The following algorithm examine the contents of each element of the array at most once.
1. Start with the first even position and the first odd position. 2. Search forward through the even positions until you reach the end of the array or find a 1, whichever comes first. 3. Search forward through the odd positions until you reach the end of the array or find a 0, whichever comes first. 4. If you reached the end of the array in either of the above searches, you are finished. 5. Otherwise, set the even position to 0 and the odd position to 1. 6. Repeat steps 2-6. Dave On Nov 13, 7:39 am, geekko <[EMAIL PROTECTED]> wrote: > you are given an array of integers containing only 0s and 1s. You have > to place all the 0s in even positions and 1s in odd position. And if > suppose, no. of 0s exceeds no. of 1s or vice versa keep them > untouched. Do in ONE PASS without taking extra memory.(modify array in > place) --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
