MATLAB: Index exeeds matrix dimensions

arraymatrix array

for n = 1:k
    %HS or LE
    prompt = 'Enter individual tasks memory location \n High Speed = 1 \n Low Energy = 0 \n';
    fprintf('M(%d) = \n',n);
    M(n) = input (prompt);
    if M(n) == 1
        h = h+1;
        prompt = 'Enter the hiding value of high speed tasks \n';
        fprintf('V(%d) = \n',h);
        Vh(h) = input (prompt);
        hst(h) = n;
    else
        l = l+1;
        prompt = 'Enter the hiding value of low speed tasks \n';
        fprintf('V(%d) = \n',l);
        Vl(l) = input (prompt);
        let(l) = n;
    end
end
for n = 1:k
    if M(n) == 1
        a = 'High speed';
        fprintf('M(%d) = %s \n',n,a);
    else
        b = 'Low Energy';
        fprintf('M(%d) = %s \n',n,b);
    end
end
fprintf('Number of High speed tasks = %d \n',h);
fprintf('Number of Low Energy tasks = %d \n',l);
for n = hst(1):hst(h)
    fprintf('Vh(%d) = %d \n',n,Vh(n));
end
for n = let(1):let(l)
    fprintf('Vl(%d) = %d \n',n,Vl(n));
end
Output
M(1) = High speed 
M(2) = High speed 
M(3) = High speed 
M(4) = High speed 
M(5) = Low Energy 
M(6) = Low Energy 
M(7) = Low Energy 
Number of High speed tasks = 4 
Number of Low Energy tasks = 3 
Vh(1) = -100 
Vh(2) = -1 
Vh(3) = -2 
Vh(4) = -3 
Index exceeds matrix dimensions.

Best Answer

for n = 1:h
not
for n = hst(1):hst(h)
and likewise
for n = 1 : l
not
for n = let(1):let(l)
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