Write a function max_sum that takes v a row vector of numbers & n,a positive integer as inputs.The function needs to find n consecutive elements of v whose sum is largest possible.It returns summa & index of first element of n consecutive integers.

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%Example-[summa,index]=max_sum([1 2 3 4 5 4 3 2 1],3)
% summa=13
% index=4
function [summa,index]=max_sum(v,n)
total=v(1,1);
if n>v
summa=0;
index=-1;
else
for ii=1:length(v)
jj=ii+(n-1);
if jj<=length(v);
total=[total,sum(v(ii):v(jj))];
end
[summa,index]=max(total);
end
end
  12 comentarios
Vivek Mishra
Vivek Mishra el 23 de En. de 2022
would you please explain the line : total=v(1,1); and total=[total,sum(v(ii):v(jj))];
khalid
khalid el 14 de Jul. de 2023
function [summa,index]=max_sum(v,n)
summa=1;
c=size(v,2);
p=0;
if(n<=c)
for jj=1:(c-n+1)
s=0;
for ii=jj:jj+n-1
s=s+v(1,ii);
end
p = v(1,jj);
if(s>summa)
summa=s;
index=p;
end
end
else
summa = 0;
index =-1;
end
end

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Respuesta aceptada

KSSV
KSSV el 19 de Mayo de 2020
Editada: KSSV el 19 de Mayo de 2020
v = [1 2 3 4 5 4 3 2 1] ;
n = 3 ;
N = length(v) ;
sumn = zeros(1, N - n + 1); % Pre-allocation
for i = 1:N - n + 1
sumn(i) = sum(v(i:(i+n-1))) ;
end
[val,idx] = max(sumn)
  22 comentarios
Walter Roberson
Walter Roberson el 6 de Ag. de 2021
There is a possibility that there is more than one location that is the maximum. The find() step returns all of the locations and min() of the find returns the first. Another way of doing that would be to use find(Summa==S, 1)

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Más respuestas (22)

Willynx Vixamar
Willynx Vixamar el 13 de Nov. de 2020
function [summa, index] = max_sum(v,n)
L = length(v);
S=zeros(1,L-n+1);
if n > L
summa = 0;
index = -1;
return
else
for i = 1:(L-n+1)
S(i)=sum(v(i:(i+n-1)));
end
summa = max(S);
ind = find(S == summa);
index = min(ind);
end
end
  6 comentarios
MINH
MINH el 30 de Jun. de 2023
Can somebody help me explain this part of the code :
ind = find(S == summa);
index = min(ind);
Cheers,
DGM
DGM el 30 de Jun. de 2023
Editada: DGM el 30 de Jun. de 2023
The first line returns the linear indices of S where S==summa.
The second line returns the minimum of those indices.
Both lines are unnecessary, because max() will return that index if you actually request it. Why do all this:
summa = max(S);
ind = find(S == summa);
index = min(ind);
... when you can just do:
[summa index] = max(S);

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Rik
Rik el 6 de Jul. de 2020
Editada: Rik el 30 de Jul. de 2020
You can get a speedup by using a convolution to calculate the moving sum:
v=randi(100,2000,1);
n=10;
clc
timeit(@() option1(v,n))
ans = 3.4287e-04
timeit(@() option2(v,n))
ans = 3.5765e-05
function [val,idx]=option1(v,n) % function as suggested by KSSV
if n>numel(v),val=0;idx=-1;return,end %my addition
N = length(v) ;
sumn = zeros(1, N - n + 1); % Pre-allocation
for i = 1:(N - n + 1)
sumn(i) = sum(v(i:(i+n-1))) ;
end
[val,idx] = max(sumn);
end
function [val,idx]=option2(v,n)
if n>numel(v),val=0;idx=-1;return,end
sumn=conv(v,ones(n,1),'valid');
[val,idx] = max(sumn);
end
  1 comentario
DGM
DGM el 8 de Mzo. de 2023
Editada: DGM el 8 de Mzo. de 2023
Edited only to demo the code.
An order of magnitude improvement and it's nice and concise -- what's not to like?

