Question

public static int getv(int num, List<Integer> rating, int target) { int count 0; HashMap<Integer, Integer> map = new HashMap<>(); for(int i = 0; i < num; i++){ if(!map.containsKey(rating.get(i))){ map.put(rating.-get(i), 0); } map.put(rating.get(i), map.get(rating.get(i))+1); } for(int i = 0; i < num; i++){ if(map.get(target -rating.get(i)) != null){ count += map.get(target-rating.get(i)); } if(target-rating.get(i) == rating.get(i)){ count--; } return count/2; }

What is the time and space complexity of this code with explanation?

Answer 1

In hashmap, inserting an element takes O(1) time,
getting an element from any index also takes O(1) time.

The first loop runs from 0 to num.
- map.conttainsKey takes O(1)
- rating.get(i) also takes O(1)
- map.put also takes O(1)
- map.put also takes O(1)

hence this loop takes O(num) to execute.

Second loop runs from o to num.
- map.get, rating.get() takes O(1)

So all the operations inside this loop execute in constant time(O(1))

Hence this loop also takes O(num) time.

so T(num) = O(num) + O(num)
T(num) = 2.O(num)

since O(big oh) represents the upper bound on the function,
hence we can neglect the constant terms.

T(num) = O(num)