---
# System prepended metadata

title: Leetcode 2089. Find Target Indices After Sorting Array
tags: [Leetcode]

---

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###### tags: `Leetcode`
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# Leetcode 2089. Find Target Indices After Sorting Array


[link](https://leetcode.com/problems/find-target-indices-after-sorting-array/)

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You are given a 0-indexed integer array nums and a target element target.

A target index is an index i such that nums[i] == target.

Return a list of the target indices of nums after sorting nums in non-decreasing order. If there are no target indices, return an empty list. The returned list must be sorted in increasing order.

#### Example 1:

Input: nums = [1,2,5,2,3], target = 2
Output: [1,2]
Explanation: After sorting, nums is [1,2,2,3,5].
The indices where nums[i] == 2 are 1 and 2.

#### Example 2:

Input: nums = [1,2,5,2,3], target = 3
Output: [3]
Explanation: After sorting, nums is [1,2,2,3,5].
The index where nums[i] == 3 is 3.

#### Example 3:

Input: nums = [1,2,5,2,3], target = 5
Output: [4]
Explanation: After sorting, nums is [1,2,2,3,5].
The index where nums[i] == 5 is 4.

#### Constraints:
- 1 <= nums.length <= 100
- 1 <= nums[i], target <= 100

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題意: 給定一個列表，以小到大排序後，返回以target為值的index

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After sorting the list, iterate the list and then store the index of each element to the new list if that element is equal to the target.


#### Solution 1
```python=
class Solution:
    def targetIndices(self, nums: List[int], target: int) -> List[int]:
        nums.sort()
        lst = []
        for i in range(len(nums)):
            if nums[i] == target:
                lst.append(i)

        return lst
```
O(T): O(nlogn)
O(S): O(n)