1. Given a 3 array like below NSArray *a = [1,3,4,5]; NSArray *b = [-1,3,0,9]; NSArray *c = [0,31,32,22,6]; Find the elements from the three array which existing in atleast 2 arrays. Eg: [3, 0] Because 3 is presented in array a, b and 0 is presented in array b, c respectively. -(NSArray)find2ElementsAtleastPresentIn2Arrays:(NSArray*)aList b::(NSArray*)bList c::(NSArray*)cList{ // -- your code here. }
iOS developers have a "dispatch_after(when, queue, block)" Grand Central Dispatch (GCD) function they can utilize but once it's set up, these calls can not be easily cancelled. Describe how you might implement a more convenient version of this named "cancellable_dispatch_after"
One question 3 of the interviewers asked me was "What was the most difficult thing you had to deal with.".
They asked a lot of iOS questions, and some general programming questions. The first question they asked was so obscure that I didn't even really understand it. I was probably dead from that point on. Another question was to figure out a way to combine three lists of items where an item would be placed in a destination array if it existed in any two of the source arrays.
Write a function that determines which function is a mirror number. i.e. 12321 would return YES
Given a string, find the longest substring which contains 2 unique characters. ""abcbbbbcccbdddadacb" => "bcbbbbcccb"
Can write a function which inputs two ints and check if it is 3 it should return 6 if input is 6 return 3, without using any condition.
Return the head node of the singly linked list with each pair of nodes swapped. If there is a last odd node leave it in place. Example: Input: 1 -> 2 -> 3 -> 4 -> 5 Output: 2 -> 1 -> 4 -> 3 -> 5
The same list, back to inorder binary tree
The same usual customer service questions. Same sort of questions that are on the application
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