“being trivially different”
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Let s(i),k denote the substring s(i)s(i+1)...s k. Let Opt(k) denote whether the sub-string s1,k can be segmented using the words in the dictionary, namely (k) =1 if the segmentation is possible and 0 otherwise. A segmentation of this sub-string s1,k is possible if only the last word (say si k) is in the dictionary theremaining substring s1,i can be segmented.
Therefore, we have equation:Opt(k) = max Opt(i) 0<i<k and s(i+1),kis a word in the dictionary
We can begin solving the above recurrence with the initial condition that Opt(0) =1 and then go on to comput eOpt(k) for k= 1, 2. The answer correspond-ing to Opt(n) is the solution and can be computed in Θ(n2) time.
Therefore, we have equation:Opt(k) = max Opt(i) 0<i<k and s(i+1),kis a word in the dictionary
We can begin solving the above recurrence with the initial condition that Opt(0) =1 and then go on to comput eOpt(k) for k= 1, 2. The answer correspond-ing to Opt(n) is the solution and can be computed in Θ(n2) time.