Doing a 2sum algorithm and I'm confused on python syntax-CodePudding

so I followed a tutorial to learn how to complete 2 sum and I understand what ever line means but I dont understand why "diffs[list[i]]" returns the 0 index. I went through the the algorithm with the current arguments and i comes out to 3 when it returns the indexes

    diffs = {}  # Make a hash map to store values

    for i in range(len(list)):  # Iterate through list
        if list[i] in diffs:  # If  the number you are on is in the has map
            return [diffs[list[i]], i]  # return indexes

        else:
            diffs[target - list[i]] = i


print(twosum([2, 11, 15, 7], 9))```

CodePudding user response：

def twosum(nums, target): 
    diffs = {}  # Make a hash map to store values  
 
    for i in range(len(nums)):  # Iterate through list 
        if nums[i] in diffs:  # If  the number you are on is in the has map 
            return [diffs[nums[i]], i]  # return indexes 
        else:   
            diffs[target - nums[i]] = i 
            print(i, diffs)

In [4]: print(twosum([2, 11, 15, 7], 9))                                                                               
0 {7: 0}
1 {7: 0, -2: 1}
2 {7: 0, -2: 1, -6: 2}
[0, 3]

As you can see from the above output, 7 has index 0 in the dictionary. It's the first element that is added to the dictionary. The reason is that you are saving the differences in the dictionary. target - nums[0] is 7 in this case because nums[0] is 2.

Then, when you reach the last element, namely 7, you find it in the dictionary with the index 0. That is diffs[nums[3]] == diffs[7] == 0. And you return [diffs[nums[i]], i] == [diffs[nums[3]], 3] == [diffs[7], 3] == [0, 3].

Also, don't use the name list to name your variables.

CodePudding user response：

On the first run of the loop. diffs[target - list[i]] = i sets diffs to {7: 0}. That's how diffs[list[3]] evaluates to 0. Eg.

diffs[list[3]]
diffs[7]
0