Given an integer array arr
, count how many elements x
there are, such that x + 1
is also in arr
. If there are duplicates in arr
, count them separately.
Example 1:
Input: arr = [1,2,3] Output: 2 Explanation: 1 and 2 are counted cause 2 and 3 are in arr.
Example 2:
Input: arr = [1,1,3,3,5,5,7,7] Output: 0 Explanation: No numbers are counted, cause there's no 2, 4, 6, or 8 in arr.
Example 3:
Input: arr = [1,3,2,3,5,0] Output: 3 Explanation: 0, 1 and 2 are counted cause 1, 2 and 3 are in arr.
Example 4:
Input: arr = [1,1,2,2] Output: 2 Explanation: Two 1s are counted cause 2 is in arr.
Example 5:
Input: arr = [1,1,2] Output: 2 Explanation: Both 1s are counted because 2 is in the array.
Constraints:
1 <= arr.length <= 1000
0 <= arr[i] <= 1000
class Solution:
def countElements(self, arr: List[int]) -> int:
s = set(arr)
res = 0
for num in arr:
if num + 1 in s:
res += 1
return res
class Solution {
public int countElements(int[] arr) {
Set<Integer> s = new HashSet<>();
for (int num : arr) {
s.add(num);
}
int res = 0;
for (int num : arr) {
if (s.contains(num + 1)) {
++res;
}
}
return res;
}
}
class Solution {
public:
int countElements(vector<int>& arr) {
unordered_set<int> s;
for (int num : arr) s.insert(num);
int res = 0;
for (int num : arr)
if (s.count(num + 1)) ++res;
return res;
}
};
func countElements(arr []int) int {
s := make(map[int]bool)
for _, num := range arr {
s[num] = true
}
res := 0
for _, num := range arr {
if s[num+1] {
res++
}
}
return res
}