CareerCup

1)Form a BST using first Array...

2)Now for each element in 2nd array...search for correponding element in BST....

3)if that element is present delete it and continue for the whole array....

let the lengths of the two arrays be m and n and m<n
sort the small array--O(mlogm)
binary search all the n elements of bigger array in the small array.--O(nlogm)
total complexity--O((n+m)logm)

let the lengths of the two arrays be m and n and m<n
create a hashmap from the smaller array --> O(m).
For every element in the larger array check if the element exists in the hashmap. if it does then push into a vector --> O(n)
Time Complexity: O(m) + O(n) ==> O(n) {since m<n}
Space Complexity: O(m) + O(m) ==> O(m)

1) Create an array of size 256 and of type int to hold all ASCII values.
2) Initialize the whole array to 0
3) Iterate through the first array and set the respected index in array according to their ASCII values to 1.
4) Iterate through the second array and check which elements have 1 in the array according to their ASCII.

Time complexity:
O(256) + O(m) + O(n) = O(n)