1366. Rank Teams by Votes - Detailed Explanation

Free Coding Questions Catalog

Boost your coding skills with our essential coding questions catalog. Take a step towards a better tech career now!

Problem Statement

You are given an array of strings, votes, where each string represents a ballot ranking for a set of teams. Every string is a permutation of the same set of teams (represented by uppercase letters). The goal is to determine the overall ranking of the teams according to the following rules:

Higher Vote Counts in Higher Positions: A team is ranked higher if it has more first-place votes. If teams tie on first-place votes, then compare second-place votes, then third-place, and so on.
Lexicographical Order: If two teams have exactly the same vote counts for all positions, the team with the lexicographically smaller name (i.e., earlier in the alphabet) comes first.

Example Inputs & Outputs

Example 1:
- Input: ["ABC", "ACB", "ABC", "ACB", "ACB"]
- Output: "ACB"
- Explanation:
  - Team A receives 5 first-place votes.
  - For teams B and C, since neither gets any first-place votes, we look at the second-place votes.
  - Team C has more second-place votes than team B, so the final ranking is A, then C, then B.
Example 2:
- Input: ["WXYZ", "XYZW"]
- Output: "XWYZ"
- Explanation:
  - The vote counts position-by-position lead to team X having the best overall standing.
  - When comparing the remaining teams, the order is determined by the number of votes in each rank and then by alphabetical order if needed.
Example 3:
- Input: ["ABC"]
- Output: "ABC"
- Explanation:
  - With only one vote, the ranking is exactly the same as the vote order.

Constraints

$(1 \leq \text{votes.length} \leq 1000)$
$(1 \leq \text{votes}[i].\text{length} \leq 26)$
Each votes[i] is a permutation of uppercase English letters.
All votes[i] contain the same set of characters.

Hints

Counting Votes:
Think about how you can count the number of votes each team receives at every ranking position. Consider using a dictionary (or array) where each team maps to an array of counts corresponding to its votes in each position.
Custom Sorting:
Once you have the counts, you need to sort the teams. The sorting should compare the counts for each rank from first to last. If the counts are the same for all positions, then sort the teams alphabetically (lexicographically).

Approach 1: Counting and Sorting (Optimal Approach)

Step-by-Step Walkthrough

Initialize a Count Structure:
Create a dictionary (in Python) or a HashMap (in Java) to store each team and an array of integers representing the count of votes for each position.
Count the Votes:
For each ballot in votes, iterate over the string:
- Increment the count for the team at the corresponding ranking position.
Sorting the Teams:
- Convert the teams (found in the first vote) into a list.
- Sort the teams using a custom comparator:
  - First, compare the vote counts for each position (starting from the first position).
  - If two teams have the same count at every position, compare them lexicographically.
Construct the Final Ranking:
Concatenate the sorted teams to form the final ranking string.

Visual Example

Let’s use Example 1: ["ABC", "ACB", "ABC", "ACB", "ACB"]

Step 1: Initialization
Teams: A, B, C
Initialize counts as:
- A: [0, 0, 0]
- B: [0, 0, 0]
- C: [0, 0, 0]
Step 2: Counting Votes
Process each vote:
- For "ABC":
  - A gets a vote at position 0 → A: [1, 0, 0]
  - B gets a vote at position 1 → B: [0, 1, 0]
  - C gets a vote at position 2 → C: [0, 0, 1]
- For "ACB":
  - A: [2, 0, 0]
  - C: [0, 1, 1]
  - B: [0, 1, 1]
- Continue processing all votes. Final counts:
  - A: [5, 0, 0]
  - B: [0, 2, 3]
  - C: [0, 3, 2]
Step 3: Sorting
- A clearly ranks first because it has 5 first-place votes.
- Compare B and C:
  - For the second position: B has 2 votes and C has 3 votes.
  - Therefore, C ranks above B.
Step 4: Final Ranking
The sorted order is "A", "C", "B", so the final ranking is "ACB".

Python Code

Python3

. . . .

Java Code

Java

. . . .

Complexity Analysis

Time Complexity:
- Counting votes takes $(O(V \times T))$ where (V) is the number of votes and (T) is the number of teams.
- Sorting the teams takes $(O(T \log T \times T))$ in the worst-case due to comparing arrays of length (T). However, since $(T \leq 26)$ , this part is effectively constant.
Space Complexity:
- We use extra space for the vote counts, which is $(O(T \times T))$ (each team has an array of size (T)). Again, since (T) is at most 26, this is very efficient.

Common Mistakes & Edge Cases

Common Mistakes

Ignoring Tie-Breakers:
Failing to compare all ranking positions before falling back to lexicographical order can lead to incorrect results.
Improper Count Initialization:
Not correctly initializing the count for each team or assuming teams might not appear in every vote.
Incorrect Sorting Order:
Mixing up the order of sorting (for example, sorting lexicographically first) rather than comparing vote counts properly.

Edge Cases

Single Vote:
When there is only one vote, the output should exactly match the order of that vote.
All Teams Tied:
If every team gets the same number of votes for all positions, the teams should be sorted in alphabetical order.
Empty Votes List:
Although constrained by the problem, always consider handling cases where the input might be empty.

Alternative Approaches & Variations

Alternative Approaches

Using Bucket Sort:
Since the number of teams is very small (at most 26) and the vote counts for each position are bounded, bucket sort can be used for each ranking position. However, the custom sort approach described above is simpler and highly efficient.
Custom Comparator with Multiple Criteria:
Some languages allow more concise custom comparators that handle multiple levels of comparison in a single sort call.

Variations of the Problem

Weighted Votes:
Instead of equal weight for every position, votes might have different weights (e.g., first-place votes count more than second-place votes). Adjusting the count structure to account for weights would be required.
Partial Rankings:
If not all voters rank every team, the algorithm must handle missing positions carefully.

Top K Frequent Elements:
Practice counting and sorting elements by frequency.
Group Anagrams:
Use hashing and sorting to group similar strings.
Longest Substring Without Repeating Characters:
Focus on window sliding techniques and frequency counting.
Merge Intervals:
Learn to sort and merge overlapping intervals.
Valid Anagram:
An exercise in comparing character counts.