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Bruno Luong
Bruno Luong el 29 de Jul. de 2020
Editada: Bruno Luong el 29 de Jul. de 2020
N=3;
v = [1 2 3 4 5 4 3 2 1];
[s,index] = maxsum(v, N)
Using this function
%% NOTE this function returns empty outputs if N>length(v)
% might be more sensitive to numerical error
function [s,index] = maxsum(v, N);
c=cumsum([0; v(:)]);
[s,index]=max(c(N+1:end)-c(1:end-N));
end
Result
s =
13
index =
4
>>
  4 comentarios
Rik
Rik el 30 de Jul. de 2020
Thanks for the correction. I tested it for random lengths random vectors with random N against your code, and it seems to match now.
The fix doesn't affect the timing.
Bruno Luong
Bruno Luong el 30 de Jul. de 2020
Editada: Bruno Luong el 30 de Jul. de 2020
The faster timing of CONV for large array can be perhaps explained by multi-thread of the engine, that cannot be carrierd out with CUMSUM, which is sequential calculation.
TMW comes from far, the CONV in the earlier years suffered from performance. Now they have improved it greatly.
In any case a factor of 0.7 1.5 between two methods are to me essentially ... 1.
In any case your code get a vote from me.

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Bruno Luong
Bruno Luong el 1 de Ag. de 2020
function [s,index] = maxmovsum(v, N)
c = movsum(v,N);
[s,index] = max(c(1+floor(N/2):end-floor((N-1)/2)));
end

tharak infinity
tharak infinity el 10 de Nov. 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
  3 comentarios
OUSSAMA El GABBARI
OUSSAMA El GABBARI el 24 de En. de 2022
Could you explain the summa = -inf; line ?
Is -inf a built in keyword or something ?
Rik
Rik el 24 de En. de 2022
inf is the keyword for infinity. Negative infinity is guaranteed to be the lowest value possible.

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Sahil Deshpande
Sahil Deshpande el 17 de Dic. de 2020
function [summa,index] = max_sum(v,n)
l=length(v);m=0;t=0;pointer=1;index=0;
%When n is larger than the length of vector v
if n>l
summa=0;
index=-1;
else
%When n is smaller than the length of vector
for a = 1:n
m=m+v(a);
end
for i = 1:(l-(n-1))
k=i;t=0;
for j = 1:n
t = t+v(k);
k=k+1;
end
if t>m
m=t;
pointer=i;
end
end
summa=m;
index=pointer;
end
I have written this code without using any inbuilt functions, just using loops. It works great!
I know it looks kinda ugly, can anyone help me optimize this code and make it shorter?
  1 comentario
Rik
Rik el 17 de Dic. de 2020
Have you looked at the other solutions in this thread?
Also, you already used a builtin function on your first line: length.

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Shantanu Nighot
Shantanu Nighot el 22 de Dic. de 2020
% Here is the working code for This Problem statement
function [summa index] = max_sum(v,n)
ii = 1;
jj = n;
% Initialize summa to zero
summa = 0;
% if n is greater than number of elements in v the return summa = 0, index = -1
if n>numel(v)
summa = 0;
index = -1;
% pro tip
% If n is equal to number of elements in v then index will always be 1 and summa = sum(v(ii:jj))
elseif n==numel(v)
summa = sum(v(ii:jj));
index = 1;
else
% while loop for calculating the sum of n elements of subsequence of v
while jj<=numel(v)
% If sum of n elements of subsequence of v is greater than the previous one
% then that sum is stored in summa and index of first element is stored in index
if sum(v(ii:jj))>summa
summa = sum(v(ii:jj));
index = ii;
end
if sum(v(ii:jj))<0 && summa == 0
summa = sum(v(ii:jj));
index = ii;
end
ii = ii + 1;
jj = jj + 1;
end
end

Ali Mohammadi
Ali Mohammadi el 21 de Oct. de 2021
and this is my solution for it:
function [summa, index] = max_sum(v, n)
%creat an empty vector for cunsecutive sums, and ultimately determine the
%highest value and its index with max(w)
w = [];
ii = 1;
if n > length(v)
summa = 0
index = -1
return;
else
while n <= length (v)
w(ii) = sum(v(ii:n));
ii = ii + 1;
n = n + 1;
end
[summa, index] = max(w)
end
%I have checked, it works :)

