What is the time complexity of Minimum Index Sum of Two Lists?

The optimal solution has a time complexity of O(n+m).

What data structures are used to solve Minimum Index Sum of Two Lists?

This problem uses Array, Hash Table, String. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Minimum Index Sum of Two Lists?

Insight 1: Using a hash table (dictionary or map) to store the strings and their indices from one list allows for efficient lookup in the second list. Insight 2: Iterating through the second list and checking if each string exists in the hash table allows for finding common strings and calculating their index sum.

Which companies ask LeetCode 599. Minimum Index Sum of Two Lists in interviews?

This problem is frequently asked at Yelp.

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Minimum Index Sum of Two Lists - LeetCode 599 Solution

Minimum Index Sum of Two Lists - Complete Solution Guide

Minimum Index Sum of Two Lists is LeetCode problem 599, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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","Burg

Detailed Explanation

The problem requires finding common strings between two lists and returning those that have the minimum sum of their indices in the respective lists. The input consists of two string arrays, `list1` and `list2`. The output should be a list of strings that are present in both input lists, such that the sum of their indices (index in `list1` + index in `list2`) is the smallest among all common strings. We are guaranteed that at least one common string exists.

Solution Approach

The solution involves creating a hash table from one of the lists (typically the first list) that stores each string as a key and its index as the value. Then, it iterates through the second list. For each string in the second list, it checks if the string is present in the hash table. If it is, it calculates the index sum (index in the second list + index from the hash table). If the current index sum is less than the minimum index sum seen so far, it updates the minimum index sum and resets the result list to contain only the current string. If the current index sum is equal to the minimum index sum, it appends the current string to the result list. Finally, it returns the result list.

Step-by-Step Algorithm

Step 1: Create a hash table (dictionary or map) called `dict1` to store strings from `list1` as keys and their corresponding indices as values.
Step 2: Initialize `min_sum` to a large value (e.g., infinity in Python, Integer.MAX_VALUE in Java, INT_MAX in C++) to represent the minimum index sum found so far.
Step 3: Initialize an empty list called `res` to store the common strings with the minimum index sum.
Step 4: Iterate through `list2` with index `i` and string `item`.
Step 5: Check if `item` exists as a key in `dict1` (meaning it's a common string).
Step 6: If `item` is in `dict1`, calculate the current index sum `curr_sum = i + dict1[item]`.
Step 7: If `curr_sum` is less than `min_sum`, update `min_sum` to `curr_sum`, clear the `res` list, and add `item` to `res`.
Step 8: If `curr_sum` is equal to `min_sum`, append `item` to `res`.
Step 9: After iterating through `list2`, return the `res` list.

Key Insights

Insight 1: Using a hash table (dictionary or map) to store the strings and their indices from one list allows for efficient lookup in the second list.
Insight 2: Iterating through the second list and checking if each string exists in the hash table allows for finding common strings and calculating their index sum.
Insight 3: Keeping track of the minimum index sum encountered so far and storing the corresponding strings enables us to identify the common strings with the smallest index sum.

Complexity Analysis

Time Complexity: O(n+m)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, String.

Companies

Asked at: Yelp.

Minimum Index Sum of Two Lists | LeetCode 599

Easy

Array Hash Table String

Time: O(n+m)Space: O(n)

Solution Code

Available in:

class Solution {
    public String[] findRestaurant(String[] list1, String[] list2) {
        Map<String, Integer> map = new HashMap<>();
        for (int i = 0; i < list1.length; i++) {
            map.put(list1[i], i);
        }
        List<String> result = new ArrayList<>();
        int minSum = Integer.MAX_VALUE;
        for (int i = 0; i < list2.length; i++) {
            if (map.containsKey(list2[i])) {
                int sum = map.get(list2[i]) + i;
                if (sum < minSum) {
                    minSum = sum;
                    result.clear();
                    result.add(list2[i]);
                } else if (sum == minSum) {
                    result.add(list2[i]);
                }
            }
        }
        return result.toArray(new String[0]);
    }
}

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Problem Info

Description

Problem 599: Minimum Index Sum of Two Lists - Detailed explanation and solution approaches.

Topics

Array Hash Table String

Companies

Yelp

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:599

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Create a hash table (dictionary or map) called `dict1` to store strings from `list1` as keys and their corresponding indices as values.
2
Step 2: Initialize `min_sum` to a large value (e.g., infinity in Python, Integer.MAX_VALUE in Java, INT_MAX in C++) to represent the minimum index sum found so far.
3
Step 3: Initialize an empty list called `res` to store the common strings with the minimum index sum.
4
Step 4: Iterate through `list2` with index `i` and string `item`.
5
Step 5: Check if `item` exists as a key in `dict1` (meaning it's a common string).
6
Step 6: If `item` is in `dict1`, calculate the current index sum `curr_sum = i + dict1[item]`.
7
Step 7: If `curr_sum` is less than `min_sum`, update `min_sum` to `curr_sum`, clear the `res` list, and add `item` to `res`.
8
Step 8: If `curr_sum` is equal to `min_sum`, append `item` to `res`.
9
Step 9: After iterating through `list2`, return the `res` list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n+m)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the number of operations required to build the hash table and iterate through the lists. The space complexity is determined by the amount of memory used to store the hash table and the result list.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Iterate through all possible pairs of indices (one from each list) and check if the strings at those indices are equal. Keep track of the minimum index sum and the corresponding strings.

Time: O(n * m), where n is the length of `list1` and m is the length of `list2`.Space: O(1), excluding the space to store the result list, which in the worst-case is O(min(n, m)).

Trade-offs: The brute force approach is simpler to implement but has a higher time complexity than the hash table approach. It is less efficient and only feasible for small input lists. It uses constant extra space, compared to O(n) with the hash table.

✨

Key Insights

1Insight 1: Using a hash table (dictionary or map) to store the strings and their indices from one list allows for efficient lookup in the second list.
2Insight 2: Iterating through the second list and checking if each string exists in the hash table allows for finding common strings and calculating their index sum.
3Insight 3: Keeping track of the minimum index sum encountered so far and storing the corresponding strings enables us to identify the common strings with the smallest index sum.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty lists or when either list1 or list2 is empty. The constraints specify that the lists are not empty and contain at least one common element, so there's no need to explicitly handle these scenarios in most of the provided solutions (except for C which requires manual memory management).
Edge case 2: The common string appears at the very beginning of both lists. The algorithm correctly handles this case because the initial `min_sum` is a large value, so the first common string will always update `min_sum` and `res`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to initialize `min_sum` to a sufficiently large value. If `min_sum` is initialized to 0 or a small value, the algorithm may not correctly find the minimum index sum.

✗Common mistake 2: Not handling the case where multiple strings have the same minimum index sum. The algorithm needs to append strings to the result list when the current index sum is equal to the minimum index sum, not just when it is less than the minimum index sum.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Restaurant recommendation systems where the lists represent restaurant preferences of two people and the algorithm finds restaurants that both people like and minimize the 'effort' (represented by the index) of each person finding their preferred restaurant.

🏦Real-world scenario 2: Collaborative filtering, where lists could represent product preferences of users and the algorithm finds products that are commonly liked and appear early in both users' preference lists.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 30, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Index Sum of Two Lists"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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