What is the time complexity of Rank Teams by Votes?

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

What is the space complexity of Rank Teams by Votes?

The space complexity is O(m*m).

What data structures are used to solve Rank Teams by Votes?

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

What are the key insights for solving Rank Teams by Votes?

Insight 1: We need to count the number of times each team appears in each ranking position. Insight 2: We need a way to sort the teams based on their rank counts, giving higher positions more weight.

Which companies ask LeetCode 1366. Rank Teams by Votes in interviews?

This problem is frequently asked at Atlassian, Coursera, Flipkart, eBay.

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Rank Teams by Votes - LeetCode 1366 Solution

Rank Teams by Votes - Complete Solution Guide

Rank Teams by Votes is LeetCode problem 1366, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

In a special ranking system, each voter gives a rank from highest to lowest to all teams participating in the competition. The ordering of teams is decided by who received the most position-one votes. If two or more teams tie in the first position, we consider the second position to resolve the conflict, if they tie again, we continue this process until the ties are resolved. If two or more teams are still tied after considering all positions, we rank them alphabetically based on their team lett

Detailed Explanation

The problem asks us to rank teams based on a specific voting system. Each voter provides a ranked list of all teams. The team with the most first-place votes ranks highest. If there's a tie, we compare the number of second-place votes, then third-place, and so on. If teams are still tied after considering all positions, we break the tie alphabetically. The input is an array of strings, where each string represents a voter's ranking. The output should be a string representing the teams sorted according to the described ranking system.

Solution Approach

The solution involves counting the rank occurrences for each team at each position and then sorting the teams based on these counts. A dictionary (or hash map) is used to store the rank counts. The teams are then sorted using a custom comparison function that prioritizes higher ranking positions and uses alphabetical order for ties.

Step-by-Step Algorithm

Step 1: Initialize a data structure (e.g., a dictionary or hash map) to store the counts of each team at each rank. The keys will be the teams (characters), and the values will be arrays representing the rank counts.
Step 2: Iterate through each vote in the `votes` array.
Step 3: For each vote, iterate through the teams in the vote string.
Step 4: Update the rank counts in the data structure. For each team at a specific rank in a vote, increment the corresponding count.
Step 5: Create a list of teams to be sorted. This can be obtained from the first vote, as it's guaranteed to contain all teams.
Step 6: Sort the list of teams using a custom comparison function. The comparison function compares the rank counts of two teams, position by position. If the counts at a position differ, the team with the higher count ranks higher. If the counts are equal for all positions, compare the teams alphabetically.
Step 7: Build a string by concatenating the sorted teams and return the resulting string.

Key Insights

Insight 1: We need to count the number of times each team appears in each ranking position.
Insight 2: We need a way to sort the teams based on their rank counts, giving higher positions more weight.
Insight 3: Alphabetical tie-breaking is crucial when teams have identical ranking profiles.

Complexity Analysis

Time Complexity: O(n*m*log(m))

Space Complexity: O(m*m)

Topics

This problem involves: Array, Hash Table, String, Sorting, Counting.

Companies

Asked at: Atlassian, Coursera, Flipkart, eBay.

Rank Teams by Votes | LeetCode 1366

Medium

Array Hash Table String Sorting Counting

Time: O(n*m*log(m))Space: O(m*m)

Solution Code

Available in:

import java.util.*;

class Solution {
    public String rankTeams(String[] votes) {
        if (votes == null || votes.length == 0) {
            return "";
        }

        int numPositions = votes[0].length();
        Map<Character, int[]> rankCounts = new HashMap<>();

        for (String vote : votes) {
            for (int i = 0; i < vote.length(); i++) {
                char team = vote.charAt(i);
                rankCounts.putIfAbsent(team, new int[numPositions]);
                rankCounts.get(team)[i]++;
            }
        }

        List<Character> teamsList = new ArrayList<>();
        for (char team : votes[0].toCharArray()) {
            teamsList.add(team);
        }

        Collections.sort(teamsList, (team1, team2) -> {
            for (int i = 0; i < numPositions; i++) {
                int rank1 = rankCounts.getOrDefault(team1, new int[numPositions])[i];
                int rank2 = rankCounts.getOrDefault(team2, new int[numPositions])[i];
                if (rank1 != rank2) {
                    return rank2 - rank1; // Sort in descending order of rank count
                }
            }
            return Character.compare(team1, team2); // If ranks are equal, sort alphabetically
        });

        StringBuilder result = new StringBuilder();
        for (char team : teamsList) {
            result.append(team);
        }

        return result.toString();
    }
}

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Problem Info

Description

Problem 1366: Rank Teams by Votes - Detailed explanation and solution approaches.

Topics

Array Hash Table String Sorting Counting

Companies

AtlassianCourseraFlipkarteBay

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:1366

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

7 steps

1
Step 1: Initialize a data structure (e.g., a dictionary or hash map) to store the counts of each team at each rank. The keys will be the teams (characters), and the values will be arrays representing the rank counts.
2
Step 2: Iterate through each vote in the `votes` array.
3
Step 3: For each vote, iterate through the teams in the vote string.
4
Step 4: Update the rank counts in the data structure. For each team at a specific rank in a vote, increment the corresponding count.
5
Step 5: Create a list of teams to be sorted. This can be obtained from the first vote, as it's guaranteed to contain all teams.
6
Step 6: Sort the list of teams using a custom comparison function. The comparison function compares the rank counts of two teams, position by position. If the counts at a position differ, the team with the higher count ranks higher. If the counts are equal for all positions, compare the teams alphabetically.
7
Step 7: Build a string by concatenating the sorted teams and return the resulting string.

💡

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*log(m))

💾

Space Complexity

O(m*m)

📝

Why?

The time complexity is dominated by the sorting step. Counting votes is O(n*m) where n is number of votes and m is number of teams. Sorting m teams with a custom comparator that takes O(m) results in O(m*log(m) * m) or O(m^2*log(m)). The space complexity comes primarily from the `rank_counts` map which stores ranking information for each team at each position i.e. m teams and m positions which leads to O(m^2) space.

📖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

Using a Priority Queue

Instead of sorting the teams list, we could use a priority queue. The priority queue would be initialized with all the teams, and the custom comparison function would be used to determine the priority of each team. This could potentially improve performance in some cases, but the overall time complexity would likely remain the same.

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

Trade-offs: The priority queue approach may have slightly different constant factors, but the overall time complexity is similar. The choice between sorting and using a priority queue depends on the specific characteristics of the input data.

✨

Key Insights

1Insight 1: We need to count the number of times each team appears in each ranking position.
2Insight 2: We need a way to sort the teams based on their rank counts, giving higher positions more weight.
3Insight 3: Alphabetical tie-breaking is crucial when teams have identical ranking profiles.

📚

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!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty `votes` array. The code correctly handles this case by returning an empty string.
Edge case 2: Only one vote. The code correctly returns the vote string as the sorted order.
Edge case 3: Tie between all teams across all positions. The code handles this by using alphabetical order as a tie-breaker.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle ties by using alphabetical order. This will result in incorrect rankings when teams have identical ranking profiles.

✗Common mistake 2: Incorrectly calculating the rank counts. Ensure that the counts are updated correctly for each team at each position.

✗Common mistake 3: Not handling the case where the `votes` array is empty.

✅

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: Ranking candidates in an election based on ranked-choice voting.

🏦Real-world scenario 2: Ranking items in a recommendation system based on user preferences (e.g., movies, products). Each user's preference list can be treated as a vote.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: September 15, 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

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

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Rank Teams by Votes"?

For this medium-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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