leetcode/hash-table/TwoSum.java

There is a better solution for the TwoSum problem than the hashmap solution. Sure hashmap is fine and meets the desired output. however it can be completely eliminated as such. The techinque employs a pointers to track the argument array i.e. one from the begining aka head and the other one from the end aka tail. A single sweep on the array and adjusting the two pointers does the job rather nicely.

       //input array must be sorted 
       int head =0;  int tail = arr.length -1;  int k = 11;  //target sum to find

        while(head < tail) {
           int sum = arr[head] + arr[tail];
           if(sum == k)  return true; //found it !!
           else if(sum < k) ++head;
           else --tail; 
        }

[vin0010]

May be this is not completely related to the discussion. But with hash map and tuple together, we can solve this problem easily.

def get_count_map(nums):
        count_map = dict()
        for i in range(0, len(nums)):
            if nums[i] in count_map:
                temp = count_map[nums[i]]
                temp[1].append(i)
                temp = (temp[0]+1, temp[1])
                count_map[nums[i]] = temp
            else:
                count_map[nums[i]] = (1, [i])
        return count_map


    def print_indices(count_map, k):
        for i in count_map:
            if 2*i == k:
                return [count_map[i][1][0], count_map[i][1][1]]
                break
            else:
                if k-i in count_map:
                    return [count_map[i][1][0], count_map[k-i][1][0]]
                    break

https://github.com/vin0010/Competitve-Programming/blob/master/python/leetcode/TwoSum.py

[IAmPramod]

@sragha45 Yes, you are correct. But O(NlogN) is the worst case scenario when we have a very poor implementation of hash code which will map all entries to same bucket.

Have a look at the performance of hashmap in average case. https://dzone.com/articles/hashmap-performance