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 score and age of the ith
player, respectively, return the highest overall score of all possible basketball teams.
Example 1:
Input: scores = [1,3,5,10,15], ages = [1,2,3,4,5] Output: 34 Explanation: You can choose all the players.
Example 2:
Input: scores = [4,5,6,5], ages = [2,1,2,1] Output: 16 Explanation: It is best to choose the last 3 players. Notice that you are allowed to choose multiple people of the same age.
Example 3:
Input: scores = [1,2,3,5], ages = [8,9,10,1] Output: 6 Explanation: It is best to choose the first 3 players.
Constraints:
1 <= scores.length, ages.length <= 1000
scores.length == ages.length
1 <= scores[i] <= 106
1 <= ages[i] <= 1000
Related Topics:
Dynamic Programming
First, sort the players such that they are sorted in ascending order of age, then in ascending order of score.
Let dp[i]
be the maximum score we can get if we choose from 0
th to i
th player and must pick i
th player.
For dp[i]
, we can check each player j
(0 <= j < i
) whose age must be the same or less than player i
, and if their ages are the same, player j
's score must be smaller.
So as long as player j
's score is smaller or equal to player i
's score, we can extend the team with player i
based on the optimal solution of dp[j]
.
dp[i] = max( dp[j] | 0 <= j < i && scores[j] <= scores[i] ) + scores[i]
// OJ: https://leetcode.com/problems/best-team-with-no-conflicts/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N)
// Ref: https://leetcode.com/problems/best-team-with-no-conflicts/discuss/899475/Fairly-easy-DP
class Solution {
public:
int bestTeamScore(vector<int>& scores, vector<int>& ages) {
vector<pair<int, int>> A;
int N = scores.size(), ans = 0;
vector<int> dp(N);
for (int i = 0; i < N; ++i) A.emplace_back(ages[i], scores[i]);
sort(begin(A), end(A));
for (int i = 0; i < N; ++i) {
auto [age, score] = A[i];
for (int j = 0; j < i; ++j) {
if (A[j].second <= score) dp[i] = max(dp[i], dp[j]);
}
ans = max(ans, dp[i] += score);
}
return ans;
}
};