What is the time complexity of Best Team With No Conflicts?

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

What data structures are used to solve Best Team With No Conflicts?

This problem uses Array, Dynamic Programming, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Best Team With No Conflicts?

Insight 1: Sorting the players by age allows us to process them in increasing age order, simplifying the conflict detection process. We can iterate through the players, ensuring that each selected player doesn't create a conflict with previously selected players. Insight 2: Dynamic programming (DP) can be used to store the maximum team score achievable up to a certain player, considering the conflict constraint. Each player can either be included or excluded from the team.

Which companies ask LeetCode 1626. Best Team With No Conflicts in interviews?

This problem is frequently asked at Morgan Stanley.

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Best Team With No Conflicts - LeetCode 1626 Solution

Best Team With No Conflicts - Complete Solution Guide

Best Team With No Conflicts is LeetCode problem 1626, 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

You are the manager of a basketball team. For the upcoming tournament, you want to choose the team with the highest overall score. The score of the team is the sum of scores of all the players in the team. However, the basketball team is not allowed to have conflicts . A conflict exists if a younger player has a strictly higher score than an older player. A conflict does not occur between players of the same age. Given two lists, scores and ages , where each scores[i] and ages[i] represents the

Detailed Explanation

The problem requires you to select a basketball team from a pool of players, maximizing the total score of the team while avoiding conflicts. A conflict occurs if a younger player has a strictly higher score than an older player. The input consists of two arrays: `scores` and `ages`, representing the score and age of each player, respectively. The output is the maximum possible team score.

Solution Approach

The solution uses a dynamic programming approach optimized with a Binary Indexed Tree (BIT). First, players are sorted by age. This ensures that when we consider a player, all players with lower or equal age have already been processed. For each player, we query the BIT to find the maximum team score achievable with players whose scores are less than or equal to the current player's score. We then add the current player's score to this maximum score and update the BIT at the current player's score with this new team score. Finally, we query the BIT for the maximum score among all scores to find the overall best team score.

Step-by-Step Algorithm

Step 1: Create an array of `Player` objects (or pairs) where each element contains the age and score of a player. Sort this array by age (ascending order). If ages are the same, it doesn't matter how you sort. For easier debugging sort by scores.
Step 2: Initialize a Binary Indexed Tree (BIT) with a size slightly larger than the maximum score to store maximum team scores.
Step 3: Iterate through the sorted players. For each player:
Step 4: Query the BIT to find the maximum team score among players with scores less than or equal to the current player's score. This ensures that adding the current player will not introduce any conflicts.
Step 5: Calculate the potential team score by adding the current player's score to the maximum score obtained from the BIT.
Step 6: Update the BIT at the index corresponding to the current player's score with the potential team score, making sure to store the maximum value at that index.
Step 7: After processing all players, query the BIT at the maximum possible score index. The value returned will be the highest overall score of all possible conflict-free teams.

Key Insights

Insight 1: Sorting the players by age allows us to process them in increasing age order, simplifying the conflict detection process. We can iterate through the players, ensuring that each selected player doesn't create a conflict with previously selected players.
Insight 2: Dynamic programming (DP) can be used to store the maximum team score achievable up to a certain player, considering the conflict constraint. Each player can either be included or excluded from the team.
Insight 3: A Binary Indexed Tree (BIT) or Fenwick Tree can be used to efficiently query the maximum team score among players with scores less than or equal to the current player's score. This optimizes the DP approach.

Complexity Analysis

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

Space Complexity: O(m)

Topics

This problem involves: Array, Dynamic Programming, Sorting.

Companies

Asked at: Morgan Stanley.

Best Team With No Conflicts | LeetCode 1626

Medium

Array Dynamic Programming Sorting

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

Solution Code

Available in:

import java.util.Arrays;
import java.util.Comparator;

class Solution {
    public int bestTeamScore(int[] scores, int[] ages) {
        int n = scores.length;
        int[][] players = new int[n][2];
        for (int i = 0; i < n; i++) {
            players[i][0] = ages[i];
            players[i][1] = scores[i];
        }

