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Write a function called max_sum that takes v, a row vector of numbers, and n, a positive integer as inputs. The function needs to find the n consecutive elements of v whose sum is the largest possible.

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In other words, if v is [1 2 3 4 5 4 3 2 1] and n is 3, it will find 4 5 and 4 because their sum of 13 is the largest of any 3 consecutive elements of v. If multiple such sequences exist in v, max_sum returns the first one. The function returns summa, the sum as the first output argument and index, the index of the first element of the n consecutive ones as the second output.
I tried the follwing code but when there are two occurences for the same number, they both get removed in the same iteration and thus causes trouble.
function [summa, index] = max_sum(v,n)
summa = 0;
i = 0;
j = v;
m = [];
while i < n
summa = summa + max(j);
b = find(v==max(j));
m = [m b];
j = j(j<max(j));
i = i + 1 ;
end
t = sort(m);
index = t(1,1);
end
  12 comentarios
Anika Kader
Anika Kader el 4 de Jun. de 2020
can u explain these two lines?
ii = 1:length(v)-n+1
currentV = v(ii:(ii+n-1));

Respuesta aceptada

Guillaume
Guillaume el 13 de Feb. de 2019
Don't use the question title to post the content of your assignment as that get truncated.
Assuming that your assigment is to find the sequence of length n of consecutive numbers with the highest sum, then I don't see how your algorithm even attempts that. You don't care about the maximum of the vector, it's completely irrevelant to the sequence sum. You want to calculate a sliding sum (i.e. first sum(v(1:n)), then sum(v(2:n+1)), etc. and find the maximum of these.
If you're allowed to use movsum, it can be trivially done in just two lines.
  19 comentarios
Maximilian Schmidt
Maximilian Schmidt el 1 de Mayo de 2019
@Guillaume: Thank you for your post. I was looking for hints on the same assignment and came across your comment. It seems to be doing the job fine:
h = [6 45 9 67 -36 -34 99 64 67 8]; n = 4;
>> [summa, index] = max(movsum(h, n, 'Endpoints', 'discard'))
summa =
238
index =
7
What I don't understand is how Matlab knows what to do with 'index'. In the documentation for movsum there was no hint about what the funciton returns if you ask for two output arguments. To check that 'index' wasn't some kind of keyword; I tried the above again but used the names 'a' and 'b' instead of summa and index. As I assumed, the output was the same.
So my question: What part of the code tells matlab to output the index of the starting point of the sum in the second output argument?
Stephen23
Stephen23 el 1 de Mayo de 2019
Editada: Stephen23 el 1 de Mayo de 2019
@Maximilian Schmidt: movesum is called inside the function max, and it is max is called with two output arguments. So you need to read the max documentation.
A function called inside another function, like movesum inside max in your example, returns exactly one output argument as an input argument to the wrapper function, so
your example:
[summa, index] = max(movsum(h, n, 'Endpoints', 'discard'))
is exactly equivalent to
tmp = movsum(h, n, 'Endpoints', 'discard');
[summa, index] = max(tmp)

Más respuestas (37)

lalit choudhary
lalit choudhary el 17 de Abr. de 2019
After hours of brainstorming, i finally figured it out
try this code-
function [summa, index]=max_sum(b,n)
y=length(b)
if n>y
summa=0;
index=-1;
else
[summa, index] = max(movsum(b, n, 'Endpoints', 'discard'));
end
end
  6 comentarios
Shandilya Kiran Bhatt
Shandilya Kiran Bhatt el 11 de Mayo de 2020
Can you explain nx-n+1 term? i did not understand why are you making a vector containing zeros of 1x nx-n+1 dimension
Walter Roberson
Walter Roberson el 11 de Mayo de 2020
When you have a vector of length nx to be taken in full windows of length n, sliding along 1 at a time, then you have nx - n + 1 full windows. For example, data = [1 2 3 4 5 6 7 8 9 10], n = 7, then nx = 10, nx - n + 1 is 10 - 7 + 1 = 4, corresponding to the windows 1:7, 2:8, 3:9, 4:10
This codes is specifically for the case where you only want full windows -- which is what the 'discard' option of movsum is intended to indicate, that you want to discard any sum that was made with a window that was not full (such as 5:10 not being the full length of 7)

