3 Approach || JAVA || Beginner Friendly || Easyyy
*PLEASE UPVOTE IF YOU LIKE IT, THANK YOU*\n# Code\n\nTIME COMPLEXITY : O(n * log(k))\n\nimport java.math.BigInteger;\nclass Solution {\n    public String kthLar
****PLEASE UPVOTE IF YOU LIKE IT, THANK YOU****\n# Code\n```\nTIME COMPLEXITY : O(n * log(k))\n\nimport java.math.BigInteger;\nclass Solution {\n    public String kthLargestNumber(String[] arr, int k) {\n    PriorityQueue<BigInteger> pq = new PriorityQueue<>();\n        for(int i = 0; i < arr.length; i++){\n            if(i < k){\n                pq.add(new BigInteger(arr[i]));\n            }else {\n                BigInteger val = new BigInteger(arr[i]);\n                if(val.compareTo(pq.peek()) > 0){\n                    pq.remove();\n                    pq.add(val);\n                }\n            }\n        }\n        return pq.remove().toString();\n    }\n}\n\n/*\nTIME COMPLEXITY : O(n * log(n))\n\nimport java.math.BigInteger;\nclass Solution {\n    public String kthLargestNumber(String[] arr, int k) {\n    PriorityQueue<BigInteger> pq = new PriorityQueue<>(Collections.reverseOrder());\n        for (String s : arr) pq.add(new BigInteger(s));\n        \n        while (k != 1) {\n            pq.remove();\n            k--;\n        }\n        return pq.remove()+"";\n    }\n}\n\n/*\nTIME COMPLEXITY : O(n * log(n)\n\nimport java.math.BigInteger;\nclass Solution {\n    public String kthLargestNumber(String[] nums, int k) {\n      BigInteger[] arr = new BigInteger[nums.length];\n        for(int i=0;i<arr.length;i++){\n           arr[i]=new BigInteger(nums[i]);\n        }\n        Arrays.sort(arr);\n        BigInteger result= arr[arr.length-k];\n        return result.toString();\n    }\n}\n*/\n```
['Heap (Priority Queue)', 'C++', 'Java']