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1626. Best Team With No Conflicts
Description
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 <= 1000scores.length == ages.length1 <= scores[i] <= 1061 <= ages[i] <= 1000
Solutions
Solution 1
LIS.
Solution 2
Similar to method 1, we can sort the players according to their scores from small to large. If the scores are the same, they can be sorted by their age from small to large.
Next, we use a Fenwick tree to maintain the maximum score for players not older than the current player’s age. Each time, we just need to find the maximum score for a player not older than the current player’s age within $O(\log m)$ time, and then add the current player’s score to that score to update the maximum score for the current player’s age.
Finally, we return the maximum score.
Time complexity $O(n \times (\log n + \log m))$, space complexity $O(n + m)$. Among them, $n$ and $m$ are the maximum number of players and the age of the players respectively.
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class Solution { public int bestTeamScore(int[] scores, int[] ages) { int n = ages.length; int[][] arr = new int[n][2]; for (int i = 0; i < n; ++i) { arr[i] = new int[] {scores[i], ages[i]}; } Arrays.sort(arr, (a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]); int[] f = new int[n]; int ans = 0; for (int i = 0; i < n; ++i) { for (int j = 0; j < i; ++j) { if (arr[i][1] >= arr[j][1]) { f[i] = Math.max(f[i], f[j]); } } f[i] += arr[i][0]; ans = Math.max(ans, f[i]); } return ans; } } // Solution 2 class BinaryIndexedTree { private int n; private int[] c; public BinaryIndexedTree(int n) { this.n = n; c = new int[n + 1]; } public void update(int x, int val) { while (x <= n) { c[x] = Math.max(c[x], val); x += x & -x; } } public int query(int x) { int s = 0; while (x > 0) { s = Math.max(s, c[x]); x -= x & -x; } return s; } } class Solution { public int bestTeamScore(int[] scores, int[] ages) { int n = ages.length; int[][] arr = new int[n][2]; for (int i = 0; i < n; ++i) { arr[i] = new int[] {scores[i], ages[i]}; } Arrays.sort(arr, (a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]); int m = 0; for (int age : ages) { m = Math.max(m, age); } BinaryIndexedTree tree = new BinaryIndexedTree(m); for (int[] x : arr) { tree.update(x[1], x[0] + tree.query(x[1])); } return tree.query(m); } } -
class Solution { public: int bestTeamScore(vector<int>& scores, vector<int>& ages) { int n = ages.size(); vector<pair<int, int>> arr(n); for (int i = 0; i < n; ++i) { arr[i] = {scores[i], ages[i]}; } sort(arr.begin(), arr.end()); vector<int> f(n); for (int i = 0; i < n; ++i) { for (int j = 0; j < i; ++j) { if (arr[i].second >= arr[j].second) { f[i] = max(f[i], f[j]); } } f[i] += arr[i].first; } return *max_element(f.begin(), f.end()); } }; // Solution 2 class BinaryIndexedTree { public: BinaryIndexedTree(int _n) : n(_n) , c(_n + 1) {} void update(int x, int val) { while (x <= n) { c[x] = max(c[x], val); x += x & -x; } } int query(int x) { int s = 0; while (x) { s = max(s, c[x]); x -= x & -x; } return s; } private: int n; vector<int> c; }; class Solution { public: int bestTeamScore(vector<int>& scores, vector<int>& ages) { int n = ages.size(); vector<pair<int, int>> arr(n); for (int i = 0; i < n; ++i) { arr[i] = {scores[i], ages[i]}; } sort(arr.begin(), arr.end()); int m = *max_element(ages.begin(), ages.end()); BinaryIndexedTree tree(m); for (auto& [score, age] : arr) { tree.update(age, score + tree.query(age)); } return tree.query(m); } }; -
class Solution: def bestTeamScore(self, scores: List[int], ages: List[int]) -> int: arr = sorted(zip(scores, ages)) n = len(arr) f = [0] * n for i, (score, age) in enumerate(arr): for j in range(i): if age >= arr[j][1]: f[i] = max(f[i], f[j]) f[i] += score return max(f) # Solution 2 class BinaryIndexedTree: def __init__(self, n): self.n = n self.c = [0] * (n + 1) def update(self, x, val): while x <= self.n: self.c[x] = max(self.c[x], val) x += x & -x def query(self, x): s = 0 while x: s = max(s, self.c[x]) x -= x & -x return s class Solution: def bestTeamScore(self, scores: List[int], ages: List[int]) -> int: m = max(ages) tree = BinaryIndexedTree(m) for score, age in sorted(zip(scores, ages)): tree.update(age, score + tree.query(age)) return tree.query(m) -
func bestTeamScore(scores []int, ages []int) int { n := len(ages) arr := make([][2]int, n) for i := range ages { arr[i] = [2]int{scores[i], ages[i]} } sort.Slice(arr, func(i, j int) bool { a, b := arr[i], arr[j] return a[0] < b[0] || a[0] == b[0] && a[1] < b[1] }) f := make([]int, n) for i := range arr { for j := 0; j < i; j++ { if arr[i][1] >= arr[j][1] { f[i] = max(f[i], f[j]) } } f[i] += arr[i][0] } return slices.Max(f) } // Solution 2 type BinaryIndexedTree struct { n int c []int } func newBinaryIndexedTree(n int) *BinaryIndexedTree { c := make([]int, n+1) return &BinaryIndexedTree{n, c} } func (this *BinaryIndexedTree) update(x, val int) { for x <= this.n { this.c[x] = max(this.c[x], val) x += x & -x } } func (this *BinaryIndexedTree) query(x int) int { s := 0 for x > 0 { s = max(s, this.c[x]) x -= x & -x } return s } func bestTeamScore(scores []int, ages []int) int { n := len(ages) arr := make([][2]int, n) m := 0 for i, age := range ages { m = max(m, age) arr[i] = [2]int{scores[i], age} } sort.Slice(arr, func(i, j int) bool { a, b := arr[i], arr[j] return a[0] < b[0] || a[0] == b[0] && a[1] < b[1] }) tree := newBinaryIndexedTree(m) for _, x := range arr { tree.update(x[1], x[0]+tree.query(x[1])) } return tree.query(m) } -
function bestTeamScore(scores: number[], ages: number[]): number { const arr = ages.map((age, i) => [age, scores[i]]); arr.sort((a, b) => (a[0] == b[0] ? a[1] - b[1] : a[0] - b[0])); const n = arr.length; const f = new Array(n).fill(0); for (let i = 0; i < n; ++i) { for (let j = 0; j < i; ++j) { if (arr[i][1] >= arr[j][1]) { f[i] = Math.max(f[i], f[j]); } } f[i] += arr[i][1]; } return Math.max(...f); } -
/** * @param {number[]} scores * @param {number[]} ages * @return {number} */ var bestTeamScore = function (scores, ages) { const arr = ages.map((age, i) => [age, scores[i]]); arr.sort((a, b) => (a[0] == b[0] ? a[1] - b[1] : a[0] - b[0])); const n = arr.length; const f = new Array(n).fill(0); for (let i = 0; i < n; ++i) { for (let j = 0; j < i; ++j) { if (arr[i][1] >= arr[j][1]) { f[i] = Math.max(f[i], f[j]); } } f[i] += arr[i][1]; } return Math.max(...f); };