1985. Find the Kth Largest Integer in the Array

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Description

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.

Solutions

Solution 1: Sorting or Quickselect

We can sort the strings in the $\textit{nums}$ array in descending order as integers, and then take the $k$-th element. Alternatively, we can use the quickselect algorithm to find the $k$-th largest integer.

The time complexity is $O(n \times \log n)$ or $O(n)$, where $n$ is the length of the $\textit{nums}$ array. The space complexity is $O(\log n)$ or $O(1)$.

Python3

class Solution:
    def kthLargestNumber(self, nums: List[str], k: int) -> str:
        return nlargest(k, nums, key=lambda x: int(x))[k - 1]

Java

class Solution {
    public String kthLargestNumber(String[] nums, int k) {
        Arrays.sort(
            nums, (a, b) -> a.length() == b.length() ? b.compareTo(a) : b.length() - a.length());
        return nums[k - 1];
    }
}

C++

class Solution {
public:
    string kthLargestNumber(vector<string>& nums, int k) {
        nth_element(nums.begin(), nums.begin() + k - 1, nums.end(), [](const string& a, const string& b) {
            return a.size() == b.size() ? a > b : a.size() > b.size();
        });
        return nums[k - 1];
    }
};

Go

func kthLargestNumber(nums []string, k int) string {
	sort.Slice(nums, func(i, j int) bool {
		a, b := nums[i], nums[j]
		if len(a) == len(b) {
			return a > b
		}
		return len(a) > len(b)
	})
	return nums[k-1]
}