# Difference between revisions of "Talk:TADM2E 4.4"

From Algorithm Wiki

(Created page with "Why break up the pairs and use buckets? It seems like you simply need three passes: First pass - get the reds and insert them in order into a new array of size n (assuming w...") |
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Each pass is O(n) since we have to visit each element, so the entire algorithm is O(n). | Each pass is O(n) since we have to visit each element, so the entire algorithm is O(n). | ||

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+ | EDIT: | ||

+ | Why three passes ? .Two passes can do it | ||

+ | index_red=-1 | ||

+ | for index =0 to a.length-1 | ||

+ | if a[index].color == "Red" | ||

+ | swap(a,++index_red,index) | ||

+ | index_blue=index_red | ||

+ | for index = index_blue+1 to a.length-1 | ||

+ | if a[index].color=="blue" | ||

+ | swap(a,++index_blue,index) |

## Revision as of 22:03, 29 October 2015

Why break up the pairs and use buckets? It seems like you simply need three passes: First pass - get the reds and insert them in order into a new array of size n (assuming we can use additional space) Second pass - get the blues and insert them in order into the array Third pass - get the yellows and insert them in order into the array

Each pass is O(n) since we have to visit each element, so the entire algorithm is O(n).

EDIT:
Why three passes ? .Two passes can do it

index_red=-1 for index =0 to a.length-1 if a[index].color == "Red" swap(a,++index_red,index) index_blue=index_red for index = index_blue+1 to a.length-1 if a[index].color=="blue" swap(a,++index_blue,index)