How do you solve Best Team With No Conflicts on LeetCode?

Best Team With No Conflicts (LeetCode #1626) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Sorting players by age and score helps in efficiently checking for conflicts.

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

The optimal solution for Best Team With No Conflicts runs in O(n²) time and uses O(n) space. The sorting step takes O(n log n), and the nested loop for the DP solution takes O(n²), making the overall complexity O(n²).

What is the brute force solution for Best Team With No Conflicts?

The brute force approach for Best Team With No Conflicts is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible combinations of players to find the maximum score without conflicts. This method is straightforward but inefficient as it considers every subset of players. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Best Team With No Conflicts?

Best Team With No Conflicts uses the following concepts: Array, Dynamic Programming, Sorting. The recommended patterns to study are: Dynamic Programming, Sorting.

What are the interview tips for Best Team With No Conflicts?

Always explain your thought process clearly; it helps interviewers understand your approach. Consider edge cases, such as all players having the same age or score. Practice writing clean and efficient code, as readability is important in interviews.

What are common mistakes when solving Best Team With No Conflicts?

Failing to sort players correctly before applying dynamic programming. Not considering the case where players of the same age can be included.

#1626

Best Team With No Conflicts

Medium

ArrayDynamic ProgrammingSortingDynamic ProgrammingSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(n)

The optimal approach involves sorting the players by age and then using dynamic programming to build the maximum score while ensuring no conflicts. This method is efficient and leverages the sorted order to simplify conflict checks.

⚙️

Algorithm

4 steps

1Step 1: Pair each player's score with their age and sort the list by age (and score to break ties).
2Step 2: Initialize a DP array where dp[i] represents the maximum score achievable including the i-th player.
3Step 3: For each player, check all previous players to see if they can be included without conflict, updating the dp array accordingly.
4Step 4: The result will be the maximum value in the dp array.

solution.py13 lines

1# Full working Python code
2
3def best_team(scores, ages):
4    players = sorted(zip(ages, scores))
5    n = len(players)
6    dp = [0] * n
7    for i in range(n):
8        dp[i] = players[i][1]  # Start with the player's own score
9        for j in range(i):
10            if players[j][1] <= players[i][1]:  # No conflict
11                dp[i] = max(dp[i], dp[j] + players[i][1])
12    return max(dp)
13

ℹ

Complexity note: The sorting step takes O(n log n), and the nested loop for the DP solution takes O(n²), making the overall complexity O(n²).

1Sorting players by age and score helps in efficiently checking for conflicts.
2Dynamic programming allows us to build solutions incrementally, maximizing scores while adhering to constraints.

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