0
What else can I help you with?
When was Max O. Lorenz born?
Max O. Lorenz was born in 1876.
When did Max O. Lorenz die?
Max O. Lorenz died in 1959.
When was Max O. Miller - director - born?
Max O. Miller - director - was born in 1918.
When did Max O. Miller - director - die?
Max O. Miller - director - died in 1992.
What has the author Max J O Strutt written?
Max J. O Strutt has written: 'Semiconductor devices'
Is it o or 0 in action replay max?
it is a 0 not a o
How do you pronounce Amaxophobia?
A/max/o/foe/bee/a
What actors and actresses appeared in O dia que Max desceu a rua - 2010?
The cast of O dia que Max desceu a rua - 2010 includes: Filipa Camacho
What is 305 take o whey 42 and 2?
305 - (42 +2) = 305 - 44 = 261
Who plays gypsy joe on max and paddy?
brendan o carroll
How much does it cost to hire Jay Sean?
19,000 pounds a night ; 4 hours max. i hired him for my 16th in sheffield.
What will be the difference in running time of the heap sort algorithm if you start to apply heap sort with an array elements rather than max heap elements?
The heap sort algorithm is as follows: 1. Call the build_max_heap() function. 2. Swap the first and last elements of the max heap. 3. Reduce the heap by one element (elements that follow the heap are in sorted order). 4. Call the sift_down() function. 5. Goto step 2 unless the heap has one element. The build_max_heap() function creates the max heap and takes linear time, O(n). The sift_down() function moves the first element in the heap into its correct index, thus restoring the max heap property. This takes O(log(n)) and is called n times, so takes O(n * log(n)). The complete algorithm therefore equates to O(n + n * log(n)). If you start with a max heap rather than an unsorted array, there will be no difference in the runtime because the build_max_heap() function will still take O(n) time to complete. However, the mere fact you are starting with a max heap means you must have built that heap prior to calling the heap sort algorithm, so you've actually increased the overall runtime by an extra O(n), thus taking O(2n * log(n)) in total.