Shubham Pandey
Shubham Pandey el 3 de Abr. de 2020
function [s,i] = max_sum(v,n)
s = 0; i = 0;
len = length(v);
l = len-n+1;
if n>len
s = 0;
i = -1;
else
for j=1:l
if j==1
s = sum(v(j:n));
i=1;
end
if s<sum(v(j:n))
s = sum(v(j:n));
i = j;
end
if n<len
n=n+1;
end
end
end
end
  3 comentarios
Vipin Parthan
Vipin Parthan el 29 de Mayo de 2020
Editada: Vipin Parthan el 29 de Mayo de 2020
if j==1
s = sum(v(j:n));
i=1;
Could you explain how this works?

Prasad Reddy
Prasad Reddy el 14 de Abr. de 2020
function [summa,index] = max_sum(v,n) % defining a function with name "max_sum", input augements are (v,n) and out put augemnts are [summa,index]
[r,e]=size(v); % reading the size of vector "v" to "r,e"
summa=0; % asaining 0 to summa
index=-1; % assaining -1 to index, as they asked in the question
sums=zeros(1,e-(n-1)); % creating a zro vector of requirted size. you can try itby taling two or th trial cases
if n>e % checking if n is greatr than th length of the vector so that summa and index can be rturnd as
summa=summa; % 0 and -1 respectively as askeed in the question.
index=index;
else % if n is less than the length of vctor then
for i=1:e-(n-1) % for loop to run across the length of the vector
for j=0:n-1 % for loop to run across the required number of elements to be summed
sums(i)=sums(i)+v(i+j); % elments of sums which are initially zeros are being updated step by step with th sum of 'n' numbers in vector
end
end
[s,i]=max(sums); % reading th maximum valu and its indx from sums to 's' and 'i'
summa=s;
index=i;
end
end

asad jaffar
asad jaffar el 4 de Abr. de 2019
jan and walter take a look at this code this is giving me correct answers except for negative elements array
function [summa, index]=max_sum(b,n)
y=length(b)
q=movsum(b,n)
[maxsum,maxidx]=max(q)
summa=maxsum
index=(maxidx-floor(n/2))
if n==y
j=1
end
end
command window
max_sum([ 79 15 -18 -28 -30 52 -81 31 -74 4 57 -96],4)
y =
12
maxsum=
94 76 48 -61 -24 -87 -28 -72 -120 18 -109 -35
summa =
94
index =
1
index =
1
but i think my code is giving the right value plz take a look and tell me what to do
  6 comentarios
Jan
Jan el 5 de Abr. de 2019
Editada: Jan el 5 de Abr. de 2019
@asad: Again: Read the documentation of movsum:
doc movsum
Then try it by your own in the command windows, e.g.:
x = rand(1, 6)
movsum(x, 2)
movsum(x, 2, 'Endpoints', 'discard')
You asked: "are guys even do coursera" - no, of course we do not solve a course for beginners. we have done this about 20 or 30 years ago (a bold but maybe matching guess). We are not going to do exactly what you are doing only to answer very easy programming questions.
"i am waiting" - this is definitely the wrong approach. Exhaustive hints and working code have been posted some hours earlier already.
Some comments to your code:
function [summa, index]=max_sum(b,n)
y=length(b) % This is not useful
q=movsum(b,n)
[maxsum,maxidx]=max(q)
summa=maxsum % Why not directly: [summa,maxidx]=max(q)
index=(maxidx-floor(n/2))
if n==y % While j is not used anywhere, omit this
j=1
end
end
A nicer version:
function [summa, index] = max_sum(b, n)
q = movsum(b, n);
[summa, maxidx] = max(q);
index = maxidx - floor(n / 2);
end
But the result is not correct: The first and last two elements of the sum (or in general: floor(n / 2) elements) are not the sum of n elements of the input vector. See: doc movsum. A solution:
function [summa, index] = max_sum(b, n)
q = movsum(b, n);
m = floor(n / 2);
[summa, index] = max(q(m+1:end-m+1));
end
Try to understand, why q is cropped here. Instead of cropping it is smarter to let movsum exclude the marginal elements already with setting 'Endpoints' to 'discard'.
Vikas Kumar
Vikas Kumar el 11 de Jun. de 2019
function [summa, index] = max_sum(A,n)
if length(A)<n
summa = 0;
index = -1;
else
B = maxk(A,n)
summa = sum(B);
z = zeros(1,length(B));
m = [];
for i = 1:length(B)
z = find(A==B(i));
m = [m z]
end
T = sort(m)
index = T(1,1)
end
end

