LeetCode #1626 — MEDIUM

Best Team With No Conflicts

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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 i^th 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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[1,3,5,10,15]
[1,2,3,4,5]

Example 2

[4,5,6,5]
[2,1,2,1]

Example 3

[1,2,3,5]
[8,9,10,1]

Step 02

Core Insight

What unlocks the optimal approach

First, sort players by age and break ties by their score. You can now consider the players from left to right.
If you choose to include a player, you must only choose players with at least that score later on.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1626: Best Team With No Conflicts
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;
    }
}

// Accepted solution for LeetCode #1626: Best Team With No Conflicts
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)
}

# Accepted solution for LeetCode #1626: Best Team With No Conflicts
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)

// Accepted solution for LeetCode #1626: Best Team With No Conflicts
struct Solution;

impl Solution {
    fn best_team_score(scores: Vec<i32>, ages: Vec<i32>) -> i32 {
        let n = scores.len();
        let mut players = vec![];
        for i in 0..n {
            players.push((ages[i], scores[i]));
        }
        players.sort_unstable();
        let mut dp = vec![0; n];
        for i in 0..n {
            dp[i] = players[i].1;
            for j in 0..i {
                if players[j].1 <= players[i].1 {
                    dp[i] = dp[i].max(players[i].1 + dp[j]);
                }
            }
        }
        dp.into_iter().max().unwrap()
    }
}

#[test]
fn test() {
    let scores = vec![1, 3, 5, 10, 15];
    let ages = vec![1, 2, 3, 4, 5];
    let res = 34;
    assert_eq!(Solution::best_team_score(scores, ages), res);
    let scores = vec![4, 5, 6, 5];
    let ages = vec![2, 1, 2, 1];
    let res = 16;
    assert_eq!(Solution::best_team_score(scores, ages), res);
    let scores = vec![1, 2, 3, 5];
    let ages = vec![8, 9, 10, 1];
    let res = 6;
    assert_eq!(Solution::best_team_score(scores, ages), res);
}

// Accepted solution for LeetCode #1626: Best Team With No Conflicts
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);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.