Finding Lucky Numbers in a List Using Python: Step‑by‑Step with map, zip, filter and a One‑Liner

The article introduces a LeetCode problem called "Lucky Numbers in a List" where a number is considered lucky if its frequency in the list equals the number itself (e.g., in [1, 2, 2, 3] the lucky numbers are 1 and 2).

Step 1 – Get unique elements : The simplest way to remove duplicates is to convert the list to a set. Example:

arr = [3,5,2,7,3,8,1,2,4,8,9,3]
unique = set(arr)
print(unique)  # {1, 2, 3, 4, 5, 7, 8, 9}

Step 2 – Count occurrences : Each element’s count can be obtained with the list method count(). The article first shows a manual loop, then replaces the loop with map to produce a generator of counts.

<code>pairs = []
for i in unique:
    pairs.append((i, arr.count(i)))
print(pairs)  # [(1, 1), (2, 2), (3, 3), (4, 1), (5, 1), (7, 1), (8, 2), (9, 1)]

m = map(arr.count, unique)          # generator of counts
z = zip(unique, m)                  # pairs each element with its count

Step 3 – Filter lucky numbers : Instead of iterating, the filter function together with a lambda expression selects only those pairs where the element equals its count. The result is then sorted.

f = filter(lambda x: x[0] == x[1], z)
s = sorted(f, key=lambda x: x[0])
print('Lucky numbers are:', [item[0] for item in s])
# Output: Lucky numbers are: [1, 2, 3]

One‑liner version : All the steps can be collapsed into a single expression that directly prints the lucky numbers.

print('Lucky numbers are:', [item[0] for item in sorted(filter(lambda x: x[0]==x[1], zip(set(arr), map(arr.count, set(arr)))), key=lambda x:x[0])])
# Output: Lucky numbers are: [1, 2, 3]

The author reflects that while the one‑liner looks clever, a more readable multi‑step solution is often preferable for clarity.