Write a function three positive integer scalar inputs year, month, day.
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Write a function called valid_date that takes three positive integer scalar inputs year, month, day. If these three represent a valid date, return a logical true, otherwise false. The name of the output argument is valid. If any of the inputs is not a positive integer scalar, return false as well. Note that every year that is exactly divisible by 4 is a leap year, except for years that are exactly divisible by 100. However, years that are exactly divisible by 400 are also leap years. For example, the year 1900 was not leap year, but the year 2000 was.
%Code to call your function
valid = valid_date(2018,4,1)
valid = valid_ date(2018,4,31)
I only got 1 of 6 which is random leap years.
Here is my code.
function valid = valid_date(year,month,day)
if ~isscalar(year) || year <1 || year ~= fix(year);
valid = false;
end
if ~isscalar(month) || month <1 || month ~= fix(month) || month > 12;
valid = false;
end
if ~isscalar(day) || day <1 || day ~= fix(day) || day > 31;
valid = false;
end
if year/4 == 0 && year/100 == 0 && year/400 == 0;
if month == [2,4,6,9,11];
if month == 2 && day > 29;
valid = false;
elseif month ~= 2 && day > 30;
valid = false;
else
valid = true;
end
else month ~= [2,4,6,9,11];
if day > 31;
valid = false;
else
valid = true;
end
end
elseif year/4 == 0 && year/100 ~= 0 && year/400 ~= 0;
if month == [2,4,6,9,11];
if month == 2 && day > 29;
valid = false;
elseif month ~= 2 && day > 30;
valid = false;
else
valid = true;
end
else month ~= [2,4,6,9,11];
if day > 31;
valid = false;
else
valid = true;
end
end
elseif year/4 == 0 && year/100 == 0 && year/400 ~= 0;
if month == [2,4,6,9,11];
if month == 2 && day > 28;
valid = false;
elseif month ~= 2 && day > 30;
valid = false;
else
valid = true;
end
else month ~= [2,4,6,9,11];
if day > 31;
valid = false;
else
valid = true;
end
end
else year/4 ~= 0 && year/100 ~= 0 && year/400 ~= 0;
if month == [2,4,6,9,11];
if month == 2 && day > 28;
valid = false;
elseif month ~= 2 && day > 30;
valid = false;
else
valid = true;
end
else month ~= [2,4,6,9,11];
if day > 31;
valid = false;
else
valid = true;
end
end
end
4 Comments
Abhishek Kumar
on 30 May 2021
%This is the working code
function [valid] = valid_date(y, m, d)
t = isscalar(y) && isscalar(m) && isscalar(d)
if t ~= 1
valid = false;
else nargin == 3 && (y && m > 0)
p = rem(y,4);
q = rem(y,100);
r = rem(y,400);
if (p == 0 && q ~= 0) || r == 0
if (m == 1 || m == 3 || m == 5 || m == 7 || m == 8 || m == 10 || m == 12) && (d > 0 && d <= 31)
valid = true;
elseif (m == 4 || m == 6 || m == 9 || m == 11) && (d > 0 && d <= 30)
valid = true;
elseif m == 2 && (d > 0 && d <= 29)
valid = true;
else
valid = false;
end
else
if (m == 1 || m == 3 || m == 5 || m == 7 || m == 8 || m == 10 || m == 12) && (d > 0 && d <= 31)
valid = true;
elseif (m == 4 || m == 6 || m == 9 || m == 11) && (d > 0 && d <= 30)
valid = true;
elseif m == 2 && (d > 0 && d <= 28)
valid = true;
else
valid = false;
end
end
end
Answers (3)
Subhadeep Koley
on 1 Feb 2020
Refer the code below. Hope this helps!
function valid = validateDate(year, month, day)
year1 = year/4;
leap1 = year1 == round(year1); % Condition 1
year2 = year/100;
leap2 = year2 == round(year2); % Condition 2
year3 = year/400;
leap3 = year3 == round(year3); % Condition 3
if leap1
leapYear = 1;
elseif leap2
leapYear = 0;
elseif leap3
leapYear = 1;
else
leapYear = 0;
end
if isscalar(year) && year >= 1 && isscalar(month) && month >= 1 && month <= 12 && isscalar(day) && day >= 1 && day <= 31
if leapYear
if month == 2
valid = day <= 29;
elseif month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12
valid = day <= 31;
else
valid = day <= 30;
end
else
if month == 2
valid = day <= 28;
elseif month == 1 || month ==3 || month ==5 || month ==7 || month == 8 || month == 10 || month == 12
valid = day <= 31;
else
valid = day <= 30;
end
end
else
valid = false;
end
end
0 Comments
Ahmed BENTALEB
on 10 May 2023
Edited: Ahmed BENTALEB
on 10 May 2023
My solution, it passed all tests and for sure it could be optimized more. hope it helps
function valid = valid_date(year, month, day)
valid = false;
if checkinput(year) && checkinput(month) && checkinput(day)
if isleap(year)
valid = (month31(month) && day<=31) || (month30(month) && day<=30) || (month==2 && day<=29);
else
valid = (month31(month) && day<=31) || (month30(month) && day<=30) || (month==2 && day<=28);
end
end
% check correct input
function check = checkinput(in)
check = isscalar(in) && in >=1 && in == fix(in);
% check for leap year to handel month 2
function leap = isleap(y)
leap = (mod(y,4)==0 && ~mod(y,100)==0) || mod(y,400)==0;
% check for lmonths where days are 31 or less
function is31 = month31(month)
is31 = (month==1 || month==3 || month==5 || month==7 || month==8 || month==10 || month==12);
% check for lmonths where days are 30 or less
function is30 = month30(month)
is30 = (month==4 || month==6 || month==9 || month==11);
Edit: another solution combined with the solution from cours
function valid = valid_date(y, m, d)
valid = false;
% check all input conditions one time
if checkinput(y) && (checkinput(m) && m<=12) && (checkinput(d) && d<=31)
% days of month in a year
daysInMonth = [31 28 31 30 31 30 31 31 30 31 30 31];
if isleap(y)
daysInMonth(2) = 29; % change month 2 days to 29
end
valid = d<=daysInMonth(m);
end
% check correct input
function check = checkinput(in)
check = isscalar(in) && in >=1 && in == fix(in);
% check for leap year to handel month 2
function leap = isleap(y)
leap = (mod(y,4)==0 && ~mod(y,100)==0) || mod(y,400)==0;
0 Comments
Paul
on 15 May 2023
Edited: DGM
on 23 Aug 2023
function valid = valid_date(y,m,d);
if ~isscalar(y)||~isscalar(m)||~isscalar(d)||y~=fix(y)||m~=fix(m)||d~=fix(d);
valid = false;
else
if y > 0 && (m == 12 || m == 1 || m == 3 || m == 5|| m == 7 || m == 8 || m == 10) && d >= 1 && d <= 31;
valid = true;
elseif (m== 4 || m == 6 || m ==9 || m == 11) && d >=1 && d <= 30;
valid = true;
elseif m==2 && d>=1 && d<=28;
valid = true;
elseif ((mod(y,4)==0 && mod(y,100)~=0)||mod(y,400)==0) && m==2 && d>=1 && d<=29;
valid = true;
else
valid = false;
end
end
end
1 Comment
DGM
on 23 Aug 2023
Same comment as before:
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