Abstract / Summary

Choose a company name from a list that has the highest number of positive points and does not have any negative points.

Note

The full problem description, including helpful examples, can be found on LeetCode.

This problem is ranked as

and has an acceptance rate of about 47% (100k+ submissions).

Typical patterns to solve this problem:

• Array
• Bit Manipulation
• Enumeration
• Hash Table
• String

Every single line of code was written by me, 100% from scratch, between August 2023 and August 2024.
However, two years later I set up this website and used various AI tools to generate most of the walkthrough / explanations below.

See this dedicated page for more details.

Overview

The provided Solution class implements a method distinctNames that counts the number of valid name pairs that can be formed by swapping the first letters of two distinct ideas. A pair is considered valid when, after swapping the initial characters, both newly formed strings are not present in the original list of ideas. The algorithm relies on set operations to achieve fast membership checks and to compute the contribution of each pair of first-letter groups.

Data Preparation

The first step creates a set containing all input strings. This set, named all, enables O(1) containment tests for any generated candidate name.

all = set(ideas)  # faster lookups

By converting the list to a set once, the code avoids repeated linear scans when later checking whether a swapped name already exists.

Grouping by Prefix

The next phase groups the ideas by their initial character. For each name, the code extracts the first character as the key and stores the remainder of the string in a defaultdict of sets. This structure, called prefix, maps each possible first letter to the collection of suffixes that appear with that letter.

prefix = defaultdict(set)
for name in ideas:
    prefix[name[0]].add(name[1:])

Using a set for each group eliminates duplicate suffixes that could arise from repeated ideas and guarantees that subsequent set operations work on unique elements.

Pairwise Computation

The core counting logic iterates over every ordered pair of distinct first-letter groups. For each pair (p1, p2), the algorithm computes the number of suffixes that appear exclusively in p1 and not in p2, and vice versa. The product of these two exclusive counts represents the number of valid swaps between the two groups, because each exclusive suffix from p1 can be combined with each exclusive suffix from p2 to form a new pair that does not clash with any original idea.

result = 0
for p1 in sorted(prefix):
    for p2 in sorted(prefix):
        if p1 != p2:  # must be different first letter
            result += len(prefix[p1] - prefix[p2]) * len(prefix[p2] - prefix[p1])

The subtraction prefix[p1] - prefix[p2] yields the set of suffixes unique to p1. Multiplying the sizes of the two unique sets directly accumulates the total number of admissible name swaps for that pair of letters.

Edge Cases Handled

When the input list contains ideas that share the same suffix but differ in their first letter, the subtraction operation correctly isolates the suffixes that are exclusive to each group, ensuring that only truly distinct swaps are counted.

If a particular first letter appears only once in the entire input, its suffix set will have size one, and the subtraction will either produce an empty set or a singleton set depending on the presence of the same suffix in another group. The multiplication then yields zero for any pair where no exclusive suffix exists, naturally excluding invalid combinations.

The algorithm also gracefully handles the situation where all ideas start with the same letter. In that case, the outer loops still iterate, but the condition p1 != p2 prevents any pair from being processed, resulting in a final count of zero, which matches the definition of a valid swap requiring distinct initial letters.

When the input list is empty, the initial set conversion produces an empty set, the prefix dictionary remains empty, and the nested loops never execute, leading to a return value of zero without error.

Performance Optimizations

The conversion of ideas to a set (all) provides constant-time membership checks, which is essential for any later validation of swapped names (though the current implementation does not explicitly query all after the initial grouping).

Storing suffixes in sets within prefix eliminates duplicate entries early, reducing the amount of work performed during the pairwise subtraction.

The use of set subtraction (prefix[p1] - prefix[p2]) leverages Python’s highly optimized C-level implementation for set operations, making the calculation of exclusive suffixes fast even for large groups.

Sorting the keys of prefix before iteration (sorted(prefix)) does not affect the correctness of the count but ensures deterministic traversal order, which can be beneficial for reproducibility and debugging.

By computing the product of the sizes of the exclusive suffix sets rather than enumerating each possible combination, the algorithm avoids an explicit inner loop over individual suffix pairs, thereby reducing the computational workload from quadratic in the size of the groups to linear in the number of distinct suffixes per group.

Overall, the combination of set-based storage, set subtraction, and multiplicative aggregation yields an implementation that scales efficiently with the number of ideas and the diversity of their first letters.

Imagine having a big box of name tags for a party. Each name tag starts with a specific letter, like a name starting with A or B.

First, the code builds a little organizer to sort all the name tags into piles based on their first letter. One pile has all the names starting with A, another pile has all the names starting with B, and so on. Inside each pile, the organizer keeps the rest of the name that comes after the first letter.

Next, the code looks at two different piles at the same time to see which names they could swap. If a name in the first pile is not in the second pile, and a name in the second pile is not in the first pile, those two names are very special. They can be traded to make two brand new, unique names.

The code counts how many of these special tradeable pairs exist by comparing every single pile with every other pile. It does this by taking the names that are only in the first pile and multiplying that number by the names that are only in the second pile. By adding all those little counts together, the code finds the total number of ways to create new, unique names for the party.

Several LLMs ranked my code quality on a scale from 1 (very poor) to 10 (outstanding):

not analyzed yet

I wrote a basic crawler to find other solutions to that problem.
Obviously, I did that after I solved the problem myself.

In general, AlgoMonster and WalkCCC solutions are high-quality and always worth a look.
I haven't checked most of the other solutions, but they might be helpful as well.

And of course, the LeetCode forum is always a good place (or even the best) to look for alternative approaches.

(click on a column header to sort the table)

	author
author	solutions
www	AlgoMonster		Java	C++	Python		TS
	WalkCCC		Java	C++	Python
	leetcode.ca		Java	C++	Python	Go	TS
	Progiez		Java	C++	Python
github	charles-wangkai		Java
	everthis	JS
	fsq			C++
	JoshCrozier	JS
	lkwq007				Python
	varunu28		Java
	wisdompeak			C++
AlgoMonster	C++, Python, Java, TypeScript
WalkCCC	C++, Python, Java
leetcode.ca	C++, Go, Java, Python, TypeScript
Progiez	C++, Java, Python
charles-wangkai	Java
everthis	JavaScript
fsq	C++
JoshCrozier	JavaScript
lkwq007	Python
varunu28	Java
wisdompeak	C++

There are a few extra pages giving you some insights into my setup:

• Tech Stack - my development environment and tools
• AI - how AI helped me create this website (NOT solving the problems)
• Progress - progression over time
• Quality - code quality