Find the Kth Largest Integer in the Array

Problem

You are given an array of strings nums and an integer k. Each string in nums represents an integer without leading zeros.

Return **the string that represents the *kth* largest integer in **nums.

Note: Duplicate numbers should be counted distinctly. For example, if nums is ["1","2","2"], "2" is the first largest integer, "2" is the second-largest integer, and "1" is the third-largest integer.

Example 1:

Input: nums = ["3","6","7","10"], k = 4
Output: "3"
Explanation:
The numbers in nums sorted in non-decreasing order are ["3","6","7","10"].
The 4th largest integer in nums is "3".

Example 2:

Input: nums = ["2","21","12","1"], k = 3
Output: "2"
Explanation:
The numbers in nums sorted in non-decreasing order are ["1","2","12","21"].
The 3rd largest integer in nums is "2".

Example 3:

Input: nums = ["0","0"], k = 2
Output: "0"
Explanation:
The numbers in nums sorted in non-decreasing order are ["0","0"].
The 2nd largest integer in nums is "0".

Constraints:

1 <= k <= nums.length <= 10^4
1 <= nums[i].length <= 100
nums[i] consists of only digits.
nums[i] will not have any leading zeros.

Solution (Java)

class Solution {
    public String kthLargestNumber(String[] nums, int k) {
        Arrays.sort(nums, (n1, n2) -> compareStringInt(n2, n1));
        return nums[k - 1];
    }

    private int compareStringInt(String n1, String n2) {
        if (n1.length() != n2.length()) {
            return n1.length() < n2.length() ? -1 : 1;
        }
        for (int i = 0; i < n1.length(); i++) {
            int n1Digit = n1.charAt(i) - '0';
            int n2Digit = n2.charAt(i) - '0';
            if (n1Digit > n2Digit) {
                return 1;
            } else if (n2Digit > n1Digit) {
                return -1;
            }
        }
        return 0;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).