1. Two Sum

https://leetcode.com/articles/two-sum/#approach-2-two-pass-hash-table-accepted

 

Question

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

 

Solution

Approach #1 (Brute Force)

Loop through each element x and find if there is another value that equals to target - x.

 

Time complexity : O(n^2)O(n​2​​)

Space complexity : O(1)O(1).

 

Approach #2 (Two-pass Hash Table)

A simple implementation uses two iterations. In the first iteration, we add each element's value and its index to the table. Then, in the second iteration we check if each element's complement (target - nums[i]) exists in the table. Beware that the complement must not be nums[i] itself!

• Time complexity : O(n). We traverse the list containing nn elements exactly twice. Since the hash table reduces the look up time to O(1), the time complexity is O(n).

• Space complexity : O(n). The extra space required depends on the number of items stored in the hash table, which stores exactly nn elements.

 

Approach #3 (One-pass Hash Table) 

 While we iterate and inserting elements into the table, we also look back to check if current element's complement already exists in the table. If it exists, we have found a solution and return immediately.

 

 

 

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        seen = {} # use dictionary data structure
        for i in range(len(nums)): # i is the position/key of the num
            num2 = target - nums[i]  # num2 is the value
            if num2 in seen: # num2 is the key
                return [seen[num2], i]
            else:
                seen[nums[i]] = i
        return 'Not found'

 

Complexity Analysis:

• Time complexity : O(n). We traverse the list containing nn elements only once. Each look up in the table costs only O(1)time.

• Space complexity : O(n). The extra space required depends on the number of items stored in the hash table, which stores at most n elements.

 

 

Plus,

 

If the question only asks us to find out the two elements themselves, rather their indices, the set() data structure in python can help us out.

 

"

For this problem we will take advantage of a set data structure. We will make a single pass through the list of integers, treating each element as the first integer of our possible sum.

At each iteration we will check to see if there is a second integer which will allow us hit the target number, adn we will use a set to check if we've already seen it in our list.

We will then update our seen set by adding the current number in the iteration to it.

"

 

def solution(lst,target):
    
    # Create set to keep track of duplicates
    seen = set()
    
    # We want to find if there is a num2 that sums with num to reach the target
    
    for num in lst:
        
        num2 = target - num
        
        if num2 in seen:
            return [num2, num]
        
        seen.add(num)
        
    # If we never find a pair match which creates the sum
    return False

def twosum(nums,target):
    seen = {}
    for i in range(len(nums)):
        num2 = target - nums[i]
        if num2 in seen:
            return  [seen[num2] +1 , i + 1]
        seen[nums[i]] = i
    return 'Not Found'
    
    
nums = [4, 2, 1, 5, 7]
target = 12
print twosum(nums,target

 

Note:

 

1 Seen is a dictionary, use {} not [] 

 

2 The problem requires us to output the index(position) of elements. When using the dictionary data structure, the key should be the element, and the value should the index. Also note we campare if the element is in the dictionary seen, that's what keys do.

 

We search keys to get values.