Eric
Eric el 22 de Jul. de 2020
Editada: Eric el 22 de Jul. de 2020
I came up with a function that, I think, accomplishes this task. It runs fine in my MATLAB (2020a). However, it doesn't seem to be acceptable in the Assignment Submission. Anyone know why this could be the case?
function[summa,index]=max_sum(v,n)
if n>length(v)
index=-1;
summa=0;
else
sumn=movsum(v,n);
[summa,index]=max(sumn);
index=index-1;
end
end
  5 comentarios
Eric
Eric el 23 de Jul. de 2020
Sorry, not logical indexing. That would be "normal" indexing, right? [summa,index]=max(sumn) gives the maximum and index of that (middle) value in the vector sumn. The next line changes that index to the first value in that rolling sum. When running the code with my debugger, it seems to work with a range of values.
TANUJA GUPTA
TANUJA GUPTA el 29 de Jul. de 2020
@Eric, Actually when we type for example [summa,index] = max(sum); it gives the maximum value which is passed to summa, but in index it passes the index of the first no.
[Y,I] = max(X) returns the indices of the maximum values in vector I.
If the values along the first non-singleton dimension contain more
than one maximal element, the index of the first one is returned.

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Capulus_love
Capulus_love el 11 de Ag. de 2020
Editada: Capulus_love el 11 de Ag. de 2020
%why second problem doesn't run?
function [summa,index] = max_sum(v,n)
s = size(v); b = 0; c = 0; index = 0;
if s(2) < n
index = s(2) - n;
summa = 0;
return
end
for i = 1:s(2)-n+1
c = sum(v(i:i+n-1));
if c > b
b = c;
index = i;
else
b = b;
end
end
summa = b;
end
%Variable ind has an incorrect value.
% max_sum([ -66 31 61 60 -70 7 9 0 8 90 -54 72 12 ...
% 89 -50 4 86 -3 83 -52 ], 23) returned sum = 0 and index = -3 which is incorrect...
  3 comentarios
Capulus_love
Capulus_love el 11 de Ag. de 2020
in my matlab program,, it returns
summa =0
index = -3
with upper problem.
I'm not sure which part is not solved.
Did I not understand the problem?
Rik
Rik el 11 de Ag. de 2020
Put a breakpoint on the first line. Then use the debugger to execute your function line by line. When do you see the index go negative? When do your variable get unexpected values?
When you do that you will notice that your assumption of a column vector input isn't enforced anywhere.

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LEE MUN LING
LEE MUN LING el 18 de Dic. de 2020
function [m k]=max_sum(v,n)
m=0;
if n>length(v)
k=-1;
m=0;
return
else
for k=1:((length(v)-n)+1);
while m<max(sum(v(k:((n-1)+k))));
m=max(sum(v(k:((n-1)+k))));
m=m;
k=k;
end
end
end
  2 comentarios
LEE MUN LING
LEE MUN LING el 18 de Dic. de 2020
Can you guys help me check for this,
I try several time inserting the break statement in the loop but the iteration still continues, it is suppose to stop at k=4 when my input argument is ([1 2 3 4 5 4 3 2 1],3)
Rik
Rik el 18 de Dic. de 2020
m=m; and k=k; will not do anything, so why are they in your code? Also, you posted this as an answer, but it is a question. (and you forgot to format your code as code)
Are you aware that break will only stop one loop, not all? And have you looked at the other solutions in this thread?

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Rahil Ginwala
Rahil Ginwala el 23 de Dic. de 2020
function [summa, index] = max_sum(v,n)
summa = 0;
i = 0;
j = v;
m = [];
c = 1;
if n>v
summa=0;
index=-1;
else
while i < n
c = numel(v(v==max(j))); % number of times the element is in the array
summa = summa + max(j)*c; % multiplying by c
b = find(v==max(j)); % find the max elemnets of v
m = [m b];% array of positions with max element
j = j(j<max(j));% remove the max number from array
i = i + 1;
end
t = sort(m);
index = t(1,1)
end
% I am not able to stop i from goigng to value 3 which gives the error for sum and index