Jaimin Motavar
Jaimin Motavar el 29 de Jun. de 2019
function [summa,index]=max_sum(a,b)
n=length(a);
summa=0;
total=0;
if b>n
summa=0;
index=-1;
return
end
for i=1:(n-b+1)
for j=i:(b-1+i)
total=total+a(1,j);
end
if total>summa
summa=total;
index=i;
end
total=0;
end
end
  2 comentarios
Jan
Jan el 30 de Jun. de 2019
Some simplifications:
function [summa, index] = max_sum(a, b)
n = length(a);
summa = 0;
index = -1;
if b <= n
for i = 1:(n - b + 1)
total = sum(a(i:b - 1 + i));
if total > summa
summa = total;
index = i;
end
end
end
end

Roshan Singh
Roshan Singh el 15 de Sept. de 2019
function [summa, index]=max_sum(v,n)
L=length(v);
p=L-(n-1);
k=0;
t=0;
for i=1:p
z=n+k;
k=k+1;
s=0;
for d=i:z
s=s+v(d);
end
t(i)=s;
end
if (n<=L)
[summa index]=max(t);
else
summa=0;
index=-1;
end
end

Hussain Bhavnagarwala
Hussain Bhavnagarwala el 13 de Mzo. de 2020
function [summa,index] = max_sum(v,n)
if n>length(v)
summa = 0;
index = -1;
return
end
x = (length(v) - n) + 1;
tot =[];
t=0;
total = 0;
for k = 1:x
for i =1:n
total = total + v(t+i);
end
tot(k) = total;
t=t+1;
total = 0;
end
summa = max(tot);
ind= find(tot==max(tot));
index = ind(1);
end

Shashidhar Rai
Shashidhar Rai el 15 de Mzo. de 2020
Editada: Walter Roberson el 15 de Mzo. de 2020
I have written the code but it's showing error. Can anyone tell me what's wrong in it.
function [summa, index] = max_sum(v, m)
ii=1;
if length(v)
index=-1;
summa=0;
elseif length(v) ==n
index=1;
summa=sum(v);
else
summa=0;
start=1;
for ii =start : start +n-1
if summa < sum (v [start : start+n-1] )
summa=sum(v[ start : start +n-1] ) ;
index = start;
end
start =start +1;
if start
length(v) - n+1
break;
end
end
end
  2 comentarios
Walter Roberson
Walter Roberson el 15 de Mzo. de 2020
What is your expectation of what this code will do:
sum (v [start : start+n-1] )
?
Guillaume
Guillaume el 15 de Mzo. de 2020
To add to the problem Walter pointed out:
  • the function has an input m that is never used, but use the undefined variable n instead.
  • there's a ii loop that never uses ii.
  • that loops will only ever do one iteration because the if start is always going to be true and break out of it.

