How do you solve Minimum Index Sum of Two Lists on LeetCode?

Minimum Index Sum of Two Lists (LeetCode #599) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a HashMap significantly reduces the time complexity.

What is the brute force solution for Minimum Index Sum of Two Lists?

The brute force approach for Minimum Index Sum of Two Lists is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every string in list1 against every string in list2 to find common elements and their index sums. This is straightforward but inefficient for larger lists. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Index Sum of Two Lists?

Minimum Index Sum of Two Lists uses the following concepts: Array, Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Index Sum of Two Lists?

Always explain your thought process clearly. Consider edge cases, like empty lists or no common elements. Practice writing clean and efficient code.

What are common mistakes when solving Minimum Index Sum of Two Lists?

Not considering all common strings with the same minimum index sum. Overlooking the need to clear the list when a new minimum is found.

#599

Minimum Index Sum of Two Lists

Easy

ArrayHash TableStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using a HashMap allows us to store the indices of strings in list1, making it easy to find common strings and their index sums in a single pass through list2.

⚙️

Algorithm

4 steps

1Step 1: Create a HashMap to store the strings from list1 with their indices.
2Step 2: Initialize variables to track the minimum index sum and a list for common strings.
3Step 3: Iterate through list2, checking if each string exists in the HashMap. If it does, calculate the index sum.
4Step 4: Update the list of common strings based on the minimum index sum found.

solution.py13 lines

1def findRestaurant(list1, list2):
2    index_map = {s: i for i, s in enumerate(list1)}
3    common = []
4    min_sum = float('inf')
5    for j, s in enumerate(list2):
6        if s in index_map:
7            index_sum = index_map[s] + j
8            if index_sum < min_sum:
9                min_sum = index_sum
10                common = [s]
11            elif index_sum == min_sum:
12                common.append(s)
13    return common

ℹ

Complexity note: The time complexity is O(n) because we make a single pass through list1 to build the HashMap and another pass through list2 to find common strings. The space complexity is O(n) due to the HashMap storing the strings from list1.

1Using a HashMap significantly reduces the time complexity.
2Finding common elements can be optimized by storing indices.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.