Abhishek Sharma
Abhishek Sharma el 28 de En. de 2021
% tbh, I used some help from these forum answers too
function [summa,index]=max_sum(v,n)
% first checking whether the size of row vector v is less than n or not
% if yes then print according to the given statement
if n>length(v);
summa=0;
index=-1;
else
% so you are thinking about why i took (-n+1) right?
% think of a value of size of row vector v and n
% you will get to know the answer by yourself ,example take size of v= 10 & n=10
for i=1:(length(v)-n+1);
mysum(i)=sum(v(i:i+n-1));
end
summa=max(mysum);
% I couldn't find this one, hence i used the help of this forum
x=find(mysum==summa);
index=min(x);
end

Daniel Anaya
Daniel Anaya el 9 de Abr. de 2021
%i dont know what to do. my works pretty good with positive integers. I dont know wht it has errors with negative integers
function [summa, index] = max_sum(v,n)
summa = 0;
ii = 1;
jj = n;
total = [];
if n > numel(v)
summa = 0;
index = -1;
elseif n == numel(v)
summa = sum(v(ii:jj));
index = v(1,1);
elseif n < numel(v)
for ii = 1:numel(v)
jj = ii + (n - 1);
if jj <= numel(v);
total = [total, sum(v(ii:jj))];
end
end
summa = max(total);
x = find(total == summa);
index2 = min(x);
index = v(index2);
end
  3 comentarios
Daniel Anaya
Daniel Anaya el 9 de Abr. de 2021
Editada: DGM el 22 de En. de 2023
Thanks for the heads up. I already have the answer. This code works fine now.
function [summa, index] = max_sum(v,n)
summa = 0;
ii = 1;
jj = n;
total = [];
if n > numel(v)
summa = 0;
index = -1;
elseif n == numel(v)
summa = sum(v(ii:jj));
index = 1;
elseif n < numel(v)
for ii = 1:numel(v)
jj = ii + (n - 1);
if jj <= numel(v);
total = [total, sum(v(ii:jj))];
end
end
summa = max(total);
x = find(total == summa);
index = min(x);
end
Jasmine
Jasmine el 18 de En. de 2023
Editada: Jasmine el 18 de En. de 2023
it worked for me thank you
i.e the reply above me

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Shun Yan
Shun Yan el 16 de Abr. de 2021
function [a,b]=max_sum(A,B)
n=1;
C=0;
for ii = 1:(size(A,2)-B+1)
C(n)=sum(A(n:(B+n-1)));
n=n+1;
end
if B>size(A,2)
a=0;
b=-1;
else
a=max(C);
b=find(C==max(C));
end
I don't get what is variable ind, and how's sum 4 index9 wrong; can someone pls help? Thanks!
  2 comentarios
Rik
Rik el 16 de Abr. de 2021
What happens when you try this input to your function?
[a,b]=max_sum([1 2 3 4 5 4 3 2 1],2)
a = 9
b = 1×2
4 5
Is that correct? How can the index (the second output) be multiple values?
function [a,b]=max_sum(A,B)
n=1;
C=0;
for ii = 1:(size(A,2)-B+1)
C(n)=sum(A(n:(B+n-1)));
n=n+1;
end
if B>size(A,2)
a=0;
b=-1;
else
a=max(C);
b=find(C==max(C));
end
end

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Giancarlo milon
Giancarlo milon el 29 de Jun. de 2021
This is my answer
function [summa,index] = max_sum(v,n)
b=1;
summa = -1000000;
%i start it with -1000000 because you ALWAYS want to overwrite your summa
index = 0;
if n > size(v)
summa = 0;
index = -1;
return
else
%this if checks for size condition if n is bigger than the size summa is 0
%and index -1
while n<=length(v);
%this will break when n = to lenght of the vector
if sum(v(b:n)) > summa
%this condition sets the sumatory of the position from b to n.
%In this case b starts in 1 and n is the input, if the sum is
%bigger than the summa it overwrites it, thats why we wanted
%always to overwrite the value summa and we started it with
%-1000000
summa = sum(v(b:n));
index = b;
end
b = b+1;
n = n+1;
%adds +1 to b and n so my positions go from 1:n to 2:n+1 to 3:n+2
%till n = the lenght and it stops
end
end
  3 comentarios
Giancarlo milon
Giancarlo milon el 23 de Jul. de 2021
the variable double has a max min number, idc wich is it. But u can change the -1000000 for that max min number that the memory can hold in the variable if u want. that should solve it. if ur input is that max min number u can add a counter of times the summa it equal to the starting number, save that position and return that position
Rik
Rik el 23 de Jul. de 2021
Your function should handle such a case. If the user needs to know how your function works and might need to edit your function, then it isn't working well.
Such limitations should be mentioned in the documentation of your function (which it currently lacks).
If you want to overwrite a variable with the lowest possible value, you should replace it with -inf.
%the number you were looking for is this:
-realmax
ans = -1.7977e+308