Irfan Hussain
Irfan Hussain el 31 de Mzo. de 2020
function [summa, index] = max_sum(v,n)
leng = length(v);
if leng < n
summa = 0;
index = -1;
else
w = [];
summa = 0;
for i = 1:leng
if summa < sum(v(i:n));
summa = sum(v(i:n));
index = i;
% x = sum(v(i:n));
%w = [w ,x];
if n < leng
n = n + 1;
%else
% summa = max(w);
% index = find(x == );
% end
end
end
end
end
  1 comentario
Shubham Pandey
Shubham Pandey el 3 de Abr. de 2020
this code has some error, wont work correctly for this case:
[summa index]=max_sum([ 45 -32 -25 27 78 -9 32 -1 -85 2 -32 -81 -32 -14 -31 46 -70 87 94 ], 15)

Jaimin Patel
Jaimin Patel el 7 de Abr. de 2020
function [X,Y] = max_sum(v,n) % X and Y are output arguments
X = -inf; %take -inf value of X for getting minus sum
p = n;
Y = 1;
if n > length(v)
X = 0;
Y = -1;
end
for i = 1:(length(v)-(n-1))
a = sum(v(i:p));
p = p + 1;
if a > X %condition for taking maximum sum of consecutive elements of vector
X = a;
Y = i;
end
end
simple code for solution...

Emine Ertugrul
Emine Ertugrul el 13 de Abr. de 2020
function [summa,index]=max_sum(v,n)
m=length(v);
if n>m
summa=0;
index=-1;
else
M = movsum(v,n,'Endpoints', 'discard')
[a,b]=max(M)
summa = a
index = b
end
end
  1 comentario
Walter Roberson
Walter Roberson el 13 de Abr. de 2020
I think the point of the assignment was not to use movsum() or other similar library functions, and to write the algorithm in more basic MATLAB.

Garvit Kukreja
Garvit Kukreja el 14 de Abr. de 2020
I have written this code. Can anyone tell me what's wrong in it.
Error : Variable summa has an incorrect value. max_sum([ -91 -57 -45 -8 -21 8 75 -80 89 -36 59 -71 -57 14 3 88 -79 -68 -6 ], 5) returned sum = 325 and index = 7 which is incorrect...
function [s,i]=max_sum(a,b)
B = maxk(a,b)
[n m]=size(B);
[y z] = size(a);
f=[]; %empty matrix
kk=0; %counter
if z>=b; %no. of element in 'a' more than or equal to'b'.
s=sum(B)
for ii=1:z;
for jj=1:m;
if B(jj)==a(ii);
kk=kk+1;
f(kk)=ii ;
end
end
end
i=min(f)
else
s=0
i= -1
end
  2 comentarios
Walter Roberson
Walter Roberson el 14 de Abr. de 2020
The sum that has to be returned is the deduced sum of the subset, not the sum of the entire vector.
Garvit Kukreja
Garvit Kukreja el 14 de Abr. de 2020
Editada: Garvit Kukreja el 14 de Abr. de 2020
Thanks for your response.
it is sum of deduced matrix 'B',
where B=Maxk(a,b) ( If a is a vector, then maxk returns a vector containing the k largest elements of a)
program worked for this command : [summa, index] = max_sum([1 2 3 4 5 4 3 2 1],3)

Ahmed J. Abougarair
Ahmed J. Abougarair el 21 de Abr. de 2020
function [summa, index]=max_sum(v,n)
L=length(v);
p=L-(n-1);
k=0;
t=0;
for i=1:p
z=n+k;
k=k+1;
s=0;
for d=i:z
s=s+v(d);
end
t(i)=s;
end
if (n<=L)
[summa index]=max(t);
else
summa=0;
index=-1;
end
end

Ahmed J. Abougarair
Ahmed J. Abougarair el 21 de Abr. de 2020
function [summa, index] = max_sum(v,n)
k =length(v);
if k==n
summa = sum(sum(v));
index =1;
return
elseif k<n
summa = 0;
index =-1;
return
else
z = length(v);
summa = 0;
index = -1;
if n <= z
for i = 1:(z - n + 1)
total = sum(v(i:n - 1 + i));
if total > summa
summa = total;
index = i;
end
end
end
end

Roshan Barnwal
Roshan Barnwal el 27 de Abr. de 2020
function [summa, index] = maxsum(v,n)
summa = sum(v(1:n-1));
for j = 2:length(v) -2;
r = sum(v(j:j+n-1));
if summa<r
summa=r;
index = j;
end
end
%% this works well but in few cases it asks for some error. can any one help me to short out the problem?

