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See original GitHub issue

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; 

Issue Analytics

  • State:open
  • Created 5 years ago
  • Comments:6 (1 by maintainers)

github_iconTop GitHub Comments

vin0010commented, Nov 19, 2018

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 = (temp[0]+1, temp[1])
                count_map[nums[i]] = temp
                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]]
                if k-i in count_map:
                    return [count_map[i][1][0], count_map[k-i][1][0]]

IAmPramodcommented, Oct 16, 2018

@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.

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