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Ujwal Dhakal
Ujwal Dhakal el 6 de En. de 2022
function [summa, index] = max_sum(v,n)
a=length(v);
sumvar=zeros(1,a-(n-1)); %preallocating a vector (you can skip this for now_go ahead and read the rest of the code)
%for invalid input
if n>a
summa=0;
index=-1;
return;
end
%assume a matrix on your own and try running i and j using the formula below, you'll get it
for i=1:(a-(n-1))
sumtemp=0;
for j=i:(n+i-1)
sumtemp=sumtemp+v(j);
end
sumvar(i)=sumtemp;%takes the sum and catenates into a vector sumvar _ that was preallocated to save some computational time
end
[summa,index_temp]=max(sumvar(:)); % if you give two output arguments to a max function, the second argument returns index of that element
index=index_temp; % the index of the maximum number in the vector sumvar happens to be also the index of our first number in the sequence
end

Yifan He
Yifan He el 27 de Jul. de 2022
function [summa,index] = max_sum(v,n)
if n > length(v)
summa = 0;
index = -1;
end
if n <= length(v)
summa = sum(v(1:n));
index = 1;
for i = 1:length(v)-n
if sum(v(i+1:i+n)) > summa
summa = sum(v(i+1:i+n));
index = i+1;
else
continue;
end
end
end

Muhammad
Muhammad el 30 de Jul. de 2022
function [summa,index]=max_sum(v,n)
Z=length(v);
summ=(1:Z-n+1);
if n>Z
summa=0;
index=-1;
return;
else
for i=1:(Z-n+1)
summ(i)=sum(v(i:(i+n-1)));
end
end
[summa,index]=max(summ);

Nyeche
Nyeche el 23 de Oct. de 2022
function [summa, index] = max_sum (v, n)
if n < 0 || n ~= fix(n) || ~isrow(v)
error('Input must be an integer');
elseif n > length(v)
summa = 0;
index = -1;
end
total = 0;
l = length(v);
for i = 1:l-(n-1)
total(i) = sum(v(i:i+(n-1)));
summa = max(total);
index_1 = find(total == summa);
index = min(index_1);
end

Boran Yigit Usta
Boran Yigit Usta el 16 de Nov. de 2022
function [summa,index] = max_sum(v, n)
b = zeros(1,n);
a = find(v == max(v));
for ii = 1:n
b(ii) = max(max(v));
if find(v==b(ii),1) < a
a = find(v==b(ii),1);
end
v(find(v==max(v),1)) = 0;
end
if n > length(v)
summa = 0;
index = -1;
else
summa = sum(b);
index = a;
end
  1 comentario
Boran Yigit Usta
Boran Yigit Usta el 16 de Nov. de 2022
i tested it it with variety of inputs and it works but still i get this error;
Assessment result: incorrectrandom vectors
Variable summa has an incorrect value. max_sum([ 53 -40 60 -76 -18 31 -61 67 60 75 -60 -2 60 51 -54 -10 -4 -48 -39 80 90 96 53 88 ], 5) returned sum = 429 and index = 10 which is incorrect...

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昱安 朱
昱安 朱 el 5 de Mzo. de 2023
Editada: DGM el 5 de Mzo. de 2023
function [summa,index]=max_sum(v,n)
len=length(v); %v的長度
if(n>length(v)) %n大於v的長度
summa=0;
index=-1;
return;
else
temp_sum=zeros(1,len-n+1);
for x=1:len-n+1
temp_sum(x)=sum(v(x:x+n-1));
end
[summa,index]=max(temp_sum);
end
end % end of function
  1 comentario
DGM
DGM el 5 de Mzo. de 2023
Editada: DGM el 5 de Mzo. de 2023
Still very samey ...
But I'll grant you that you've made some minor improvements, including the addition of a few comments. Next try adding formatting.
If you want to post answers, it's probably best to start with a thread that still has room for unique answers.

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