2053. Kth Distinct String in an Array

2053. Kth Distinct String in an Array

A distinct string is a string that is present only once in an array.

Given an array of strings arr, and an integer k, return the kth distinct string present in arr. 
If there are fewer than k distinct strings, return an empty string "".

Note that the strings are considered in the order in which they appear in the array.

 

Example 1:

Input: arr = ["d","b","c","b","c","a"], k = 2
Output: "a"
Explanation:
The only distinct strings in arr are "d" and "a".
"d" appears 1st, so it is the 1st distinct string.
"a" appears 2nd, so it is the 2nd distinct string.
Since k == 2, "a" is returned. 

Example 2:

Input: arr = ["aaa","aa","a"], k = 1
Output: "aaa"
Explanation:
All strings in arr are distinct, so the 1st string "aaa" is returned.
Example 3:

Input: arr = ["a","b","a"], k = 3
Output: ""
Explanation:
The only distinct string is "b". Since there are fewer than 3 distinct strings, 
we return an empty string "".
 

Constraints:

1 <= k <= arr.length <= 1000
1 <= arr[i].length <= 5
arr[i] consists of lowercase English letters.

难度 : Easy

思路

先用HashMap来记录每个string出现的次数然后遍历数组得到第k个distinct的string
注意edge case: 如果k非常大直接返回””.

class Solution {
    public String kthDistinct(String[] arr, int k) {
        if (k > arr.length) {
            return "";
        }
        Map<String, Integer> counts = new HashMap<>();
        for (String str : arr) {
            int cnt = counts.getOrDefault(str, 0) + 1;            
            counts.put(str, cnt);
        }
        int index = 1;
        for (int i = 0; i < arr.length && index <= k; i++) {
            if (counts.get(arr[i]) == 1) {
                if (index == k) {
                    return arr[i];
                }
                index++;
            }
            
        }
        return "";
    }
}