Count Elements Whose Next Element is Also in the Array in Python

Suppose we have a list of numbers say nums, we have to find the number of elements x in the array, such that x + 1 also exists in the array.

So, if the input is like nums = [4, 2, 3, 3, 7, 9], then the output will be 3, because 2+1 = 3 is present, 3+1 = 4 is present and another 3 is present so total 3.

To solve this, we will follow these steps −

• answer := 0

• c := a list containing frequencies of each elements present in nums

• dlist := a list from list of all keys of c

• for each i in dlist, do

  • if c[i + 1] > 0, then

    • answer := answer + c[i]

• return answer

Example

Let us see the following implementation to get better understanding

from collections import Counter
def solve(nums):
   answer = 0
   c = Counter(nums)
   dlist = list(c.keys())
   for i in dlist:
      if c[i + 1] > 0:
         answer += c[i]

   return answer

nums = [4, 2, 3, 3, 7, 9]
print(solve(nums))

Input

[4, 2, 3, 3, 7, 9]

Output

3