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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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Ajai Kannan il 13 Feb 2019
Chiuso: Rik il 22 Giu 2020
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 CommentiMostra 11 commenti meno recentiNascondi 11 commenti meno recenti
Walter Roberson il 4 Giu 2020

### Risposta accettata

Guillaume il 13 Feb 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 CommentiMostra 18 commenti meno recentiNascondi 18 commenti meno recenti
Stephen23 il 1 Mag 2019
Modificato: Stephen23 il 1 Mag 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
[summa, index] = max(movsum(h, n, 'Endpoints', 'discard'))
is exactly equivalent to
tmp = movsum(h, n, 'Endpoints', 'discard');
[summa, index] = max(tmp)

### Più risposte (37)

lalit choudhary il 17 Apr 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 CommentiMostra 5 commenti meno recentiNascondi 5 commenti meno recenti
Walter Roberson il 11 Mag 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 il 3 Apr 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 CommentiMostra 2 commenti meno recentiNascondi 2 commenti meno recenti
Vipin Parthan il 29 Mag 2020
Modificato: Vipin Parthan il 29 Mag 2020
if j==1
s = sum(v(j:n));
i=1;
Could you explain how this works?

Prasad Reddy il 14 Apr 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

asad jaffar il 4 Apr 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 CommentiMostra 5 commenti meno recentiNascondi 5 commenti meno recenti
Vikas Kumar il 11 Giu 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 il 29 Giu 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 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
Irfan Hussain il 31 Mar 2020
what about negative values, take summa = -inf;

Roshan Singh il 15 Set 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

Hussain Bhavnagarwala il 13 Mar 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

Shashidhar Rai il 15 Mar 2020
Modificato: Walter Roberson il 15 Mar 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 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
Guillaume il 15 Mar 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 il 31 Mar 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 CommentoMostra NessunoNascondi Nessuno
Shubham Pandey il 3 Apr 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 il 7 Apr 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...
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

Emine Ertugrul il 13 Apr 2020
function [summa,index]=max_sum(v,n)
m=length(v);
if n>m
summa=0;
index=-1;
else
[a,b]=max(M)
summa = a
index = b
end
end
##### 1 CommentoMostra NessunoNascondi Nessuno
Walter Roberson il 13 Apr 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 il 14 Apr 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 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
Garvit Kukreja il 14 Apr 2020
Modificato: Garvit Kukreja il 14 Apr 2020
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 il 21 Apr 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

Ahmed J. Abougarair il 21 Apr 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

Roshan Barnwal il 27 Apr 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?
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

Salman P H il 30 Apr 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

Olel Arem il 30 Apr 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

Prithvi Shams il 5 Mag 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

anuj chaudhari il 5 Mag 2020
Modificato: anuj chaudhari il 5 Mag 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

Shandilya Kiran Bhatt il 11 Mag 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

Tahsin Oishee il 17 Mag 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 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
Walter Roberson il 17 Mag 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 il 18 Mag 2020
Modificato: khyathi beeram il 18 Mag 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

vighnesh rana il 19 Mag 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

Taif Ahmed BIpul il 20 Mag 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

utkarsh singh il 21 Mag 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

sai teja il 28 Mag 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

Kumar Shubham il 2 Giu 2020
Modificato: Kumar Shubham il 2 Giu 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

Sumit Kumar Sharma il 2 Giu 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

KAVITI BHARGAV RAM NAIDU il 3 Giu 2020
Modificato: KAVITI BHARGAV RAM NAIDU il 5 Giu 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))
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

Methil Muley il 4 Giu 2020 ##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

zineb britel il 4 Giu 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 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
zineb britel il 7 Giu 2020
thank you, I haven't paid attention to 'consecutif'

Ujjawal Barnwal il 8 Giu 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

AYUSH MISHRA il 13 Giu 2020
Modificato: AYUSH MISHRA il 13 Giu 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

Create a row vector named x that starts at 1, ends at 10, and contains 5 elements.
##### 1 CommentoMostra NessunoNascondi Nessuno
Walter Roberson il 18 Giu 2020
This does not appear to be an Answer to the current Question ?

Sneha Paul il 19 Giu 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)
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

MICHAEL il 19 Giu 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
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

Sakib Javed il 22 Giu 2020
Modificato: Sakib Javed il 22 Giu 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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