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Linear partition in Coffeescript (Javascript)
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# Linear partition | |
# Partitions a sequence of non-negative integers into k ranges | |
# Based on Óscar López implementation in Python (http://stackoverflow.com/a/7942946) | |
# Also see http://www8.cs.umu.se/kurser/TDBAfl/VT06/algorithms/BOOK/BOOK2/NODE45.HTM | |
# Dependencies: UnderscoreJS (http://www.underscorejs.org) | |
# Example: linear_partition([9,2,6,3,8,5,8,1,7,3,4], 3) => [[9,2,6,3],[8,5,8],[1,7,3,4]] | |
linear_partition = (seq, k) => | |
n = seq.length | |
return [] if k <= 0 | |
return seq.map((x) -> [x]) if k > n | |
table = (0 for x in [0...k] for y in [0...n]) | |
solution = (0 for x in [0...k-1] for y in [0...n-1]) | |
table[i][0] = seq[i] + (if i then table[i-1][0] else 0) for i in [0...n] | |
table[0][j] = seq[0] for j in [0...k] | |
for i in [1...n] | |
for j in [1...k] | |
m = _.min(([_.max([table[x][j-1], table[i][0]-table[x][0]]), x] for x in [0...i]), (o) -> o[0]) | |
table[i][j] = m[0] | |
solution[i-1][j-1] = m[1] | |
n = n-1 | |
k = k-2 | |
ans = [] | |
while k >= 0 | |
ans = [seq[i] for i in [(solution[n-1][k]+1)...n+1]].concat ans | |
n = solution[n-1][k] | |
k = k-1 | |
[seq[i] for i in [0...n+1]].concat ans |
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