Hello, coding enthusiast! Welcome to a new challenge in your journey to mastering programming and problem-solving. Today, we're diving into combinatorial problems that involve working with large datasets and multiple pairs of numbers. We'll learn to solve significant problems efficiently using smart data structures like PHP associative arrays, which will help us avoid expensive operations such as iterating over large arrays. Are you ready? Let's dive in!

In this unit's task, you will be given a large array composed of pairs of distinct, positive integers, with up to 1,000,000 elements. Your challenge is to write a PHP function to count the number of indices (i, j) (i not equals j) where the i-th pair does not share a common element with the j-th pair. A crucial point to remember is that a pair (a, b) is considered identical to (b, a), meaning the order of elements in a pair is irrelevant in this case. It is guaranteed that no two pairs are element-wise equal.

For example, given the array [[2, 5], [1, 6], [3, 2], [4, 2], [5, 1], [6, 3]], the output should be 8. The required index pairs are the following: (0, 1), (0, 5), (1, 2), (1, 3), (2, 4), (3, 4), (3, 5), (4, 5).

At the core of our solution, we'll leverage combinatorial logic and a clever way of keeping track of occurrences to solve this problem efficiently.

The central idea is to calculate the total number of pairs and then subtract from this total the number of pairs that share a common element. This will give us the count of pairs that do not share a common element, which is what we're after.

Firstly, we will calculate the total number of pairs possible in the array using the formula $n \sdot (n - 1) / 2$, where each element can pair with every other element, and we divide by 2 because the order of pairs doesn't matter (i.e., pair (a, b) is identical to pair (b, a)).

Secondly, we'll count the number of pairs that have at least one common element. To do this, we utilize PHP associative arrays to track each number's appearance in the pairs and calculate how many pairs it appears in.

The first step in our solution is to store the occurrence of each number in the pairs using PHP associative arrays. Next, we calculate the total number of pairs using the formula $n \sdot (n - 1) / 2$. We'll need this for our final calculation.

php
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1<?php
2
3function nonCommonPairs($pairs) {
4    $indices = [];
5    $totalPairs = count($pairs) * (count($pairs) - 1) / 2;
6    
7    // (The rest of the code will be added in subsequent steps)
8    return $totalPairs; // Temporary return to avoid errors
9}
10
11// Testing code will be added later
12
13?>

Next, we populate the associative array by iterating over the array of pairs. For each pair, we'll examine its two elements and either append the current index to the list of indices for this number (if it’s already in the associative array) or start a new list for it (if it isn't).

php
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1<?php
2
3function nonCommonPairs($pairs) {
4    $indices = [];
5    $totalPairs = count($pairs) * (count($pairs) - 1) / 2;
6    
7    foreach ($pairs as $index => $pair) {
8        foreach ($pair as $num) {
9            if (!isset($indices[$num])) {
10                $indices[$num] = [];
11            }
12            $indices[$num][] = $index;
13        }
14    }
15    
16    // (The rest of the code will be added in the next step)
17    return $totalPairs; // Temporary return to avoid errors
18}
19
20?>

Finally, with all the data in place, we calculate the total pairs of indices that share at least one common element. For each number, we count how many pairs it appears in, using the previously filled associative array. We then subtract these common pairs from the total pairs to get our answer — the count of pairs without a common number.

php
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1<?php
2
3function nonCommonPairs($pairs) {
4    $indices = [];
5    $totalPairs = count($pairs) * (count($pairs) - 1) / 2;
6    
7    foreach ($pairs as $index => $pair) {
8        foreach ($pair as $num) {
9            if (!isset($indices[$num])) {
10                $indices[$num] = [];
11            }
12            $indices[$num][] = $index;
13        }
14    }
15
16    $commonPairs = 0;
17    foreach ($indices as $list) {
18        $size = count($list);
19        $commonPairs += $size * ($size - 1) / 2;
20    }
21    
22    return $totalPairs - $commonPairs;
23}
24
25// Test the function with a sample array
26$pairs = [[2, 5], [1, 6], [3, 2], [4, 2], [5, 1], [6, 3]];
27echo "Count of non-common pairs: " . nonCommonPairs($pairs);
28
29?>

Excellent work! Today's challenge was a tough one, but you successfully navigated through it. You efficiently tracked occurrences within a large dataset using PHP associative arrays and applied combinatorial reasoning to subtract the opposite case from the total possibilities. Consequently, you came up with a solution that operates efficiently.

This knowledge will serve you well in solving similar complex problems in the future. Remember, the best way to handle large data is to apply clever techniques that sidestep unnecessary computations, just like we did today.

Now, it's time to solidify your understanding. Up next are practice problems related to this lesson. Start working on them to reinforce these concepts. Happy coding!