LeetCode in Kotlin
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 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 <= 1041 <= nums[i].length <= 100nums[i]consists of only digits.nums[i]will not have any leading zeros.
Solution
class Solution {
fun kthLargestNumber(nums: Array<String>, k: Int): String {
nums.sortWith { n1: String, n2: String -> compareStringInt(n2, n1) }
return nums[k - 1]
}
private fun compareStringInt(n1: String, n2: String): Int {
if (n1.length != n2.length) {
return if (n1.length < n2.length) -1 else 1
}
for (i in n1.indices) {
val n1Digit = n1[i].code - '0'.code
val n2Digit = n2[i].code - '0'.code
if (n1Digit > n2Digit) {
return 1
} else if (n2Digit > n1Digit) {
return -1
}
}
return 0
}
}