599. Minimum Index Sum of Two Lists
Easy
TopicsCompanies
Given two arrays of strings list1 and list2, find the common strings with the least index sum.
A common string is a string that appeared in both list1 and list2.
A common string with the least index sum is a common string such that if it appeared at list1[i] and list2[j] then i + j should be the minimum value among all the other common strings.
Return all the common strings with the least index sum. Return the answer in any order.
 
Example 1:
Input: list1 = ["Shogun","Tapioca Express","Burger King","KFC"], list2 = ["Piatti","The Grill at Torrey Pines","Hungry Hunter Steakhouse","Shogun"]
Output: ["Shogun"]
Explanation: The only common string is "Shogun".


Example 2:
Input: list1 = ["Shogun","Tapioca Express","Burger King","KFC"], list2 = ["KFC","Shogun","Burger King"]
Output: ["Shogun"]
Explanation: The common string with the least index sum is "Shogun" with index sum = (0 + 1) = 1.


Example 3:
Input: list1 = ["happy","sad","good"], list2 = ["sad","happy","good"]
Output: ["sad","happy"]
Explanation: There are three common strings:
"happy" with index sum = (0 + 1) = 1.
"sad" with index sum = (1 + 0) = 1.
"good" with index sum = (2 + 2) = 4.
The strings with the least index sum are "sad" and "happy".


 
Constraints:
1 <= list1.length, list2.length <= 1000
1 <= list1[i].length, list2[i].length <= 30
list1[i] and list2[i] consist of spaces ' ' and English letters.
All the strings of list1 are unique.
All the strings of list2 are unique.
There is at least a common string between list1 and list2.
Q-




—> we have two arrays of strings “list1 and “list2” , find the common stings with the least index sum 


Q- what is the defintion of a common string ? 


A common string is a string that appeared in both list1 and list2 

Q- what is the constrain of this exercise 

We need to find the common string with the least index sum possible 

Q- do we have use a greedy approach here ? 

Yes we can use a greedy approach here where the minimum possible index sum is if we find the common string in quickes index i , jtraversing the two list from the left to right ? 

Q- how to deal if both list of words has deffrente length ? 

We can use two pointers i and j  each one tranver the lenth n , m respecticvly 
Q- what the hell n and m represent ? 

N: represent the length of the first length 
M: reprenent the length of the second length 
Q-  do we need to use while or for loop and why ? 

We can use both but i think while loop is more appropriate here ? 
Really i do not have the strongest argument why but can you help me ? with god argument ? 



Q- what is the input & output? 


Input : List[str] , List[str] 

Output: int 


Q- did my approach so far true yes or non ? 

et's address your questions and clarify your approach to solving the Minimum Index Sum of Two Lists problem.

Clarifying Input and Output
Q: What is the input & output?
Input:
list1: List[str] — An array of strings.
list2: List[str] — Another array of strings.
Output:
List[str] — A list of common strings with the smallest index sum.
Correction:
In your previous response, you mentioned the output as int, but actually, the problem requires returning a list of strings (List[str]) that are common to both list1 and list2 and have the minimal index sum.

Evaluating Your Approach
Q: Did my approach so far true yes or no?
Your Approach So Far:
Greedy Strategy:
You suggested using a greedy approach by finding the common string with the smallest index sum by traversing both lists from left to right.
Handling Different Lengths:
You considered using two pointers, i and j, to traverse each list separately.
Assessment:
Your approach has the right intuition but can be optimized for efficiency and clarity. Here's why:
Greedy Approach Consideration:
Pros:
Traversing from left to right can help in identifying strings with smaller indices early.
Cons:
Simply traversing both lists sequentially with two pointers might not account for all possible pairs efficiently, especially since common strings can appear anywhere in either list.
Using Two Pointers (i and j):
This method is more suited for problems where both lists are sorted or have some order based on the indices.
In this problem, since the lists are not guaranteed to be sorted in any particular way concerning the common strings, using two pointers might not be the most efficient strategy.



Q- oky one approach here is to convert one list of hashamp 

The key is the value and the index is the index 

It help to make a quick lockup when traversing the second list oky 


Q- then traverse the second list word and index at the same time oky ? 

For each word check it presence in the hashmap 
If yes store the word in a list of in a hashamp in
Where the key is the value of this word and the value it it’s sum 

Q- after breaking from the loop we can to return a list of key “words” by filter only the min value here oky 

Q- example of my workflow of this 

“Input: list1 = ["happy","sad","good"], list2 = ["sad","happy","good"]
Output: ["sad","happy"]
“
Throw this example 

Step 1 : converter it into a hashmap oky 
{“happy”:0,”sad”:1,”good”:2} 
Step 2 : traverse the second list word by word with inex 

In python we can leverage the enumerate() 
For i , word in enumerate(lists2) 

Step 3 : check the present of the word in the hashmap 
	If word in hashmap: 
		dic[word] =i+hashmap[word] 
Step 4 : get list okys by filtering based on the keys based on the min value possibl 

dic={“sad”:1,”happy”:1} 
The min value is 1 
result=dic.filter(lambda x:x is min(dic.value) 

Q- first think i want to give your honest feadback about my approach and if it’s right yes or non oky ? 

Q- use methodological approach to guide me to the end implementation and do give the actual solution