+            while top_size < tgt:
+                # move best from bottom to top
+                item = pop_valid_bottom()
+                if not item:
+                    break
+                f, v = item
+                # move v to top
+                in_top[v] = True
+                heapq.heappush(top_heap, (f, v, v))
+                top_size += 1
+                sum_top += f * v
+            # 2) Shrink top if needed (some elements died or D decreased)
+            while top_size > tgt:
+                item = pop_valid_top()
+                if not item:
+                    break
+                f, v = item
+                # move v to bottom (or remove if freq==0)
+                in_top[v] = False
+                top_size -= 1
+                sum_top -= f * v
+                if freq.get(v, 0) > 0:
+                    heapq.heappush(bottom_heap, (-freq[v], -v, v))
+            # 3) Ensure ordering invariant: every top element >= every bottom element
+            while True:
+                top_peek = peek_valid_top()
+                bottom_peek = peek_valid_bottom()
+                if not top_peek or not bottom_peek:
+                    break
+                ft, vt = top_peek
+                fb, vb = bottom_peek
+                # if bottom's best is better than top's worst -> swap
+                # compare (fb, vb) > (ft, vt)
+                if fb > ft or (fb == ft and vb > vt):
+                    # pop both and swap membership
+                    pop_valid_bottom()  # removes bottom best
+                    pop_valid_top()     # removes top worst
+                    # move bottom best to top
+                    in_top[vb] = True
+                    heapq.heappush(top_heap, (fb, vb, vb))
+                    sum_top += fb * vb
+                    # move top worst to bottom (if still >0 freq)
+                    in_top[vt] = False
+                    sum_top -= ft * vt
+                    if freq.get(vt, 0) > 0:
+                        heapq.heappush(bottom_heap, (-freq[vt], -vt, vt))
+                    # top_size unchanged
+                    continue
+                else:
+                    break
+
+        # add one value to the window
+        def add_val(v):
+            nonlocal D, sum_top
+            prev = freq.get(v, 0)
+            freq[v] = prev + 1
+            if prev == 0:
+                # new distinct
+                D += 1
+                # start in bottom
+                in_top[v] = False
+                heapq.heappush(bottom_heap, (-freq[v], -v, v))
+            else:
+                # existing
+                if in_top.get(v, False):
+                    # its contribution in top increases by v
+                    sum_top += v
+                    heapq.heappush(top_heap, (freq[v], v, v))
+                else:
+                    heapq.heappush(bottom_heap, (-freq[v], -v, v))
+
+        # remove one value from the window
+        def remove_val(v):
+            nonlocal D, top_size, sum_top
+            prev = freq.get(v, 0)
+            if prev == 0:
+                return
+            curr = prev - 1
+            freq[v] = curr
+            if in_top.get(v, False):
+                # element was in top: reduce contribution by v
+                sum_top -= v
+                if curr == 0:
+                    # element removed entirely
+                    in_top[v] = False
+                    top_size -= 1
+                    D -= 1
+                    # do not push any tuple
+                else:
+                    # still present and still considered in top (we'll push updated tuple)
+                    heapq.heappush(top_heap, (curr, v, v))
+            else:
+                # element was in bottom
+                if curr == 0:
+                    D -= 1
+                    # nothing to push
+                else:
+                    heapq.heappush(bottom_heap, (-curr, -v, v))
+
+        # initialize first window
+        for i in range(k):
+            add_val(nums[i])
+        rebalance()
+        ans = [sum_top]
+
+        # slide
+        for i in range(k, n):
+            add_val(nums[i])
+            remove_val(nums[i - k])
+            rebalance()
+            ans.append(sum_top)
+
+        return ans
+
+# Example usage wrapper for LeetCode-style:
+class SolutionWrapper:
+    def findXsum(self, nums: List[int], k: int, x: int) -> List[int]:
+        return Solution().xSum(nums, k, x)
+
+# If you want, you can test with the examples:
+if __name__ == "__main__":
+    s = Solution()
+    print(s.xSum([1,1,2,2,3,4,2,3], 6, 2))  # [6,10,12]
+    print(s.xSum([3,8,7,8,7,5], 2, 2))      # [11,15,15,15,12]
+```
+
+- Notes about the solution:
+  - We maintain two heaps (top and bottom) and a frequency map with lazy deletion to avoid expensive arbitrary deletions from heaps.
+  - top_heap stores tuples (freq, value, value) so the root is the worst among top (lowest freq, and lower value in tie) — we can pop the worst to move it down quickly.
+  - bottom_heap stores (-freq, -value, value) so the root is the best among bottom (highest freq, and larger value in tie) — we can pop the best to move it up quickly.
+  - in_top marks current membership; lazy entries are ignored when they become stale because their stored freq or in_top status does not match the current state.
+  - sum_top is maintained incrementally on adds/removes and during moves between heaps so we can output the x-sum in O(1) per window after rebalancing.
+  - Time complexity: O(n log m) where m is the number of distinct elements encountered (<= n). Each add/remove causes a few heap operations; lazy deletions ensure every stale entry is popped once.
+  - Space complexity: O(m) for heaps and maps.