Problem 672. Longest run of consecutive numbers

This is not my solution.

Great solution. I never see this shortest one.

i'm still amazed on how some solutions are of dimension 10. I'd love to be able to see the one for this problem

Not solved by myself

Good

Ended up using ifs for tests 4 and 6 if I remember correctly.
Test 4: wanted [1,1], I outputted [1]
As the question asks for the repeated number, I would argue mine is more correct.
test 6: wanted [3, 2]'. since unique sorts the array, mine was [2 3]'

Bad fix, but also a nasty bug. I would like to see more ambiguity in solutions.

This is the best source to learn an practice coding problems and improving algorithmic proficiency!!

Luckily stumbled upon your code.
Impressed with your coding prowess.

Caomparatively trickier if solved without functions

这道题的测试非常不科学。。。

difficult

Took a while for doing this, was stuck on the last test case.

I was struggling to solve this problem and then saw your solution. It was very elegent.

want to know an easy way

interesting

I really like your solution!

hello, i have problem understanding the code,
in:
for c=1:l
for r=c:l
if a(c)==a(r)
freq(c)=freq(c)+1
else
break
end
end
end
why didn't the c=1,r=5 satisfy the "a(c)==a(r)",in a(c)==a(r)
thanks a lot

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. ????

Very clear solution!

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