Solutions to review questions on structs
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1. TopHalf

function C = topHalf(A)
% Refer to Lecture 21.
% Suppose A is a length-50 structure array with two fields
% Assume A(k).name is a string that names a state and A(k).pop
% is an integer whose value is the states population.
% Assume that the states are ordered alphabetically.
% C is a length-25 cell array of strings that names all the
% states whose populations are above the median state population.
% The states should be ordered alphabetically in C.

% Obtain a vector of populations...
pop = zeros(50,1);
for k=1:50
   pop(k) = A(k).pop;
end

[y,idx] = sort(pop);
% Note that idx(k) is the index of the kth smallest state.
% Above-the-median states have indices idx(26),..,idx(50).
% Put these indices in a single, length-25 vector

Above = idx(26:50);
% Without the following step the "big" states wont be assembled in alphabetical
% order...

Above = sort(Above);

% This assembles the state names in C...
C = cell(25,1);
for k=1:25
   j = Above(k);
   str = A(j).name;
   C{k} = str;
end



2. BigTriplets
function N = BigTriplets(A)
% Refer to the Lectures 20 and 21.
% Suppose A is a length-50 structure array with two fields.
% Assume A(k).name is a string that names a state and A(k).pop
% is an integer whose value is the states population.
% We say that three different states form a "big triplet" if
% the sum of their populations is greater than 15 million.
% N is the number of big triplets.

N = 0;
for i=1:50
   for j=i+1:50
      for k=j+1:50
         if A(i).pop+A(j).pop+A(k).pop > 15000000
            N = N+1;
         end
      end
   end
end



3.  Point struct

% Given a structure array Pts where each structure has two fields, x and y,
% sort Pts so that the structures are in the order of 
% increasing distance from (0,0)

n = length(Pts);

% compute the distances
dist = zeros(n, 1);
for i=1:n
    dist(i)=sqrt(Pts(i).x^2+Pts(i).y^2);
end
% sort the distances
[s, idx] = sort(dist);

oldPts = Pts;

for i=1:n
    Pts(i) = oldPts(idx(i));
end