comments | difficulty | edit_url | rating | source | tags | ||||||
---|---|---|---|---|---|---|---|---|---|---|---|
true |
Medium |
1414 |
Weekly Contest 256 Q2 |
|
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.
class Solution:
def kthLargestNumber(self, nums: List[str], k: int) -> str:
def cmp(a, b):
if len(a) != len(b):
return len(b) - len(a)
return 1 if b > a else -1
nums.sort(key=cmp_to_key(cmp))
return nums[k - 1]
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];
}
}
class Solution {
public:
string kthLargestNumber(vector<string>& nums, int k) {
auto cmp = [](const string& a, const string& b) { return a.size() == b.size() ? a > b : a.size() > b.size(); };
sort(nums.begin(), nums.end(), cmp);
return nums[k - 1];
}
};
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]
}