variable to store probing location calculating the hash code

//variable to store probing location

//calculating the hash code

//if the position is empty, immediately return failure...

//...but if it\'s a match, return the data straight away...

//...otherwise, probe to the next item, looping to zero if necessary

//keep probing until data is found or entire table has been visited

//if the probed element is completely empty, return failure

//if the probed element is a match, return the data...

//...otherwise, keep probing for the next item, looping back to zero if necessary

//if nothing has been returned by now, data is not present

Solution

int probe;

//assignment is a constant time operation so O(1)

//here we are checking if table[code] == null which is also a constant time operation so O(1)

//if the position is empty, immediately return failure...

  //same here O(1)

//...but if it\'s a match, return the data straight away...

//...otherwise, probe to the next item, looping to zero if necessary

//The time complexity is O(n) for while loop since we are looping the elements and O(1) for if loop, but as we consider only the biggest term , time complexity is O(n)

//O(n) is also time complexity for whole program for the same reason that we consider only highest term

//keep probing until data is found or entire table has been visited

//same here O(1)

//if the probed element is completely empty, return failure

//same here O(1)

//if the probed element is a match, return the data...

//same here O(1), reason mentioned above

//...otherwise, keep probing for the next item, looping back to zero if necessary

return null;

//if nothing has been returned by now, data is not present