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3536628
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0
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2010-08-21T06:38:49.137
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2010-08-21T06:38:49.137
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407254
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3536409
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<p>Here's my rather anti-pythonic implementation that nevertheless leverages python's wonderful built in sets and strings.</p> <pre><code>a = 'ABCDDEGF' b = 'FPCDBDAX' best_solution = None best_solution_total_length = 0 def try_expand(a, b, a_loc, b_loc): # out of range checks if a_loc[0] < 0 or b_loc[0] < 0: return if a_loc[1] == len(a) or b_loc[1] == len(b): return if set(a[a_loc[0] : a_loc[1]]) == set(b[b_loc[0] : b_loc[1]]): global best_solution_total_length, best_solution #is this solution better than anything before it? if (len(a[a_loc[0] : a_loc[1]]) + len(b[b_loc[0] : b_loc[1]])) > best_solution_total_length: best_solution = (a_loc, b_loc) best_solution_total_length = len(a[a_loc[0] : a_loc[1]]) + len(b[b_loc[0] : b_loc[1]]) try_expand(a, b, (a_loc[0]-1, a_loc[1]), (b_loc[0], b_loc[1])) try_expand(a, b, (a_loc[0], a_loc[1]+1), (b_loc[0], b_loc[1])) try_expand(a, b, (a_loc[0], a_loc[1]), (b_loc[0]-1, b_loc[1])) try_expand(a, b, (a_loc[0], a_loc[1]), (b_loc[0], b_loc[1]+1)) for a_i in range(len(a)): for b_i in range(len(b)): # starts of the recursive expansion from identical letters in two substrings if a[a_i] == b[b_i]: # if substrings were expanded from this range before then there won't be an answer there if best_solution == None or best_solution[0][0] > a_i or best_solution[0][1] <= a_i or best_solution[1][0] > b_i or best_solution[1][1] <= b_i: try_expand(a, b, (a_i, a_i), (b_i, b_i)) print a[best_solution[0][0] : best_solution[0][1]], b[best_solution[1][0] : best_solution[1][1]] </code></pre> <p>Forgot to mention that this is obviously a fairly bruteforce approach and I'm sure there's an algorithm that runs much, much faster.</p>
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1. POGiven two strings, find the longest common bag of chars
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1. USNovikov
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