swap(&arr[i], &arr[j]);
renderComponent(controller); // Promises created, objects allocated
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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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Like the original Connections, the game is all about finding the "common threads between words." And just like Wordle, Connections resets after midnight and each new set of words gets trickier and trickier — so we've served up some hints and tips to get you over the hurdle.
"assignment": [false, false, false, false, false, false, false, false, false, false]。业内人士推荐一键获取谷歌浏览器下载作为进阶阅读