Salman P H
Salman P H el 30 de Abr. de 2020
function [summa,index] = max_sum(v,n)
if n>length(v)
summa = 0;
index = -1;
return;
end
x=1;
y=n;
great = -inf;
while y<=(length(v))
z = sum(v(1,x:y));
if z>great
great=z;
a=x;
x=x+1;
y=y+1;
elseif great==z
a=x-(x-a);
x=x+1;
y=y+1;
else
x=x+1;
y=y+1;
end
end
summa = great;
index = a;
end

Olel Arem
Olel Arem el 30 de Abr. de 2020
function [summa,index]=max_sum(v,n)
len_v=length(v);summa=0;poss=[];ii=1;jj=n;
if n>len_v
summa=0;
index=-1;
return
end
while jj<=len_v
poss(ii)= sum(v(ii:jj));
[summa,index]=max(poss);
ii=ii+1;
jj=jj+1;
end

Prithvi Shams
Prithvi Shams el 5 de Mayo de 2020
Here's my version of the code.
While the problem can be solved with movsum() in a single line, it's recommended to utilize if and for loops as a learning exercise.
function [summa, index] = max_sum(v, n)
if n > numel(v)
summa = 0;
index = -1;
else
j = n;
for i = 1:(numel(v) - n + 1)
s(i) = [sum(v(i:j))];
j = j + 1;
end
if numel(max(s)) > 1
m = max(s);
[summa index] = m(1);
else
[summa index] = max(s);
end
end

anuj chaudhari
anuj chaudhari el 5 de Mayo de 2020
Editada: anuj chaudhari el 5 de Mayo de 2020
function [a, b] = max_sum(v,n)
if ~isscalar(n)||n<0||n~=fix(n)
error('n must be a positive integer')
end
[~,c]=size(v);
d=length(v);
r = zeros(1,d-n+1);
if n>c
b=-1;
a=0;
else
for k=1:(d-n+1) %from here, this is a means of movsum
r(k)=sum(v(k:k+n-1));
[a, b]=max(r);
end
end
end

Shandilya Kiran Bhatt
Shandilya Kiran Bhatt el 11 de Mayo de 2020
You can also solve this question by using while loop if you don't know movsum function like me.Look at following code:-
function [summa , index ] = max_sum(A,n)
b = length(A);
c = zeros(1,b-n+1);
i = 1;
if b < n
summa = 0;
index = -1;
else
while n<=b && i<=b
c(1,i) = sum(A(i:n));
n = n+1;
i = i+1;
end
[summa , index ] = max(c);
end

Tahsin Oishee
Tahsin Oishee el 17 de Mayo de 2020
function [summa,index]=max_sum(v,n)
summa=0;
index=0;
w=1;
[b a]=size(v);
if n>a
index=-1
summa=0;
else
if n<=a
for i=1:(a+1-n)
sum=0;
for j=1:n
sum=sum+v(w);
w=w+1;
end
w=w-n+1;
if sum>summa;
summa=sum;
index=v(i);
end
summa
index
end
end
end
end
  2 comentarios
Walter Roberson
Walter Roberson el 17 de Mayo de 2020
You already have two loop control variables, i and j: use 2D indexing instead of doing strange things with w.
We recommend against naming a variable sum: it is very common to want to use the sum() function after having used sum as a variable.

