From 9066f45bb0f96359ca44e3fa2eb779e9e9cd0927 Mon Sep 17 00:00:00 2001
From: Jörg Sommrey <28217714+jo-37@users.noreply.github.com>
Date: Mon, 20 Nov 2023 19:35:01 +0100
Subject: Blog for week 244

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+https://github.com/manwar/perlweeklychallenge-club/blob/master/challenge-244/jo-37/blog/Blog.md
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+# Smaller Heroes
+
+## Task 1: Count Smaller
+**Submitted by: Mohammad S Anwar**
+
+---
+You are given an array of integers.
+
+Write a script to calculate the number of integers smaller than the integer at each index.
+
+### Example 1
+```
+Input: @int = (8, 1, 2, 2, 3)
+Output: (4, 0, 1, 1, 3)
+
+For index = 0, count of elements less 8 is 4.
+For index = 1, count of elements less 1 is 0.
+For index = 2, count of elements less 2 is 1.
+For index = 3, count of elements less 2 is 1.
+For index = 4, count of elements less 3 is 3.
+```
+### Example 2
+```
+Input: @int = (6, 5, 4, 8)
+Output: (2, 1, 0, 3)
+```
+### Example 3
+```
+Input: @int = (2, 2, 2)
+Output: (0, 0, 0)
+```
+---
+### Solution
+The number of integers smaller than a member of a list is just the number's statistical rank reduced by 1.
+Similar rank calculations have already been done in task 2 from week 9 and in task 1 from week 214.
+Taking a different approach here.
+
+The module `List::Rank` has a `rank` subroutine that can be used with a little post processing:
+
+ 1. Equal scores an equal sign appended to their rank that needs to be removed.
+ 2. The rank is one more than the number of smaller values in the list.
+
+This leads to a very terse implementation:
+```
+sub count_smaller {
+	map tr/=//dr - 1, rank @_;
+}
+```
+## Task 2: Group Hero
+**Submitted by: Mohammad S Anwar**
+
+---
+
+You are given an array of integers representing the strength.
+
+Write a script to return the sum of the powers of all possible combinations; power is defined as the square of the largest number in a sequence, multiplied by the smallest.
+
+### Example 1
+```
+Input: @nums = (2, 1, 4)
+Output: 141
+
+Group 1: (2) => square(max(2)) * min(2) => 4 * 2 => 8
+Group 2: (1) => square(max(1)) * min(1) => 1 * 1 => 1
+Group 3: (4) => square(max(4)) * min(4) => 16 * 4 => 64
+Group 4: (2,1) => square(max(2,1)) * min(2,1) => 4 * 1 => 4
+Group 5: (2,4) => square(max(2,4)) * min(2,4) => 16 * 2 => 32
+Group 6: (1,4) => square(max(1,4)) * min(1,4) => 16 * 1 => 16
+Group 7: (2,1,4) => square(max(2,1,4)) * min(2,1,4) => 16 * 1 => 16
+
+Sum: 8 + 1 + 64 + 4 + 32 + 16 + 16 => 141
+```
+---
+### Preliminary Considerations
+The concept of a "combination" is based on a set of distinct objects.
+So the question arises, how to deal with non-unique strengths.
+There is more the one way to do it:
+
+  1. Require unique values.
+  2. Consider the strengths as "multi-sets".
+  3. Consider a set of unique elements having a "strength" property that need not be unique.
+
+The first would be too simple. The second and third have substantial differences:
+
+Consider the multi-set `[a, a, b]`. The sub-multi-sets thereof are:
+```
+[], [a], [b], [a, a], [a, b], [a, a, b]
+```
+Compare it with a set of uniquely indexed elements `{0:a, 1:a, 2:b}` having these subsets:
+```
+{}, {0:a}, {1:a}, {2:b}, {0:a, 1:a}, {0:a, 2:b}, {1:a, 2:b}, {0:a, 1:a, 2:b}
+```
+The "sets of combinations" are basically different and would lead to different results for this task.
+The task's wording mentions "heroes" having a certain "strength" which suggests to look at it in the third way.
+
+### Solution
+At first glance it looks like we need to loop over all (nonempty) combinations of the given list of strengths, resulting in a complexity of `O(2^L)` with L as the length of the list.
+
+But this is not necessary: From each combination we need only the smallest and the largest elements.
+By picking two elements `nums[i]` and `nums[k]` (where `i <= k`) from the **sorted list**, we find all `L * (L + 1) / 2` pairs of minimum and maximum.
+Regardless of which elements come between them from the sorted list, the group power will always be the same.
+All we need to know is the number of combinations having the selected elements as "borders".
+
+If the offset `offs = k - i` is positive, then there are `offs - 1` elements between min and max, resulting in `2^(offs - 1)` combinations.
+If `i` equals `k`, then there is only one singleton combination.
+
+Thus we may find the sum over the powers in a complexity of `O(L^2)`.
+
+Consider a list of 64 elements.
+It has `2^64 (~ 1.8E19)` combinations that are practical impossible to enumerate, while it is easy to calculate the sum of the powers in `64 * 65 / 2 = 2080` iterations.
+Though the sum will become large, this is not a problem when `bigint` is in effect and it will be calculated in an instant.
+
+#### Example:
+Let `@nums = (2, 3, 5, 7)`. For the total sum of powers, each power needs to be multiplied with the number of corresponding combinations (columns "count" and "power").
+
+| min | max | offs |count| combinations                  | power |
+|-----|-----|------|-----|-------------------------------|-------|
+|   2 |   2 |    0 |   1 |(2)                            |     8 |
+|   2 |   3 |    1 |   1 |(2,3)                          |    18 |
+|   2 |   5 |    2 |   2 |(2,5),(2,3,5)                  |    50 |
+|   2 |   7 |    3 |   4 |(2,7),(2,3,7),(2,5,7),(2,3,5,7)|    98 |
+|   3 |   3 |    0 |   1 |(3)                            |    27 |
+|   3 |   5 |    1 |   1 |(3,5)                          |    75 |
+|   3 |   7 |    2 |   2 |(3,7),(3,5,7)                  |   147 |
+|   5 |   5 |    0 |   1 |(5)                            |   125 |
+|   5 |   7 |    1 |   1 |(5,7)                          |   245 |
+|   7 |   7 |    0 |   1 |(7)                            |   343 |
+
+```
+number of iterations: (4 * 5) / 2 = 10
+sum: 1 * 8 + 1 * 18 + 2 * 50 + 4 * 98 + 1 * 27 + 1 * 75 + 2 * 147 + 1 * 125 + 1 * 245 + 1 * 343 = 1627
+```
+#### Implementation
+We take the first element of the sorted list as the current minimum and loop over the whole list for the current maximum.
+Afterwards we remove the first element and continue with the next minimum.
+Maximum and offset may be retrieved simultaneously by calling `each` on the array.
+The minimum and maximum define the power of all corresponding combinations and from the offset we find the number of combinations.
+```
+sub total_power {
+    my @s = sort {$a <=> $b} @_;
+    my $power;
+    while (defined (my $min = $s[0])) {
+        while (my ($offs, $max) = each @s) {
+            $power += $min * $max**2 * ($offs ? 2**($offs - 1) : 1);
+        }
+    } continue {
+        shift @s;
+    }
+
+    $power;
+}
+
+```
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cgit