LeetCode-in-Java

1985. 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 4^th 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 3^rd 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 2^nd 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.Arrays;

@SuppressWarnings("java:S2234")
public 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;
    }
}

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