khyathi beeram
khyathi beeram el 18 de Mayo de 2020
Editada: khyathi beeram el 18 de Mayo de 2020
function [summa index]=max_sum(v,n)
a=[ ];
if n<=length(v)
for i=1:length(v)-(n-1)
sum=0;
for j=i:i+n-1
sum=sum+v(j);
end
a(i)=sum;
end
summa=max(a);
g=find(a==summa);
index=g(1,1);
else
summa=0;
index=-1;
end
end

vighnesh rana
vighnesh rana el 19 de Mayo de 2020
function [summa,index] = max_sum(A,n)
if length(A)< n
summa = 0;
index = -1;
return;
end
summa = -inf;
index = -1;
for i = 1:(length(A)-n+1)
total = sum(A(i:(i+n-1)));
if total > summa
summa = total;
index = i;
end
end
end

Taif Ahmed BIpul
Taif Ahmed BIpul el 20 de Mayo de 2020
function[summa,index]=max_sum(v,n)
if size(v,2)<n
summa=0;
index=-1;
return
end
p=size(v,2)-(n-1);
k=1;
A=zeros(1,n);
summ1=zeros(1,p);
for i=1:p
jj=i;
while jj<=(i+n-1)
while k<=n
A(k)=v(jj);
jj=jj+1;
k=k+1;
end
end
summ1(i)=sum(A);
k=1;
end
summa=max(summ1);
index1=find(summ1==max(summ1));
if size(index1,2)>1
index=index1(1);
return
else
index=index1;
end

utkarsh singh
utkarsh singh el 21 de Mayo de 2020
function [summa, index]=max_sum(m,n)
if numel(m)<n
summa=0;
index=-1;
else
summa=-inf;
for i=1:(numel(m)-n+1)
j=m(i:i+n-1);
maxsumma=sum(j);
if(maxsumma>summa)
summa=maxsumma;
index=i;
end
end
end
end

sai teja
sai teja el 28 de Mayo de 2020
function [summa, ind] = max_sum(v,n)
% If n is greater than v return the specified values
% Using return keyword exits the function so no further code is
% evaluated if n > length(v) summa = 0; ind = -1;
return;
end
% Initialize summa to -inf.
% Then work through the vector, checking if each sum is larger than the
% current value of summa summa = -inf; ind = -1;
% Once we get to length(v)-n+1
we stop moving through the vector for ii = 1:length(v)-n+1
currentV = v(ii:(ii+n-1));
currentSumma = sum(currentV);
% If currentSumma greater than summa, update summa and ind
if currentSumma > summa
summa = currentSumma;
ind = ii;
end
end
end

Kumar Shubham
Kumar Shubham el 2 de Jun. de 2020
Editada: Kumar Shubham el 2 de Jun. de 2020
function [summa, index]= max_sum(v,n)
if n > length(v)
summa = 0;
index = -1;
return;
elseif n<=length(v)
a=movsum(v,[0,n-1]);
b=a(1:length(v)-n+1);
[summa,index]= max(b);
end
end

Sumit Kumar Sharma
Sumit Kumar Sharma el 2 de Jun. de 2020
function [summa, index]=max_sum(v,n)
l=length(v);
if n>l
summa=0;
index=-1;
else
p=l-n+1;
ii=0;
total=-inf;
for j=1:p
s=sum(v(j:n+ii));
ii=ii+1;
if s>total
total=s;
index=j;
else
continue;
end
end
summa=total;
end
end

KAVITI BHARGAV RAM NAIDU
KAVITI BHARGAV RAM NAIDU el 3 de Jun. de 2020
Editada: KAVITI BHARGAV RAM NAIDU el 5 de Jun. de 2020
function [summa,index] = max_sum(v,n)
b=zeros(1,length(v)-n+1); % initializing the b vector(only zeros) of length = length(v)-n+1
% this will decrease time taken for getting answer
if n > (length(v)) % if n is larger than length of v sum = 0 and index = -1 as per question (last line)
summa = 0 ;
index = -1;
else
for p = 1:[length(v)-n+1]
b(1,p) = sum (v(p:p+n-1)); % b(1,1)= v(1,1)+v(1,2) (if n = 2)
end % giving values to vector b using sum function
[summa index] = max(b); % if n = 2 sum should be between 2 consecutive elements.
end % if p = 2 and n = 2 sum between 2 nd and 3rd element should be considered
% p+n-1 = 2+2-1 = 3 // sum (v(2:3)) means sum of 2nd and 3rd element of vector b.
% so sum(v(p:p+n-1))

