find target indices after sorting array

Okay, let's dive into the "Find Target Indices After Sorting Array" problem. I'll provide a comprehensive explanation, different approaches, a detailed code example (in Python, but easily adaptable), and discuss time/space complexity.

**Problem Statement:**

Given an integer array `nums` and an integer `target`, sort `nums` in **non-decreasing** order. You need to find all indices `i` in the sorted array such that `nums[i]` is equal to `target`. Return a list of these indices in **ascending order**.

**Example:**

**Understanding the Problem**

The core task is to:

1. **Sort the input array:** Arrange the elements in ascending order.
2. **Find target indices:** Locate all positions in the *sorted* array where the element is equal to the given `target`.
3. **Return indices in order:** Return the list of indices in ascending order.

**Approaches & Strategies**

Here are the primary approaches we can take:

1. **Sort & Linear Scan (Most Common and Recommended):**

- **Algorithm:**
1. Sort the input array `nums` using a standard sorting algorithm (e.g., merge sort, quicksort, or the built-in sorting function in your language, which is typically highly optimized).
2. Iterate through the sorted array.
3. If `nums[i] == target`, add the index `i` to the result list.
4. Return the result list.

- **Advantages:**
- Simple to understand and implement.
- Efficient (particularly if using a good sorting algorithm).

- **Disadvantages:**
- Sorting can take O(n log n) time in the worst case.

2. **Counting Sort (If Target and Elements are Within a Limited Range):**

- **Algorithm:**
1. Create a "count" array to store the frequency of each number within the range of `nums`.
2. Count the frequency of each number within nums, and store in the 'count' array
3. Iterate through the 'count' array, and count how many numbers in nums are lower than target.
4. Count the amount of target in nums, and store in a variabl ...

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