3159. Find Occurrences of an Element in an Array

#Problem

You are given an integer array nums, an integer array queries, and an integer x.

For each queries[i], you need to find the index of the queries[i]ᵗʰ occurrence of x in the nums array. If there are fewer than queries[i] occurrences of x, the answer should be -1 for that query.

Return an integer array answer containing the answers to all queries.

https://leetcode.cn/problems/find-occurrences-of-an-element-in-an-array/

Example 1:

Input: nums = [1,3,1,7], queries = [1,3,2,4], x = 1
Output: [0,-1,2,-1]
Explanation:

• For the 1ˢᵗ query, the first occurrence of 1 is at index 0.
• For the 2ⁿᵈ query, there are only two occurrences of 1 in nums, so the answer is -1.
• For the 3ʳᵈ query, the second occurrence of 1 is at index 2.
• For the 4ᵗʰ query, there are only two occurrences of 1 in nums, so the answer is -1.

Example 2:

Input: nums = [1,2,3], queries = [10], x = 5
Output: [-1]
Explanation:

• For the 1ˢᵗ query, 5 doesn’t exist in nums, so the answer is -1.

Constraints:

• 1 <= nums.length, queries.length <= 10⁵
• 1 <= queries[i] <= 10⁵
• 1 <= nums[i], x <= 10⁴

#Test Cases

class Solution:
    def occurrencesOfElement(self, nums: List[int], queries: List[int], x: int) -> List[int]:

import pytest

from solution import Solution


@pytest.mark.parametrize('nums, queries, x, expected', [
    ([1,3,1,7], [1,3,2,4], 1, [0,-1,2,-1]),
    ([1,2,3], [10], 5, [-1]),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, nums, queries, x, expected):
    assert sol.occurrencesOfElement(nums, queries, x) == expected

#Thoughts

直接按题意实现即可（这不应该是 easy 么）。

在 Python 里，直接用 list comprehension 比用 list(filter(lambda ..., nums)) 快不少。

#Code

class Solution:
    def occurrencesOfElement(self, nums: list[int], queries: list[int], x: int) -> list[int]:
        occurs = [i for i, num in enumerate(nums) if num == x]
        freq = len(occurs)
        return [occurs[j-1] if j <= freq else -1 for j in queries]