Formatted question description: https://leetcode.ca/all/2089.html

# 2089. Find Target Indices After Sorting Array (Easy)

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


Example 4:

Input: nums = [1,2,5,2,3], target = 4
Output: []
Explanation: There are no elements in nums with value 4.


Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i], target <= 100

Similar Questions:

## Solution 1. Sorting

// OJ: https://leetcode.com/problems/find-target-indices-after-sorting-array/
// Time: O(NlogN)
// Space: O(1) extra space
class Solution {
public:
vector<int> targetIndices(vector<int>& A, int target) {
sort(begin(A), end(A));
vector<int> ans;
for (int i = 0; i < A.size(); ++i) {
if (A[i] == target) ans.push_back(i);
}
return ans;
}
};


## Solution 2. Counting Sort

// OJ: https://leetcode.com/problems/find-target-indices-after-sorting-array/
// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<int> targetIndices(vector<int>& A, int target) {
int cnt = 0, rank = 0; // cnt is the frequency of target, rank is the sum of frequencies of all numbers < target
for (int n : A) {
cnt += n == target;
rank += n < target;
}
vector<int> ans;
while (cnt--) ans.push_back(rank++);
return ans;
}
};


## Discuss

https://leetcode.com/problems/find-target-indices-after-sorting-array/discuss/1599800/C%2B%2B-O(N)-Time-Counting-Sort