[Java] Two Line Solution With Explanation
```\n/\nSort the array by length, and if length are the same then we can just sort by contents. \nIf we try to sort by contents only, the strings will be compar
```\n/*\nSort the array by length, and if length are the same then we can just sort by contents. \nIf we try to sort by contents only, the strings will be compared left to right\nSo "123" will be smaller than "9" as "1" < "9", so we have to check the length FIRST\nIf the strings are the same size "123" vs "999", then we will compare char by char so \'1\' < \'9\', so "123" will come first\n\nLastly, we can just grab the kth value from the back of the array now that it\'s sorted. \n*/\npublic String kthLargestNumber(String[] nums, int k) {\n\tArrays.sort(nums, (a,b) -> (a.length() == b.length() ? a.compareTo(b) : Integer.compare(a.length(), b.length())));\n\treturn nums[nums.length-k];\n}
['Java']