canto - 1 built - 2 foist - 2 admit - 2 cedis - 1 inept - 2 stale - 2No zeros here, so we're in for some work.
Might as well start at the top. If there's a c, then there's no a, n, t, or o (canto=1). Working our way down the list, we see that we must have two of d, m, and i (admit=2 and anto=0). However, since there is a c, we know there is no e, d, i, or s (cedis=1). That means it's impossible to have two of d, m, and i, so we have a contradiction. Therefore, there is no c.
Let's try the next letter in canto. If there's an a, then there's no c, n, t, or o (canto=1). Looking at built, we know that we must have two of b, u, i, or l. Working our way down the rest of the list in a similar fashion, we have the following:
a(buil=2)(fis=2)(dmi=1)(edis=1)(iep=2)(sle)That doesn't look too helpful at first, but we can break it up into smaller units. Starting from the left, we see that there are 6 combinations from (buil=2), which is a lot. However, there are only 3 combinations from (fis=2), so let's look at that.
If it's fi, then, knocking out i from some of the combinations, we have:
a(bul=1)fi(dm=0)(eds=0)(ep=1)(sle=1)Taking the new zeros into account, we can reduce (ep=1) to p, and we can reduce (sle=1) to l. This gives us the five letters afipl. Since we're left with exactly five letters, this looks like a good opportunity to see if we can make a word out of this. Some playing around with ordering of the letters turns up the solution, pilaf. Since we've found a word, we know we can stop because this is a Jotto puzzle, which by definition only has one word that works.