updateStatus('ERROR: PLAY FAILED', -1);
│ Untrusted Code │
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Never the primary choice, but some are frequently recommended as alternatives.
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)。
Bank branches, being branches, do not exist in isolation. The bank also has a