Leetcode 1626: Best Team With No Conflicts
You are the manager of a basketball team, and you are tasked with selecting a team that maximizes the overall score. The score of the team is calculated by summing the individual scores of all the selected players. However, there is a rule: a conflict arises if a younger player has a strictly higher score than an older player. There is no conflict if players have the same age.
Problem
Approach
Steps
Complexity
Input: Two lists are given, `scores` and `ages`, where each element in `scores` corresponds to the score of a player and each element in `ages` corresponds to the age of that player.
Example: scores = [2, 4, 8, 10, 12], ages = [1, 2, 3, 4, 5]
Constraints:
• 1 <= scores.length, ages.length <= 1000
• scores.length == ages.length
• 1 <= scores[i] <= 106
• 1 <= ages[i] <= 1000
Output: Return the maximum possible score of a valid team, where the team has no conflicts.
Example: Output: 36
Constraints:
• The result is an integer representing the maximum score.
Goal: Maximize the total score of a valid team.
Steps:
• 1. Pair each player's age with their score.
• 2. Sort the list of players based on their age in descending order.
• 3. Use dynamic programming to find the best possible score by avoiding conflicts.
Goal: The number of players (scores and ages) is up to 1000, and each player's score and age are bounded within certain limits.
Steps:
• 1 <= scores.length, ages.length <= 1000
• scores.length == ages.length
• 1 <= scores[i] <= 106
• 1 <= ages[i] <= 1000
Assumptions:
• All players have a unique age or the same age, but the score is distinct for each player.
• Input: Input: scores = [2, 4, 8, 10, 12], ages = [1, 2, 3, 4, 5]
• Explanation: You can select all the players without any conflicts, resulting in a total score of 36.
Approach: The approach is to use dynamic programming to maximize the score while avoiding conflicts between players.
Observations:
• Sorting players by age ensures that the condition of having no conflicts between players can be easily handled.
• The idea is to use a dynamic programming approach to compute the maximum score for each player while ensuring no conflicts.
Steps:
• 1. Pair each player’s age with their score.
• 2. Sort the players based on age in descending order.
• 3. Use dynamic programming to maximize the team’s total score while avoiding conflicts.
Empty Inputs:
• If either the scores or ages list is empty, the output will be 0.
Large Inputs:
• For large inputs with the maximum number of players (1000), the algorithm should still perform efficiently.
Special Values:
• All players having the same score or age might result in different team configurations.
Constraints:
• Ensure the solution handles the upper bounds of input size and constraints.
int bestTeamScore(vector<int>& scores, vector<int>& ages) {
int n = ages.size();
vector<pair<int, int>> palyer;
for(int i = 0; i < n; i++)
palyer.push_back({ages[i], scores[i]});
sort(palyer.begin(), palyer.end(), greater<>());
int ans = 0;
vector<int> dp(n);
for(int i = 0; i < n; i++) {
pair<int, int> young = palyer[i];
dp[i] = young.second;
for(int j = 0; j < i; j++) {
pair<int, int> elder = palyer[j];
if(young.second <= elder.second)
dp[i] = max(dp[i], dp[j] + young.second);
}
ans = max(ans, dp[i]);
}
return ans;
}
1 : Method Definition
int bestTeamScore(vector<int>& scores, vector<int>& ages) {
Define the main method 'bestTeamScore' which takes in two vectors: scores and ages.
2 : Variable Initialization
int n = ages.size();
Initialize 'n' to be the size of the 'ages' vector, which is the total number of players.
3 : Variable Initialization
vector<pair<int, int>> palyer;
Initialize a vector of pairs 'palyer' where each pair stores the age and score of a player.
4 : Loop Constructs
for(int i = 0; i < n; i++)
Iterate through each player to add their age and score to the 'palyer' vector.
5 : Vector Manipulations
palyer.push_back({ages[i], scores[i]});
Push a new pair containing the current player's age and score into the 'palyer' vector.
6 : Sorting
sort(palyer.begin(), palyer.end(), greater<>());
Sort the 'palyer' vector in descending order based on age first, and score second (if ages are equal).
7 : Variable Initialization
int ans = 0;
Initialize 'ans' to store the maximum score found during the process.
8 : Vector Initialization
vector<int> dp(n);
Initialize a dynamic programming vector 'dp' to store the best scores at each player position.
9 : Loop Constructs
for(int i = 0; i < n; i++) {
Start an outer loop to go through each player in the sorted 'palyer' vector.
10 : Variable Initialization
pair<int, int> young = palyer[i];
Assign the current player (sorted by age and score) to the variable 'young'.
11 : Vector Operations
dp[i] = young.second;
Initialize 'dp[i]' with the score of the current player ('young').
12 : Loop Constructs
for(int j = 0; j < i; j++) {
Start an inner loop to check previous players for a valid team formation.
13 : Variable Initialization
pair<int, int> elder = palyer[j];
Assign the previous player to 'elder' to compare their score with 'young'.
14 : Conditional Statements
if(young.second <= elder.second)
Check if the score of the current player ('young') is less than or equal to the score of the previous player ('elder').
15 : Mathematical Operations
dp[i] = max(dp[i], dp[j] + young.second);
If valid, update 'dp[i]' by considering the best score combination of the current and previous player.
16 : Mathematical Operations
ans = max(ans, dp[i]);
Update the maximum score 'ans' by considering the current best score at 'dp[i]'.
17 : Return Statement
return ans;
Return the maximum score 'ans' which is the best team score.
Best Case: O(n log n) due to sorting.
Average Case: O(n^2) due to dynamic programming.
Worst Case: O(n^2) due to dynamic programming and sorting.
Description: The time complexity is dominated by the sorting step followed by the dynamic programming evaluation.
Best Case: O(n) for storing the dynamic programming table.
Worst Case: O(n) for storing the dynamic programming table.
Description: The space complexity is linear as we store the dynamic programming table.
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