        Arrays.sort(players, Comparator.comparingInt(a -> a[0]));

        int max_score = 0;
        for (int score : scores) {
            if (score > max_score) {
                max_score = score;
            }
        }

        int[] bit = new int[max_score + 1];

        
        for (int i = 0; i < bit.length; i++) {
            bit[i] = 0;
        }
        

        for (int[] player : players) {
            int age = player[0];
            int score = player[1];
            int max_prev_team_score = query(bit, score);
            int current_team_score = score + max_prev_team_score;
            update(bit, score, current_team_score);
        }

        return query(bit, max_score);
    }

    private void update(int[] bit, int idx, int val) {
        while (idx < bit.length) {
            bit[idx] = Math.max(bit[idx], val);
            idx += idx & -idx;
        }
    }

    private int query(int[] bit, int idx) {
        int res = 0;
        while (idx > 0) {
            res = Math.max(res, bit[idx]);
            idx -= idx & -idx;
        }
        return res;
    }
}

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Problem Info

Description

Problem 1626: Best Team With No Conflicts - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Sorting

Companies

Morgan Stanley

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1626

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: Create an array of `Player` objects (or pairs) where each element contains the age and score of a player. Sort this array by age (ascending order). If ages are the same, it doesn't matter how you sort. For easier debugging sort by scores.
2
Step 2: Initialize a Binary Indexed Tree (BIT) with a size slightly larger than the maximum score to store maximum team scores.
3
Step 3: Iterate through the sorted players. For each player:
4
Step 4: Query the BIT to find the maximum team score among players with scores less than or equal to the current player's score. This ensures that adding the current player will not introduce any conflicts.
5
Step 5: Calculate the potential team score by adding the current player's score to the maximum score obtained from the BIT.
6
Step 6: Update the BIT at the index corresponding to the current player's score with the potential team score, making sure to store the maximum value at that index.
7
Step 7: After processing all players, query the BIT at the maximum possible score index. The value returned will be the highest overall score of all possible conflict-free teams.

💡

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*log(m))

💾

Space Complexity

O(m)

📝

Why?

The time complexity is determined by the sorting step and the BIT operations within the loop. The sorting contributes O(n log n) and the BIT operations (query and update) contribute O(n log m). The space complexity is dominated by the size of the BIT, which depends on the maximum score.

📖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

Dynamic Programming (without BIT)

Instead of using a BIT, we can use a DP array where `dp[i]` represents the maximum team score ending with the i-th player. We iterate through all possible previous players `j` and check if there is a conflict. If not, we update `dp[i]` with `max(dp[i], dp[j] + scores[i])`.

Time: O(n^2)Space: O(n)

Trade-offs: The DP approach is simpler to implement but has a higher time complexity of O(n^2) compared to the BIT-optimized solution's O(n log m). This makes it less suitable for large input sizes.

✨

Key Insights

1Insight 1: Sorting the players by age allows us to process them in increasing age order, simplifying the conflict detection process. We can iterate through the players, ensuring that each selected player doesn't create a conflict with previously selected players.
2Insight 2: Dynamic programming (DP) can be used to store the maximum team score achievable up to a certain player, considering the conflict constraint. Each player can either be included or excluded from the team.
3Insight 3: A Binary Indexed Tree (BIT) or Fenwick Tree can be used to efficiently query the maximum team score among players with scores less than or equal to the current player's score. This optimizes the DP approach.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Sorting

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

Edge Cases

Edge case 1: All players have the same age. In this case, there are no conflicts, so the solution should select all players, and the result should be the sum of all scores. The sorting and BIT approach correctly handle this since the BIT will store the cumulative scores as it iterates through each player.
Edge case 2: All players have the same score. In this case, the solution should choose the players of largest age. The sorting by age ensures that when scores are equivalent, the larger ages will override smaller ages with the same scores, naturally taking into consideration the largest ages.
Edge case 3: When scores array is empty the code does not work and throws an error.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the players by age first. This can lead to incorrect conflict detection and a wrong team score.

✗Common mistake 2: Incorrectly implementing the Binary Indexed Tree (BIT) query and update operations, leading to incorrect maximum score calculations.

✗Common mistake 3: Forgetting to account for players of the same age, where the scores should be allowed regardless. This is usually naturally handled by the sorting. However, specific cases must be tested for edge conditions with duplicates.

✅

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: Resource allocation with dependencies. Imagine allocating projects to developers where projects have priorities ('scores') and deadlines ('ages'). A younger deadline cannot have a higher priority. This algorithm can help find the best set of projects that can be completed without violating the constraints.

🏦Real-world scenario 2: Building a software team. Consider developers with skill levels ('scores') and years of experience ('ages'). You want to assemble a team such that a junior developer (lower experience) doesn't outrank a senior developer (higher experience) in terms of skill. The algorithm can then assist in selecting the most skilled group of developers under the experience constraint.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: June 7, 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.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

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 "Best Team With No Conflicts"?

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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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