Murali wrote: > Now I dont want to create another list but just modify it in place.
Why does that matter? List copies are cheap. > SolutionA: > > for x in range(len(L)): > if L[x] == X: > L[x:x] = Y Did you run this code ? > SolutionB: > > p = L.index(X) > while p >= 0: > L[p:p] = Y > p = L.index(X) Did you run this code ? > Problem with both solutions is the efficiency. Both methods require > time O(N^2) in the worst case, where N is the length of the list. > Because L.index() and L[x:x] both take O(N) time in the worst case. Assigning a single item to L[x:x] doesn't work. Assigning M items to a slice of length M is an O(M) operation, so if you do L[x:x+1] = [Y], you get your O(1) operation. But that's just a silly way to write L[x] = Y, which I assume was what you meant. L[x] = Y is also an O(1) operation, of course. </F> -- http://mail.python.org/mailman/listinfo/python-list
