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Let us parse the expression into an array of lexical tokens, tokens[2*N:0] such that:
- tokens[2*i] i=0,N are the number values
- tokens[2*i-1], i=1,N are the operator values
For each pair of indices (i,j), 0 <= i <= j <= N we define:
• Exp(i,j) = the set of all the expressions can be built with the tokens from 2*i to 2*j
• Min(i,j) = the min value of all the expressions in Exp(i,j)
• Max(i,j) = the max value of all the expressions in Exp(i,j)
Max(0,N) is the max value for the entire expression.
We use dynamic programming to compute Min and Max, based on a few recursive equations.
We cache the computed values as two hash tables, hashMin, hashMax.
Complexity: O(N^3).- ranan.fraer April 03, 2016
There are O(N^2) pairs (i.j). For each pair we compute Min(i,j), Max(i,j) at most once, thanks to the caching. Iterating over k from i+1 to j, takes O(N), so the total is O(N^2 * N) = O(N^3)