find all pairs possible from the given array

Okay, let's dive deep into finding all possible pairs from a given array. This is a fundamental problem with various applications, and understanding different approaches is crucial. We'll cover several methods, discuss their time and space complexities, and provide code examples in Python (because it's versatile and easy to understand).

**Understanding the Problem**

The core problem is: Given an array (or list) of elements, we want to generate all unique combinations of pairs from that array. The order of elements within a pair *usually* doesn't matter (e.g., `(a, b)` is the same as `(b, a)`). We'll address situations where order *does* matter too.

**Important Considerations:**

* **Uniqueness:** Do we want to include pairs like `(x, x)` where the element is paired with itself? By default, we'll assume we *don't* want self-pairs.
* **Order:** Does the order of elements in a pair matter? If it does, we're looking for *permutations* instead of *combinations*. If it doesn't, we're looking for *combinations*.
* **Duplication:** Are there duplicate elements in the input array? If so, do we want to treat pairs with duplicate elements as distinct pairs?

**Method 1: Nested Loops (Brute Force)**

This is the simplest and most intuitive approach. We iterate through the array using nested loops to compare each element with every other element.

* **Explanation:**
* The outer loop iterates from `i = 0` to `n-1`.
* The inner loop iterates from `j = i+1` to `n-1`. Starting `j` at `i+1` ensures that we don't create self-pairs (`(arr[i], arr[i])`) and that we only include each pair once (e.g., we get `(1, 2)` but not `(2, 1)`).
* For each combination of `i` and `j`, we add the pair `(arr[i], arr[j])` to the `pairs` list.

* **Time Complexity:** O(nsup2/sup) - The nested loops iterate over all possible pairs.
* **Space Complexity:** O(m), where 'm' is the number of pairs. In the worst case, m can be O(nsup2/sup). This is for stor ...

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