Methil Muley
Methil Muley el 4 de Jun. de 2020

zineb britel
zineb britel el 4 de Jun. de 2020
%what's wrong with my code
function [summa index]=max_sum(v,n)
index=0; summa=0;
if n > length(v)
summa= 0;
index=-1;
else
a=sort(v,'descend');
summa= sum(a(1:n));
index_row= find(v>=a(n));
index= min(index_row);
end
end
  2 comentarios
KAVITI BHARGAV RAM NAIDU
KAVITI BHARGAV RAM NAIDU el 5 de Jun. de 2020
Editada: KAVITI BHARGAV RAM NAIDU el 5 de Jun. de 2020
  • By using sort function u will get all elements in descending order
  • But according to question
  • we should keep the order of elements same.
  1. case 1 : if v = [1 2 3 4 5]; n = 2;
  2. summa is maximum of (1+2),(2+3),(3+4),(4+5)
  3. index is position where we are getting maximum sum here maximum sum is 9 so index = 4
  4. case 2 : if v = [1 2 3 4 5 ];n = 3;
  5. summa is maximum of (1+2+3),(2+3+4),(3+4+5)
  6. here maximum sum is 12 so index = 3
  7. case 3 : if v = [1 2 3 4 5 5 4 3 ];n=2;
  8. summa is maximum of (1+2),(2+3),(3+4),(4+5),(5+5),(5+4),(4+3)
  9. here maximum sum is 10 so index = 5

Ujjawal Barnwal
Ujjawal Barnwal el 8 de Jun. de 2020
function [summa , index]=max_sum(v,n)
summa=sum(v(v<0));
for ii=1:(length(v)-(n-1))
t=0;
for jj=ii:(ii+(n-1))
t=t+v(jj);
end
if (t>summa)
summa=t;
index=ii;
end
end
end

AYUSH MISHRA
AYUSH MISHRA el 13 de Jun. de 2020
Editada: AYUSH MISHRA el 13 de Jun. de 2020
function [summa, index] = max_sum(v,n)
if n > length(v)
summa = 0;
index = -1;
return;
end
summa = -inf;
index = -1;
for ii = 1:length(v)-n+1
currentV = v(ii:(ii+n-1));
currentSumma = sum(currentV);
if currentSumma > summa
summa = currentSumma;
index = ii;
end
end
end

Maddireddy Harshavardhan Reddy
Maddireddy Harshavardhan Reddy el 18 de Jun. de 2020
Create a row vector named x that starts at 1, ends at 10, and contains 5 elements.

Sneha Paul
Sneha Paul el 19 de Jun. de 2020
function [summa,index]=max_sum(v,n)
z=[];
if n>length(v)
summa=0;
index=-1;
return
elseif (n<=length(v))
for i=1:length(v)-(n-1)
z(i)=sum(v(i:i+(n-1)));
end
end
[summa,index]=max(z)

MICHAEL
MICHAEL el 19 de Jun. de 2020
function [s,i] = max_sum(v,n)
s = 0; i = 0;
len = length(v);
l = len-n+1;
if n>len
s = 0;
i = -1;
else
for j=1:l
if j==1
s = sum(v(j:n));
i=1;
end
if s<sum(v(j:n))
s = sum(v(j:n));
i = j;
end
if n<len
n=n+1;
end
end
end
end

Sakib Javed
Sakib Javed el 22 de Jun. de 2020
Editada: Sakib Javed el 22 de Jun. de 2020
ffunction [summa index] = max_sum(A,n)
m=length(A);
if n > m
summa = 0;
index = -1;
return;
end
B = [];
q=m-(n-1);
for ii = 1:q
B=[B sum(A(ii:ii+n-1))];
end
[summa index] = max(B);
end

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