I don't understand why for test suite 4 :
a=[0 1 1 1 0 2 2 0 1 1 1 0];
y_correct = [1 1];
I expect y_correct = [1 1 1]
The idea seems to be to return the element that occurs in the longest run, or all such elements in case of a tie. In case 4 there are two longest runs, both with element 1.
Thanks Tim for the explanation . So as the vector [1 1 1] appears twice in test 4 with the unique value 1 , the result must be twice the unqiue value -> [ 1 1]. ok ok thanks again
Nice
I have this array as I counted the times it repeats. However, can someone give me a command to pull out the max values with the number to the left? Any advise? The first value in column on the right top doesn't matter as it should always be 1. I try the max command it only shows the max value of 3...:
1 0
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
Nice
Great problem. Thank you.
Good one
Nice problem
Is there a way to test output without submitting?
I really liked this one
This was fun.
The problem should specify that if the same number repeats the longest run of consecutive numbers more than once, it must be repeated in the output.
I really like your solution!
hy,is there a problem with the above mentioned question
l=length(a);
freq=zeros(1,l);
for c=1:l
for r=c:l
if a(c)==a(r)
freq(c)=freq(c)+1;
else
break
end
end
end
M=max(freq);
idx=find(freq==M);
val=a(idx);
function val=longrun(a)
b = zeros(size(a));
old = a(1);
cnt = 0;
for idx = 1:length(a)
if a(idx) == old
cnt = cnt +1;
else
old = a(idx);
cnt = 1;
end
b(idx)=cnt;
end
bmax = max(b);
val = a(b==bmax);
end
function val=longrun(a)
a_len=length(a); % Calculate the length of a
c=ones(a_len,1); % Initialize the consecutive times of the corresponding position
for num_a=1:(a_len-1)
n_val=1;
for next_a=(num_a+1):a_len
if a(num_a)==a(next_a) % If this number is equal to the following number, then n_val+1, and the next cycle
n_val=n_val+1;
c(num_a)=n_val;
else
c(num_a)=n_val; % If this number is not equal to the following number, assign the current n_val to the corresponding number of consecutive times, and jump out of the inner for loop
break
end
end
end
a_conti=max(c); %Find the largest number of consecutive times in the c array
max_location=find(c==a_conti); % Corresponds to the position of the maximum consecutive times
val=a(max_location); % Find out the corresponding position in a
end
Test 1 is error！
y_correct =[1 2];
This is the best solution I have seen. ????
Why is test 4 like that tho
Tricky one, but enjoyed solving it.
Tricky one.. but enjoyed it.
function val=longrun(a)
a_len=length(a); %计算a的长度
c=ones(a_len,1); %初始化对应位置的连续次数
for num_a=1:(a_len-1)
n_val=1;
for next_a=(num_a+1):a_len
if a(num_a)==a(next_a) %如果这个数字和后面的数字相等，则n_val+1，并进行下一次循环
n_val=n_val+1;
c(num_a)=n_val;
else
c(num_a)=n_val; %如果这个数字和后面数字不等，则把当前的n_val赋值给对应数字的连续次数，并跳出内层for循环
break
end
end
end
a_conti=max(c); %找出c数组中最大的连续次数
max_location=find(c==a_conti); %对应最大连续次数的位置
val=a(max_location); %找出对应在a中的位置
end
Test 1 and 4 are wrong
@Piero Cimule
Could you please comment on this method? It is very unlikely for me as a beginner in Matlab to have such a concise code. You are more than welcome to elaborate here or via my e-mail address zasevzasev42@gmail.com
Really succinct code. One of the best I saw so far in the contest. @Piero Cimule
@Jovan Krunic:
%
% find(vertcat(1,diff(a(:)),1))
% val=a(ans(diff(ans)==max(diff(ans))))
%
% the approach is:
% find the differences between the elements of the vector
% consecutive zero's mean there is a run of the same values
% get the indices of none zero elements of this vector
% get the differences of these indices (which are a measure for the length the run of consecutive zero's, the 'gap' resulting from removing the zeros)
% note that to make this work for vectors that start or end with the longest run we add a 1 to the beginning and the end of the first difference vector
% get the max value from this last result which is the max sequence length
% get the indices of these max values, that is, the first indices of the runs that have this max length
% note that this last result is always a column vector
% get the values of the first elements of the runs with max length from the original vector
% note that if the original vector was a row vector, the result will be a row vector and when it was a column vector the result is a column vector,
% as required in the problem description
function val=longrun(a)
val = [];
if length(a) > 0
% a(:) will convert a to a column vector
a_as_column_vector = a(:);
% differences between elements of a
diff_a = diff(a_as_column_vector);
% differences between elements of a enclosed in ones (to make sure first and last element are non zero)
vertcat_1_diff_a_1 = vertcat(1, diff_a, 1);
% non zero elements of vertcat_1_diff_a_1
find_vertcat_1_diff_a_1 = find(vertcat_1_diff_a_1);
% the difference between the non zero indices is equal to the length of the run of consecutive numbers
lengts_of_runs_of_consecutive_numbers = diff(find_vertcat_1_diff_a_1);
% set the elements that have the max length to one and the others to 0 - note that shift of indices as a result of diff is compensated by adding 1 at the front of the vector
indices_of_elements_that_have_max_length_run = lengts_of_runs_of_consecutive_numbers == max(lengts_of_runs_of_consecutive_numbers);
indices_of_first_elements_of_longest_run_of_consecutive_numbers = find_vertcat_1_diff_a_1(indices_of_elements_that_have_max_length_run);
values_of_first_elements_of_longest_run_of_consecutive_numbers = a(indices_of_first_elements_of_longest_run_of_consecutive_numbers);
val = values_of_first_elements_of_longest_run_of_consecutive_numbers;
end
end
Could you check this for me, please?
[ra,ca]=size(a);
if ra>ca
a=a';
end
j=1;
V=[a(1,1);1];
for i=2:size(a,2)
X=a(1,i);
if V(1,j)==X
V(2,j)=V(2,j)+1;
else
j=j+1;
V(1,j)=X;
V(2,j)=V(2,j)+1;
end
end
val=V(V(2,:)==max(V(2,:)))
if ra>ca
val=val';
end
How else could you solve in any other traditional programming language?
Its good,
It was not so easy...
very clever!
i cannot understand
312 Solvers
Permute diagonal and antidiagonal
226 Solvers
417 Solvers
262 Solvers
Switch matrix to a column vector
266 Solvers
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!