-
Notifications
You must be signed in to change notification settings - Fork 215
/
Copy pathsum_of_unique_elements.java
57 lines (46 loc) · 1.22 KB
/
sum_of_unique_elements.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
import java.util.HashMap;
/*
You are given an integer array nums. The unique elements of an array are the elements that appear exactly once in the array.
Return the sum of all the unique elements of nums.
Example 1:
Input: nums = [1,2,3,2]
Output: 4
Explanation: The unique elements are [1,3], and the sum is 4.
Example 2:
Input: nums = [1,1,1,1,1]
Output: 0
Explanation: There are no unique elements, and the sum is 0.
Example 3:
Input: nums = [1,2,3,4,5]
Output: 15
Explanation: The unique elements are [1,2,3,4,5], and the sum is 15.
Constraints:
1 <= nums.length <= 100
1 <= nums[i] <= 100
*/
class Solution {
public int sumOfUnique(int[] nums) {
HashMap<Integer,Integer> hm=new HashMap<Integer,Integer>();
int sum=0;
for(int x:nums)
{
if(hm.containsKey(x))
hm.put(x,hm.get(x)+1);
else
hm.put(x,1);
}
for(int x:nums)
{
if(hm.get(x)== 1)
sum+=x;
}
return sum;
}
}
class SumOfUniqueElements {
public static void main(String[] args) {
int arr[] = {1,2,2,3};
Solution s = new Solution();
System.out.println(s.sumOfUnique(arr));
}
}