Difference between revisions of "TADM2E 4.18"
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+ | <pre> | ||
+ | from random import choices | ||
+ | |||
+ | def group_by_color(data, start, stop, color): | ||
+ | operations_counter = 0 | ||
+ | |||
+ | while start < stop and start < len(data) and stop >= 0: | ||
+ | # data[start] is equal to examine(data[start]) except swap function call | ||
+ | |||
+ | while start < len(data) and data[start] == color: # Finding first element with incorrect color from start | ||
+ | start += 1 # Element with correct color already on right position, moving forward | ||
+ | operations_counter += 1 | ||
+ | |||
+ | while start < stop and stop >= 0 and data[stop] != color: # Finding first element with correct color from end | ||
+ | stop -= 1 # Element with incorrect color will be swapped later or keep on that position, moving backward | ||
+ | operations_counter += 1 | ||
+ | |||
+ | if start < stop: | ||
+ | data[start], data[stop] = data[stop], data[start] # Equal to swap(start, stop) | ||
+ | operations_counter += 1 | ||
+ | |||
+ | assert operations_counter <= len(data) * 2 # Check O(2*n) restriction | ||
+ | return start # Return last element with specified color | ||
+ | |||
+ | |||
+ | colors_list = choices(['red', 'white', 'blue'], k=100) | ||
+ | red_ending = group_by_color(colors_list, 0, len(colors_list) - 1, 'red') # Sort red O(n) | ||
+ | group_by_color(colors_list, red_ending + 1, len(colors_list) - 1, 'white') # Sort white O(n) | ||
+ | # Blue already sorted | ||
+ | </pre> | ||
+ | |||
+ | First pass: group red elements from 0 to k. Swapping not red elements found from begin and red elements found from end. | ||
+ | |||
+ | Second pass: group white elements from k to j. Swapping not white elements from k and white elements from end. | ||
+ | |||
+ | Blue elements already groupped from j to n. | ||
+ | |||
+ | --[[User:Bkarpov96|Bkarpov96]] ([[User talk:Bkarpov96|talk]]) 17:36, 19 June 2020 (UTC) |
Revision as of 11:20, 22 July 2020
from random import choices def group_by_color(data, start, stop, color): operations_counter = 0 while start < stop and start < len(data) and stop >= 0: # data[start] is equal to examine(data[start]) except swap function call while start < len(data) and data[start] == color: # Finding first element with incorrect color from start start += 1 # Element with correct color already on right position, moving forward operations_counter += 1 while start < stop and stop >= 0 and data[stop] != color: # Finding first element with correct color from end stop -= 1 # Element with incorrect color will be swapped later or keep on that position, moving backward operations_counter += 1 if start < stop: data[start], data[stop] = data[stop], data[start] # Equal to swap(start, stop) operations_counter += 1 assert operations_counter <= len(data) * 2 # Check O(2*n) restriction return start # Return last element with specified color colors_list = choices(['red', 'white', 'blue'], k=100) red_ending = group_by_color(colors_list, 0, len(colors_list) - 1, 'red') # Sort red O(n) group_by_color(colors_list, red_ending + 1, len(colors_list) - 1, 'white') # Sort white O(n) # Blue already sorted
First pass: group red elements from 0 to k. Swapping not red elements found from begin and red elements found from end.
Second pass: group white elements from k to j. Swapping not white elements from k and white elements from end.
Blue elements already groupped from j to n.