Program to find elements from list which have occurred at least k times in Python

Sometimes we need to find elements from a list that appear a certain number of times or more. Python provides several approaches to solve this problem using frequency counting techniques.

So, if the input is like nums = [2,5,6,2,6,1,3,6,3,8,2,5,9,3,5,1] k = 3, then the output will be [2, 5, 6, 3]

Using Counter from collections

The Counter class provides an efficient way to count element frequencies ?

from collections import Counter

def solve(nums, k):
    c = Counter(nums)
    res = []
    for n in c:
        if c[n] >= k:
            res.append(n)
    return res

nums = [2,5,6,2,6,1,3,6,3,8,2,5,9,3,5,1]
k = 3
print(solve(nums, k))

[2, 5, 6, 3]

Using Dictionary for Manual Counting

We can manually count frequencies using a dictionary ?

def solve_manual(nums, k):
    freq = {}
    # Count frequencies
    for num in nums:
        freq[num] = freq.get(num, 0) + 1
    
    # Find elements with frequency >= k
    result = []
    for num, count in freq.items():
        if count >= k:
            result.append(num)
    
    return result

nums = [2,5,6,2,6,1,3,6,3,8,2,5,9,3,5,1]
k = 3
print(solve_manual(nums, k))

[2, 5, 6, 3]

Using List Comprehension

A more concise approach using list comprehension with Counter ?

from collections import Counter

def solve_compact(nums, k):
    c = Counter(nums)
    return [num for num, count in c.items() if count >= k]

nums = [2,5,6,2,6,1,3,6,3,8,2,5,9,3,5,1]
k = 3
print(solve_compact(nums, k))

[2, 5, 6, 3]

Comparison

Conclusion

Use Counter for the most readable solution. All methods have O(n) time complexity and effectively solve the frequency counting problem. Choose based on your preference for readability versus conciseness.