Let's clarify and summarize the content from your image and text, focusing on the **Element Uniqueness Problem** and its **time complexity** analysis. --- ## Example 3: Element Uniqueness Problem **Problem Statement:** Given an array, determine if all elements are unique (i.e., are there any duplicate elements?). --- ### Algorithm Outline (Brute-force Approach) For each pair of elements in the array, compare them: 1. Loop through each element `i` from `0` to `n-2` 2. For each `i`, loop through each element `j` from `i+1` to `n-1` 3. Compare `A[i]` and `A[j]` - If any pair is equal, the array does **not** have all unique elements --- ### Best Case - **Occurs when**: The first two elements are the same. - **Number of comparisons**: Only **1** comparison is needed. --- ### Worst Case - **Occurs when**: - All elements are unique **OR** - Only the last two elements are equal - **Number of comparisons**: For each pair `(i, j)` with `i < j`: - Total comparisons = sum over all possible pairs #### Calculation: \[ \sum_{i=0}^{n-2} \sum_{j=i+1}^{n-1} 1 = \sum_{i=0}^{n-2} (n-1-i) \] This simplifies to: \[ (n-1) + (n-2) + \ldots + 1 = \frac{n(n-1)}{2} \] --- ### Asymptotic Complexity - **Worst-case time complexity:** \[ O(n^2) \] --- ### Summary Table | Case | # Comparisons | Complexity | |------------|----------------------|---------------| | Best | 1 | O(1) | | Worst | \( \frac{n(n-1)}{2} \) | \( O(n^2) \) | --- **In summary:** The brute-force approach to the Element Uniqueness Problem has a best case of O(1) and a worst case of O(n²) time complexity. If you need a more efficient approach, you can use hashing or sorting, which can reduce the time complexity to O(n log n) or O(n) depending on the method and data structure used. --- **Let me know if you need a step-by-step algorithm or code as well!**