Two Sum

# Brief taken from: https://leetcode.com/problems/two-sum/
# 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].

# Will use the example values to run
nums = list([2, 7, 11, 15])
target = 9


def slowMethod():
    for i in range(0, len(nums)):
        for j in range(0, len(nums)):
            if i == j:
                # Accounts for the two variables being the same value, whilst also allowing for repeating values
                continue
            elif nums[i] + nums[j] == target:
                targetVals = [i, j]
                # Without breaks, the code keeps going, thus increasing processing time, as well as the targetVals list will
                # be overwritten
                break
        # Could potentially make the code a function to enable a cleaner exit from the loops
        if len(targetVals) == 2:
            break

    # Checks the end states, could use either the length of the targetVals array, or test for if it exists
    if len(targetVals) == 2:
        print("The indices of the values that sum to "+str(target)+" are: "+str(targetVals[0])+","+str(targetVals[1]))
    else:
        print("No matches found")

def hashmapMethod():
    # Creates a dictionary with values and their indexes to see if a given value's partner value is in said index.
    visited = {}
    for i, num in enumerate(nums):
        pairVal = target - num
        if pairVal not in visited:
            visited[num] = i
        else:
            return [visited[pairVal], i]


print(hashmapMethod())