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

Author: jeremymanning

problems/3542/gpt5-mini.md

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+# [Problem 3542: Minimum Operations to Convert All Elements to Zero](https://leetcode.com/problems/minimum-operations-to-convert-all-elements-to-zero/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+We can pick any subarray and set all occurrences of the minimum non-negative integer inside that subarray to 0. Zeros are special: if a subarray contains a 0, the minimum is 0 and the operation does nothing, so we never want to include zeros in an operation — zeros partition the array into independent segments.
+
+Within a non-zero segment, one operation can remove all entries equal to the minimum value of that chosen subarray. Intuitively, as we move left-to-right, when we see a value larger than the previous values we may need an operation for that new "level". When we see smaller values we can "pop" higher levels that won't be needed further. This smells like a monotonic stack approach where we count pushes of new positive values (distinct levels) and pop when encountering smaller numbers. Equal values do not require a new operation.
+
+I recall a known pattern: keep an increasing stack of positive values; zeros clear the stack; whenever current value v > 0 and is strictly greater than top, push and increment answer; if v < top, pop until top <= v; if after popping top != v and v > 0, push and increment. That should produce minimal operations.
+
+## Refining the problem, round 2 thoughts
+- Zeros split the array into segments; operations never cross zeros because they'd be wasted on zeros.
+- Within a segment, operations correspond to introducing new distinct positive "levels" as we traverse, with higher values that later get overshadowed by smaller ones removed by popping.
+- Duplicate adjacent values should not add operations.
+- Complexity target: O(n) time, O(n) space for the stack.
+- Edge cases: all zeros -> 0 operations; all same value -> 1 operation (unless value is 0); alternating patterns handled by stack logic.
+
+I'll implement the monotonic stack solution described above, with careful handling of zeros (stack cleared) and equality checks.
+
+## Attempted solution(s)
+```python
+from typing import List
+
+class Solution:
+    def minOperations(self, nums: List[int]) -> int:
+        """
+        Monotonic increasing stack approach:
+        - zeros partition the array: clear stack on zero.
+        - maintain increasing stack of positive values.
+        - when seeing v > 0:
+            pop while stack top > v
+            if stack is empty or stack top < v:
+                push v and increment answer
+            if stack top == v: do nothing
+        """
+        stack = []
+        ans = 0
+        for v in nums:
+            if v == 0:
+                # zeros partition; operations touching zeros are useless
+                stack.clear()
+                continue
+            # remove strictly larger values that cannot be matched further
+            while stack and stack[-1] > v:
+                stack.pop()
+            # if top equals v, this value already accounted for
+            if stack and stack[-1] == v:
+                continue
+            # need a new operation for this new positive level
+            stack.append(v)
+            ans += 1
+        return ans
+```
+- Approach notes: We treat zeros as separators and maintain an increasing stack of active positive values. Every time we see a new positive value not represented at the stack top, it corresponds to an operation (we will need to remove that value eventually). Larger values that appear before a smaller value are popped because once a smaller value appears, future subarrays that cover both would pick the smaller as minimum and removing smaller values first will not require separate operations for those larger values in the same way — the stack popping captures this ordering.
+- Time complexity: O(n). Each element is pushed/popped at most once.
+- Space complexity: O(n) in worst case for the stack.