Re: [algogeeks] Minimal number of swaps

1 - keep a running total of the sum in an array (or reuse the same array) 2 - a[n - 1] (ie, the length of the array) - a[a[n - 1] - 1] is the answer 0 1 0 1 1 1 0 0 1 1 2 3 4 4 ans = 4 - a [ 4 - 1 ] = 2 1 0 1 0 1 1 1 2 2 3 ans = 3 - a [ 3 - 1 ] = 1 Same Os as the prev solution mentioned, e

Re: [algogeeks] Minimal number of swaps

Clever. Thanks! Le mardi 29 mars 2016 20:23:22 UTC+2, kumar raja a écrit : > > Adding time and space complexities. > > Time complexity: O(n) > Space complexity: O(1) > > > On 29 March 2016 at 23:44, kumar raja > > wrote: > >> I think it is straight forward. Correct me if i am wrong or if there is

Re: [algogeeks] Minimal number of swaps

Adding time and space complexities. Time complexity: O(n) Space complexity: O(1) On 29 March 2016 at 23:44, kumar raja wrote: > I think it is straight forward. Correct me if i am wrong or if there is > better solution. > > 1) Do one pass over the list of elements and count the number of 1's. l

Re: [algogeeks] Minimal number of swaps

I think it is straight forward. Correct me if i am wrong or if there is better solution. 1) Do one pass over the list of elements and count the number of 1's. let us say it is K 2) count = 0 from index 0 to K-1 do if array[index] != 1 count ++ end end Th

[algogeeks] Minimal number of swaps

This puzzle comes from a contribution on codingame.com (link to the puzzle ). Any idea to solve it efficiently? Given a list of 1 and 0, you must regroup all the 1 at the begin of the list in a minimum number of steps. A step is the inter