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1626 - Best Team With No Conflicts (Medium)

https://leetcode.com/problems/best-team-with-no-conflicts/

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 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

Approach 1: Dynamic Programming​

For this kind of DP questions, remember the following

  • calculate the result if you skip this candidate
  • calculate the result if you pick this candidate
  • take the max result of them then memorise it
Written by @wingkwong
class Solution {
public:
    int dp[1005][1005];
    // i is the index of g
    // age is the max age so far
    int dfs(vector<pair<int, int>>& g, int i, int age) {
        // if i reaches the end, then return 0
        if (i == g.size()) return 0;
        // if dp[i][age] is calculated before, return it directly
        if (dp[i][age] != -1) return dp[i][age];
        // we have two choices - either take this player or skip it
        // we first calculate the score if we skip this player
        int res = dfs(g, i + 1, age);
        // if we want to take this player, we need to check if there is a conflict
        // since the score is sorted, we just need to check the age
        // if we take this player, the score will be increased by g[i].first
        // then we add the result from `dfs(g, i + 1, g[i].second)`
        // since we take this player, we need ti mark the age as g[i].second
        if (g[i].second >= age) res = max(res, g[i].first + dfs(g, i + 1, g[i].second));
        // memo it
        return dp[i][age] = res;
    }
    int bestTeamScore(vector<int>& scores, vector<int>& ages) {
        // init dp with initial value -1
        memset(dp, -1, sizeof(dp));
        // we sort by score, then age
        vector<pair<int, int>> g;
        for (int i = 0; i < scores.size(); i++) {
            g.push_back({scores[i], ages[i]});
        }
        sort(g.begin(), g.end());
        return dfs(g, 0, 0);
    }
};