Number of Mondays Falls on the First of the month - linux

Question

I want a command line can display number of Monday(s) which fall(s) on the first of the month in a given year without using sed or awk commands
I have this command that display the first date of the current month
date -d "-0 month -$(($(date +%d)-1)) days"

Answer 1

With GNU date, you can read input from a file (or standard input):
printf '%s\n' 2021-{01..12}-01 | date -f- +%u | grep -c 1
This prints dates for the first of each month in a year, then formats them as "weekday" (where 1 is "Monday"), then counts the number of Mondays.
To parametrize the year, replace 2021 with a variable containing the year; wrapped in a function:
mondays() {
local year=$1
printf '%s\n' "$year"-{01..12}-01 | date -f- +%u | grep -c 1
}

Answer 2

Using a for loop, this can be accomplished as follows.
for mon in {01..12}; do date -d "2021-$mon-01" +%u; done | grep -c 1
Breakdown
We iterate through the numbers 01 to 12 representing the months.
We call date passing in the custom date value with the first date of each month in the year. We use +%u to return the day of week where 1 represents Monday.
Lastly we count the number of 1s using grep -c or grep --count
Note, the desired year has been hard coded as 2021. The current year can be used as:
for mon in {01..12}; do date -d "$(date +%Y)-$mon-01" +%u; done | grep -c 1
This can also all be put into a function and the desired year passed in as an argument:
getMondays() {
for mon in {01..12}; do date -d "$1-$mon-01" +%u; done | grep -c 1
}

Answer 3

I implemented it as:
for ((i=1,year=2021,mondays=0; i< 12; i++)) {
if [ $(date -d "$i/1/$year" +%u) -eq 1 ]
then
let "mondays++"
fi
}
echo "There are $mondays Mondays in $year."
That said, I like Mushfiq's answer. Quite elegant.