Counting elements¶
Time: O(N); Space: O(N); easy
Given an integer array arr, count how many elements x there are, such that x + 1 is also in arr.
If there’re duplicates in arr, count them seperately.
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 <= len(arr) <= 1000
0 <= arr[i] <= 1000
[1]:
class Solution1(object):
"""
Time: O(N)
Space: O(N)
"""
def countElements(self, arr):
"""
:type arr: List[int]
:rtype: int
"""
lookup = set(arr)
return sum(1 for x in arr if x+1 in lookup)
[2]:
s = Solution1()
arr = [1,2,3]
assert s.countElements(arr) == 2
arr = [1,1,3,3,5,5,7,7]
assert s.countElements(arr) == 0
arr = [1,3,2,3,5,0]
assert s.countElements(arr) == 3
arr = [1,1,2,2]
assert s.countElements(arr) == 2
arr = [1,1,2]
assert s.countElements(arr) == 2
[3]:
class Solution2(object):
"""
Time: O(NLogN)
Space: O(1)
"""
def countElements(self, arr):
"""
:type arr: List[int]
:rtype: int
"""
arr.sort()
result, l = 0, 1
for i in range(len(arr)-1):
if arr[i] == arr[i+1]:
l += 1
continue
if arr[i]+1 == arr[i+1]:
result += l
l = 1
return result
[4]:
s = Solution1()
arr = [1,2,3]
assert s.countElements(arr) == 2
arr = [1,1,3,3,5,5,7,7]
assert s.countElements(arr) == 0
arr = [1,3,2,3,5,0]
assert s.countElements(arr) == 3
arr = [1,1,2,2]
assert s.countElements(arr) == 2
arr = [1,1,2]
assert s.countElements(arr) == 2