That's not possible for your example string, is it?
I don't see why not?
If a->1, b->2, c->3, d->4, e->1, f->2, g->3, h->4, every other i->1|2, every other j->3|4; that'd would 25 each of 1, 2, 3 & 4.
Which I guess indicates a laborious way to do it:
chhbfbjjjcgaiiaachbaefeabcbjffdgaggbgccaihgjddedfheejdigcdeiacebdehfig
+egifibdabhfdihbdhifcgjfjhcaejj
3442223433311211342112112324224313323331143344142411442334111312414223
+131222412424142442233324431134
11 11 11 11 1 11 1 11 111 1 1
+1 1 1 1 11 25
222 2 2 2 2 2 22 2 2 2 2 22
+ 222 2 2 2 22 2 25
3 3 333 3 3 3 33 333 33 33 3 3
+ 3 333 3 3 25
44 4 4 4 4 4 44 4 4 44 4 4 4
+ 4 4 4 4 44 44 4 25
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