Program to count elements whose next element also in the array in Python

Given an array of numbers, we need to count how many elements x have their next consecutive number (x + 1) also present in the array. This problem involves frequency counting and checking for consecutive elements.

Problem Understanding

For the input nums = [4, 2, 3, 3, 7, 9], we need to find elements whose next consecutive number exists:

• Element 2: 2 + 1 = 3 exists in array (count: 1)

• Element 3: 3 + 1 = 4 exists in array (count: 2, since 3 appears twice)

• Element 4: 4 + 1 = 5 does not exist

• Element 7: 7 + 1 = 8 does not exist

• Element 9: 9 + 1 = 10 does not exist

Total count = 1 + 2 = 3

Solution Using Counter

We use Python's Counter to track element frequencies, then check if each element's successor exists ?

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]
result = solve(nums)
print(f"Input: {nums}")
print(f"Output: {result}")

Input: [4, 2, 3, 3, 7, 9]
Output: 3

Alternative Solution Using Set

For cases where we only need to check existence (not frequencies), a set-based approach works efficiently ?

def solve_with_set(nums):
    num_set = set(nums)
    count = 0
    
    for num in nums:
        if num + 1 in num_set:
            count += 1
    
    return count

nums = [4, 2, 3, 3, 7, 9]
result = solve_with_set(nums)
print(f"Input: {nums}")
print(f"Output: {result}")

Output: 3

Step-by-Step Algorithm

1. Create a frequency counter of all elements in the array

2. For each unique element x in the array:

   • Check if x + 1 exists in the counter

   • If yes, add the frequency of x to the answer

3. Return the total count

Comparison

Conclusion

Use Counter when element frequencies matter, as it properly handles duplicate elements. The set-based approach offers a cleaner solution when only checking for consecutive number existence.