Untitled

Deciding to use **counting sort** in this problem comes from observing the **constraints and problem requirements**. Let’s break down the reasoning and key insights:

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### Key Observations:
1. **Sorting isn't the focus**:
   - The goal is not to fully sort the array but to find the indices of elements equal to the `target` *after sorting*.
   - Counting sort avoids the overhead of sorting the array entirely, as it focuses only on frequency distribution.

2. **Nature of the Input**:
   - **Positive Integers**: The array `nums` consists of integers in the range `[1, 100]` (small range, bounded values).
   - Counting sort thrives when the range of possible values is small because it uses a frequency array.

3. **Output Requirements**:
   - The problem involves identifying indices based on frequencies, not rearranging the entire array.
   - Counting sort inherently aligns with frequency-based solutions since it accumulates counts to determine relative positions.

4. **Efficiency Focus**:
   - Sorting the array directly (e.g., via quicksort or mergesort) has a time complexity of \(O(n \log n)\).
   - Counting sort leverages the small range of values to achieve \(O(n)\), making it faster for this scenario.

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### When to Apply Counting Sort:
Counting sort is effective under these conditions:
1. **Input Range is Small**:
   - The range of input values (difference between the smallest and largest value) should be manageable.
   - For instance, here the range is \(1 \leq \text{nums[i]} \leq 100\), which is much smaller than the array size \(n\).

2. **Integer Inputs**:
   - Counting sort works best when the elements are discrete integers.
   - It doesn’t perform well with floating-point numbers or very large ranges without adjustments.

3. **Frequency-Based Problems**:
   - If the task involves counting occurrences, determining ranks, or segmenting data by value, counting sort is a natural fit.
   - Examples: Frequency histograms, ranking scores, or finding specific indices like in this problem.

4. **Sorting Isn't the End Goal**:
   - If we’re solving problems where sorting the array is incidental (e.g., finding specific positions, counts, or relationships), counting sort avoids unnecessary computations.

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### Insights from the Problem:
- **Constraints**:
  - \(1 \leq \text{nums.length} \leq 100,000\): Sorting with \(O(n \log n)\) methods may work but can be suboptimal.
  - \(1 \leq \text{nums[i]} \leq 100\): Small range of input values is ideal for counting sort.
- **Output Requirements**:
  - The output depends only on the indices of the `target` in the *sorted version*. We don’t need the entire sorted array, just the positional insights, which counting sort can derive efficiently.

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### Why Counting Sort Stands Out Here:
1. **Avoid Full Sorting**:
   - It avoids rearranging elements and focuses only on counts, saving computation time.
2. **Direct Mapping**:
   - Frequencies directly translate to indices, making it a natural fit for this problem.

By leveraging these observations, counting sort provides an elegant, efficient solution tailored to the problem's constraints.

To solve this problem efficiently, we can avoid sorting the entire array by using counting sort. Counting sort is particularly useful here because the problem involves small, positive integers that are easy to count and process.

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### Approach:

1. **Count Frequencies**:
   - Use a frequency array to count occurrences of each number in `nums`.

2. **Identify Target Indices**:
   - Calculate the starting index of each number in the sorted version of `nums` based on cumulative counts.
   - Use the count of the target element to determine its indices in the sorted array.

3. **Efficiency**:
   - Avoid explicitly sorting the array, which reduces time complexity.

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### Implementation:
```java
import java.util.*;

class Solution {
    public List<Integer> targetIndices(int[] nums, int target) {
        // Step 1: Count frequencies of each number
        int[] count = new int[101]; // Since nums[i] ∈ [1, 100]
        for (int num : nums) {
            count[num]++;
        }
        
        // Step 2: Calculate starting index of each number
        int startIndex = 0;
        for (int i = 0; i < target; i++) {
            startIndex += count[i];
        }
        
        // Step 3: Generate the result list for the target
        List<Integer> result = new ArrayList<>();
        for (int i = 0; i < count[target]; i++) {
            result.add(startIndex + i);
        }
        
        return result;
    }
}
```

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### Explanation:

#### Example 1:
**Input**: `nums = [1, 2, 5, 2, 3]`, `target = 2`

1. **Count Frequencies**:
   ```
   count = [0, 1, 2, 1, 0, 1] // Count of numbers from 1 to 5
   ```

2. **Calculate Start Index**:
   - `startIndex` for `target = 2` is the sum of counts of numbers less than 2:
     ```
     startIndex = count[1] = 1
     ```

3. **Generate Result**:
   - Indices for `2` in the sorted array are:
     ```
     [1, 2]
     ```

**Output**: `[1, 2]`

#### Example 2:
**Input**: `nums = [1, 2, 5, 2, 3]`, `target = 3`

1. **Count Frequencies**:
   ```
   count = [0, 1, 2, 1, 0, 1]
   ```

2. **Calculate Start Index**:
   - `startIndex` for `target = 3`:
     ```
     startIndex = count[1] + count[2] = 1 + 2 = 3
     ```

3. **Generate Result**:
   - Indices for `3` in the sorted array:
     ```
     [3]
     ```

**Output**: `[3]`

---

### Complexity Analysis:
1. **Time Complexity**:
   - Counting: \(O(n)\) where \(n\) is the size of `nums`.
   - Index Calculation: \(O(100) \approx O(1)\).
   - Result Construction: \(O(k)\) where \(k\) is the count of the target.

   Overall: \(O(n + k)\).

2. **Space Complexity**:
   - \(O(101) = O(1)\) for the frequency array.

This approach avoids sorting, making it more efficient for large arrays.