List months ending on the same day of the week

Question

This is an assignment. I have the following code:

#! /bin/bash

y=$1
if [ -z $1 ] # if year is not specified use the current year
    then y=(`date +%Y`)
fi

for m in {1..12}; do
    if [ $m -eq 12 ] # december exception
    then echo $(date -d $m/1/$y +%b) - $(date -d "$(($m%12+1))/1/$y" +%A)
          break 
    fi  
   echo $(date -d $m/1/$y +%b) - $(date -d "$(($m%12+1))/1/$y - 1 days" +%A) # print the last day of the week for the month
done

It lists the last day of the week for every month:

Jan - Monday
Feb - Monday
Mar - Thursday
Apr - Saturday
May - Tuesday
Jun - Thursday
Jul - Sunday
Aug - Wednesday
Sep - Friday
Oct - Monday
Nov - Wednesday
Dec - Saturday

Now I need to reverse it, so that it lists months ending on every day of the week like so:

Sunday - Jul
Monday - Jan Feb Oct
Tuesday - May
Wednesday - Aug Nov
Thursday - Mar Jun
Friday - Sep
Saturday - Apr Dec

I'm thinking of a nested loop,

for d in {1..7};

And storing months in an array?

Answer 1

#! /usr/bin/env bash

# if year is not specified use the current year
declare -r year="${1:-$(date +%Y)}"

# associative array (aka hash table)
declare -A months_per_day=()
for m in {01..12}; do
    day_month=$(LANG=C date -d "${year}-${m}-01 +1 month -1 day" +"%A %b")
    months_per_day[${day_month% *}]+=" ${day_month#* }"
done

for day in Sunday Monday Tuesday Wednesday Thursday Friday Saturday; do
    echo "${day} -${months_per_day[${day}]:-}"
done

Output:

Sunday - Jul
Monday - Jan Feb Oct
Tuesday - May
Wednesday - Aug Nov
Thursday - Mar Jun
Friday - Sep
Saturday - Apr Dec