function y = KaprekarSteps(x)
y=0;
while x<1000
x=x*10;
end
x1=floor(x/1000);
x2=floor((x-x1*1000)/100);
x3=floor((x-x1*1000-x2*100)/10);
x4=mod(x,10);
if x1==x2&&x2==x3&&x3==x4
y=Inf;
elseif x==6174
y=0;
else
x_ori=[x1,x2,x3,x4];
x_ori=sort(x_ori);
x_inv=fliplr(x_ori);
dif=0;
while dif ~=6174
x_1=1000*x_inv(1)+100*x_inv(2)+10*x_inv(3)+x_inv(4);
x_2=1000*x_ori(1)+100*x_ori(2)+10*x_ori(3)+x_ori(4);
dif=x_1-x_2;
y=y+1;
while dif<1000
dif=dif*10;
end
dif1=floor(dif/1000);
dif2=floor((dif-dif1*1000)/100);
dif3=floor((dif-dif1*1000-dif2*100)/10);
dif4=mod(dif,10);
dif_ori=[dif1,dif2,dif3,dif4];
x_ori=sort(dif_ori);
x_inv=fliplr(x_ori);
end
end
end

The test suite doesn't match the problem description - how can the answer to test two be 5?

Cheated with 1...:
1000 - 0001 = 999.
999 - 999 = 0
y = inf;

No?

How it works with x = 3, x = 691, x = 1?

This works for all but test 3 where it gives, in my opinion, the correct answer 8.

But zero is not sorted in this solution. You should get same answer for input x = 691 and 6910. (And then no need for abs())

1) 9610-0169=9441
2) 9441-1449=7992
3) 9972-2799=7173
4) 7731-1377=6354
5) 6543-3456=3087
6) 8730-0378=8352
7) 8532-2358=6174

This has been the lamest test so far. You need to pad the numbers with zeros to build it up to be a 4 digit number. Not explained in the rules.

Not all test cases seem to be correct. For example x=3 would imply that the next value should be x=3-3=0, so y_correct=Inf and not 6

An efficient lookup table solution

Interesting - a recursive approach

Could you stop doing this kind of thing? I think it would be a lot more fun if we could see the _actual_ best solution..