Why is this a problem?
思路：先对 nums2 用单调栈求每个元素的下一个更大值，存入 Map 缓存；再遍历 nums1 直接查 Map 得结果。时间复杂度 O(len1 + len2)。
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This fragmentation hurts portability. Code that performs well on one runtime may behave differently (or poorly) on another, even though it's using "standard" APIs. The complexity burden on runtime implementers is substantial, and the subtle behavioral differences create friction for developers trying to write cross-runtime code, particularly those maintaining frameworks that must be able to run efficiently across many runtime environments.