Description
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.
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 <= 1051 <= queries[i] <= 1051 <= nums[i], x <= 104
Solution
Python3
class Solution:
def occurrencesOfElement(self, nums: List[int], queries: List[int], t: int) -> List[int]:
res = []
A = []
for i, x in enumerate(nums):
if x == t:
A.append(i)
for q in queries:
q -= 1
if q < len(A):
res.append(A[q])
else:
res.append(-1)
return res