Question
Given an integer array arr, count element x such that x + 1 is also in arr.
If there're duplicates in arr, count them seperately.
Example
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.
Constraints:
- 1 <= arr.length <= 1000
- 0 <= arr[i] <= 1000
Code C
int countElements(int *arr, int arrSize)
{
int data[1002] = {0};
for (int i = 0; i < arrSize; ++i)
{
data[arr[i]] = 1;
}
int sum = 0;
for (int i = 0; i < arrSize; ++i)
{
if (data[arr[i] + 1] == 1)
{
++sum;
}
}
return sum;
}
Code Golang
func countElements(arr []int) int {
data := make([]bool, 1002)
for _, v := range arr {
data[v] = true
}
sum := 0
for _, v := range arr {
if data[v+1] {
sum = sum + 1
}
}
return sum
}
Code Python
@Ricky
class Solution:
def countElements(self, arr: List[int]) -> int:
count=0
for x in arr:
if x+1 in arr:
count += 1
return count