Minimum Index Sum of Two Lists

The main idea to solve this problem is to use a dictionary to keep track of the strings in the first list and their indices. Then we traverse the second list. If a string is found in the dictionary (meaning it’s common), we calculate its index sum. If it’s smaller than the current minimum, we update the minimum and clear the result list. If it’s equal to the minimum, we add it to the result list.

Python solution:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
class Solution:
    def findRestaurant(self, list1: List[str], list2: List[str]) -> List[str]:
        dict1 = {string: index for index, string in enumerate(list1)}
        min_index_sum = float('inf')
        result = []

        for index, string in enumerate(list2):
            if string in dict1:
                index_sum = index + dict1[string]
                if index_sum < min_index_sum:
                    min_index_sum = index_sum
                    result = [string]
                elif index_sum == min_index_sum:
                    result.append(string)

        return result

In this solution, we first create a dictionary from list1 using Python’s dictionary comprehension. We then iterate over list2 using enumerate to get both the string and its index. If the string is in the dictionary (i.e., it’s a common string), we calculate its index sum. If this sum is less than the current min_index_sum, we update min_index_sum and set result to be a list containing only this string. If the sum is equal to min_index_sum, we add the string to result. In the end, result will contain all the common strings with the least index sum.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
class Solution:
    def findRestaurant(self, list1: List[str], list2: List[str]) -> List[str]:
        words1 = {word: idx for idx, word in enumerate(list1)}

        min_sum = math.inf
        for idx2, word2 in enumerate(list2):
            if word2 in words1:
                if words1[word2] + idx2 < min_sum:
                    min_sum = words1[word2] + idx2
                    min_words = [word2]
                elif words1[word2] + idx2 == min_sum:
                    min_words.append(word2)

        return min_words

Problem Classification

This problem falls under lists/arrays and string manipulation. It also involves elements of searching and comparison, and might be relevant in contexts such as text processing, search optimizations, and database queries.

‘What’ Components:

Two lists of strings, list1 and list2.
The task is to find common strings between the two lists.
Not just any common strings, but those with the least index sum. If a common string appeared at list1[i] and list2[j], i + j should be the minimum among all other common strings.
The output should be all the common strings with the least index sum. The order in which they’re returned does not matter.

This problem can be classified as a searching and optimization problem where the goal is to find a subset of data (common strings) with a minimum property (sum of indices). Solving it efficiently would involve techniques from the field of data structure manipulation and possibly hash table usage for efficient lookups.

Language Agnostic Coding Drills

Dissect the Code:

The key concepts involved in this code are:

a. List Iteration: The code uses basic list iteration to traverse through `list1` and `list2`.
b. Dictionary Creation: The code creates a dictionary `words1` from `list1` with words as keys and their indices as values.
c. Element Lookup in Dictionary: The code checks if a word from `list2` exists in the dictionary `words1`.
d. Conditional Statements: The code uses `if` and `elif` to determine which action to take based on the condition.
e. Dictionary Value Retrieval: The code retrieves the value (index) for a word from the dictionary.
f. List Creation and Append Operation: The code creates a list `min_words` and appends elements to it.
g. Usage of Infinity (`math.inf`): The code initializes `min_sum` with `math.inf` to ensure any sum encountered will be less than the initial value.

Coding Concepts Ordered by Increasing Difficulty:
a. List Iteration (Easy): Fundamental operation of looping over each element in a list. b. Conditional Statements (Easy): Basic if and elif statements to control the flow of code. c. List Creation and Append Operation (Easy): Basic operation to create a list and add elements to it. d. Dictionary Creation (Intermediate): Creating a dictionary from a list with custom keys and values. e. Element Lookup in Dictionary (Intermediate): Checking if a key exists in a dictionary. f. Dictionary Value Retrieval (Intermediate): Retrieving the value for a key from a dictionary. g. Usage of Infinity (math.inf) (Intermediate): Conceptual understanding of using math.inf as an initial large value in comparison scenarios.
Problem-solving Approach:

The strategy employed here is one of space-time tradeoff. Instead of directly comparing each element of list1 with each element of list2 (which would be O(n^2) complexity), we create a dictionary from list1 (which requires O(n) space). This dictionary enables quick lookups in constant time, thus reducing the time complexity to linear.

The steps involved are:

a. Iterate through `list1` and construct a dictionary, mapping words to their indices.
b. Iterate through `list2`, and for each word, check if it exists in the dictionary.
c. If it does, calculate the index sum (sum of the current index in `list2` and the value from the dictionary).
d. If this sum is less than the current minimum sum, update the minimum sum and replace `min_words` with the current word.
e. If the sum is equal to the current minimum sum, append the current word to `min_words`.
f. At the end of iteration, `min_words` will contain all the words with the least index sum.

Targeted Drills in Python

Python-based Coding Drills for Identified Concepts:

a. List Iteration:

 ```python
 my_list = [1, 2, 3, 4, 5]
 for element in my_list:
     print(element)
 ```

b. Conditional Statements:

 ```python
 a = 5
 b = 10
 if a > b:
     print("a is greater")
 elif a < b:
     print("b is greater")
 else:
     print("a and b are equal")
 ```

c. List Creation and Append Operation:

 ```python
 my_list = []
 my_list.append("Hello")
 my_list.append("World")
 print(my_list)
 ```

d. Dictionary Creation:

 ```python
 my_dict = {i: i * i for i in range(1, 6)}
 print(my_dict)
 ```

e. Element Lookup in Dictionary:

 ```python
 my_dict = {"apple": 1, "banana": 2, "cherry": 3}
 if "banana" in my_dict:
     print("banana is in the dictionary")
 ```

f. Dictionary Value Retrieval:

 ```python
 my_dict = {"apple": 1, "banana": 2, "cherry": 3}
 print(my_dict["banana"])
 ```

g. Usage of Infinity (math.inf):

 ```python
 import math
 my_num = math.inf
 if 1000000 < my_num:
     print("Infinity is greater than a million")
 ```

Problem-Specific Concepts:
No unique problem-specific concepts beyond what’s mentioned above are necessary for this problem. The primary strategy is to use these concepts and integrate them cohesively.
Integration of Concepts to Solve Problem:
Once a good understanding of the above concepts is attained, they can be integrated together to solve the problem in the following manner:
a. Use the List Iteration and Dictionary Creation concepts to iterate through list1 and create a dictionary mapping words to their indices.
b. Next, use List Iteration to traverse list2, for each word, use Element Lookup in Dictionary to check if it exists in the dictionary. If it does, calculate the index sum (with Dictionary Value Retrieval).
c. Use Conditional Statements and List Creation and Append Operation to update min_words and min_sum based on the calculated sum. Utilize the Usage of Infinity (math.inf) concept for the initial comparison of min_sum.
d. At the end of this process, min_words will contain the desired output. The learned concepts have now been effectively used to solve the problem.