05 July 2024 | Challenge 276 |

# Hourly Frequencies

## Task 1: Complete Day

**Submitted by: Mohammad Sajid Anwar**

You are given an array of integers, `@hours`

.

Write a script to return the number of pairs that forms a complete day.

A complete day is defined as a time duration that is an exact multiple of 24 hours.

### Example 1

```
Input: @hours = (12, 12, 30, 24, 24)
Output: 2
Pair 1: (12, 12)
Pair 2: (24, 24)
```

### Example 2

```
Input: @hours = (72, 48, 24, 5)
Output: 3
Pair 1: (72, 48)
Pair 2: (72, 24)
Pair 3: (48, 24)
```

### Example 3

```
Input: @hours = (12, 18, 24)
Output: 0
```

### Solution

While the definition of a *complete day* seems to be clear, *the number of pairs* is not.

Two important observations can be made from the above examples:

- Pairs must be build from different elements of the given list, thought their values may be identical.
- Pairs are unordered.

The key to a consistent description is to look at *index* pairs instead of *value* pairs.
Thus a more formal description of this task might be:

Given an n-tuple \(H = (h_1,\ldots,h_n)\) of integers, find the number of index pairs \(\{i, j\}\) where \(h_i \equiv h_j \bmod{24}\)

Let us consider the index set \(I = \{1,\ldots,n\}\).
We may say that an index pair \(\{i, j\} \subset I\) forms a *complete day*, if their corresponding values do, i.e. \(h_i \equiv h_j \bmod{24}\).

The index set \(I\) can be partitioned by the equivalence relation \(h_i \equiv h_j \bmod{24}\), leading to subsets \(P_k \subset I, 0 \le k < 24\) with \(P_k = \{i \in I | h_i \equiv k \bmod{24}\}\). For the sake of convenience we permit empty subsets \(P_k\).

By construction, each pair \(\{i, j\} \subset P_k\) represents a *complete day*, while no pair from different subsets does.
The number of pairs that can be formed from a set of \(c\) elements is \(c (c - 1) / 2\).

If \(c_k\) denotes the number of elements in \(P_k\),
then the total number of pairs forming a *complete day* is:

The effort for calculating all values \(c_k\) is \(\mathcal{O}(n)\) as we just need to count the number of elements in \(H\) per remainder modulo 24. As the sum is calculated in \(\mathcal{O}(1)\), the resulting total complexity is \(\mathcal{O}(n)\).

An implementation using `PDL`

is straightforward.
For `hist`

we use 24 bins centered at the integers `0 .. 23`

.

```
use strict;
use warnings;
use PDL;
sub complete_days {
my $c = hist pdl(@_) % 24, -0.5, 23.5, 1;
sum $c * ($c - 1) / 2;
}
```

See the full solution to task 1.

## Task 2: Maximum Frequency

**Submitted by: Mohammad Sajid Anwar**

You are given an array of positive integers, `@ints`

.

Write a script to return the total number of elements in the given array which have the highest frequency.

### Example 1

```
Input: @ints = (1, 2, 2, 4, 1, 5)
Ouput: 4
The maximum frequency is 2.
The elements 1 and 2 has the maximum frequency.
```

### Example 2

```
Input: @ints = (1, 2, 3, 4, 5)
Ouput: 5
The maximum frequency is 1.
The elements 1, 2, 3, 4 and 5 has the maximum frequency.
```

### Solution

This task can be solved with a complexity of \(\mathcal{O}(n)\).
Moreover, it can be done in a *single* pass over the given elements.

There is no need to restrict the array elements to positive integers, not even to numbers.

Recording the elements’ frequencies in a hash `%freq`

and the (currently observed) maximum frequency in `$maxfreq`

, for each element `$_`

of the list there are three cases:

`$freq{$_} < $maxfreq`

: The running result in`$a`

does not change.`$freq{$_} == $maxfreq`

: The running result is incremented by`$maxfreq`

.`$freq{$_} > $maxfreq`

:`$maxfreq`

is set to the frequency of`$_`

, which gives the running result simultaneously.

We may build a list of three anonymous subroutines each of which performs one of the required actions and call the sub indexed by the result of the *spaceship operator* `<=>`

.
(Note that an index of `-1`

accesses the last element of an array or list.)

```
use strict;
use warnings;
use List::Util 'reduce';
sub max_freq {
my ($maxfreq, %freq) = 0;
reduce {
&{ ( sub {$a + $maxfreq},
sub {$maxfreq = $freq{$b}},
sub {$a}
)[++$freq{$b} <=> $maxfreq]
}()
} 0, @_;
}
```

See the full solution to task 2.

If you have a question about this post or if you like to comment on it, feel free to open an issue in my github repository.