2113. Find The Kth Largest Integer In The Array¶

LeetCode Problem View on GitHub

2113. Find the Kth Largest Integer in the Array

Medium

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 <= 104
• 1 <= nums[i].length <= 100
• nums[i] consists of only digits.
• nums[i] will not have any leading zeros.

Solution¶

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;

class Solution {
    static class customSort implements Comparator<String> {
        @Override
        public int compare(String first, String second) {
            if (first.length() != second.length()) {
                if (first.length() < second.length())
                    return 1;
                else
                    return -1;
            } else {
                for (int i = 0; i < first.length(); i++) {
                    int f = first.charAt(i) - '0';
                    int s = second.charAt(i) - '0';
                    if (f == s)
                        continue;
                    if (f < s)
                        return 1;
                    if (s < f)
                        return -1;
                }
                return 0;
            }
        }
    }
    public String kthLargestNumber(String[] nums, int k) {
        ArrayList<String> res = new ArrayList<>();
        for (int i = 0; i < nums.length; i++)
            res.add(nums[i]);
        Collections.sort(res, new customSort());
        return res.get(k - 1);
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here