#1626Medium~45 min

Best Team With No Conflicts

Time O(nloga) · Space O(n) · Official statement on LeetCode

dynamic programming sort dynamic programming segment tree

Solutions

// Time:  O(nloga)
// Space: O(n)

// optimized from Solution3
class Solution {
public:
    int bestTeamScore(vector<int>& scores, vector<int>& ages) {
        vector<pair<int, int>> players;
        for (int i = 0; i < size(scores); ++i) {
            players.emplace_back(scores[i], ages[i]);
        }
        sort(begin(players), end(players));
        set<int> sorted_ages(cbegin(ages), cend(ages));
        unordered_map<int, int> lookup;
        for (const auto& age : sorted_ages) {
            lookup[age] = size(lookup);
        }
        SegmentTree segment_tree(size(lookup));
        int result = 0;
        for (const auto& [score, age] : players) {
            segment_tree.update(lookup[age], lookup[age], segment_tree.query(0, lookup[age]) + score);
        }
        return segment_tree.query(0, size(lookup) - 1);
    }

private:
    class SegmentTree {
    public:
        SegmentTree(int N)
            : N_(N),
              tree_(2 * N),
              lazy_(N)
        {
            H_ = 1;
            while ((1 << H_) < N) {
                ++H_;
            }
        }

        void update(int L, int R, int h) {
            L += N_; R += N_;
            int L0 = L, R0 = R;
            while (L <= R) {
                if ((L & 1) == 1) {
                    apply(L++, h);
                }
                if ((R & 1) == 0) {
                    apply(R--, h);
                }
                L >>= 1; R >>= 1;
            }
            pull(L0); pull(R0);
        }

        int query(int L, int R) {
            L += N_; R += N_;
            auto result = 0;
            push(L); push(R);
            while (L <= R) {
                if ((L & 1) == 1) {
                    result = max(result, tree_[L++]);
                }
                if ((R & 1) == 0) {
                    result = max(result, tree_[R--]);
                }
                L >>= 1; R >>= 1;
            }
            return result;
        }

    private:
        int N_, H_;
        vector<int> tree_, lazy_;

        void apply(int x, int val) {
            tree_[x] = max(tree_[x], val);
            if (x < N_) {
                lazy_[x] = max(lazy_[x], val);
            }
        }

        void pull(int x) {
            while (x > 1) {
                x >>= 1;
                tree_[x] = max(tree_[x * 2], tree_[x * 2 + 1]);
                tree_[x] = max(tree_[x], lazy_[x]);
            }
        }

        void push(int x) {
            for (int h = H_; h > 0; --h) {
                int y = x >> h;
                if (lazy_[y] > 0) {
                    apply(y * 2, lazy_[y]);
                    apply(y * 2 + 1, lazy_[y]);
                    lazy_[y] = 0;
                }
            }
        }
    };
};

// Time:  O(nlogs)
// Space: O(n)
// optimized from Solution4
class Solution2 {
public:
    int bestTeamScore(vector<int>& scores, vector<int>& ages) {
        vector<pair<int, int>> players;
        for (int i = 0; i < size(scores); ++i) {
            players.emplace_back(ages[i], scores[i]);
        }
        sort(begin(players), end(players));
        set<int> sorted_scores(cbegin(scores), cend(scores));
        unordered_map<int, int> lookup;
        for (const auto& score : sorted_scores) {
            lookup[score] = size(lookup);
        }
        SegmentTree segment_tree(size(lookup));
        int result = 0;
        for (const auto& [age, score] : players) {
            segment_tree.update(lookup[score], lookup[score], segment_tree.query(0, lookup[score]) + score);
        }
        return segment_tree.query(0, size(lookup) - 1);
    }

private:
    class SegmentTree {
    public:
        SegmentTree(int N)
            : N_(N),
              tree_(2 * N),
              lazy_(N)
        {
            H_ = 1;
            while ((1 << H_) < N) {
                ++H_;
            }
        }

        void update(int L, int R, int h) {
            L += N_; R += N_;
            int L0 = L, R0 = R;
            while (L <= R) {
                if ((L & 1) == 1) {
                    apply(L++, h);
                }
                if ((R & 1) == 0) {
                    apply(R--, h);
                }
                L >>= 1; R >>= 1;
            }
            pull(L0); pull(R0);
        }

        int query(int L, int R) {
            L += N_; R += N_;
            auto result = 0;
            push(L); push(R);
            while (L <= R) {
                if ((L & 1) == 1) {
                    result = max(result, tree_[L++]);
                }
                if ((R & 1) == 0) {
                    result = max(result, tree_[R--]);
                }
                L >>= 1; R >>= 1;
            }
            return result;
        }

    private:
        int N_, H_;
        vector<int> tree_, lazy_;

        void apply(int x, int val) {
            tree_[x] = max(tree_[x], val);
            if (x < N_) {
                lazy_[x] = max(lazy_[x], val);
            }
        }

        void pull(int x) {
            while (x > 1) {
                x >>= 1;
                tree_[x] = max(tree_[x * 2], tree_[x * 2 + 1]);
                tree_[x] = max(tree_[x], lazy_[x]);
            }
        }

        void push(int x) {
            for (int h = H_; h > 0; --h) {
                int y = x >> h;
                if (lazy_[y] > 0) {
                    apply(y * 2, lazy_[y]);
                    apply(y * 2 + 1, lazy_[y]);
                    lazy_[y] = 0;
                }
            }
        }
    };
};

// Time:  O(n * a)
// Space: O(n)
// optimized from Solution5
class Solution3 {
public:
    int bestTeamScore(vector<int>& scores, vector<int>& ages) {
        vector<pair<int, int>> players;
        for (int i = 0; i < size(scores); ++i) {
            players.emplace_back(scores[i], ages[i]);
        }
        sort(begin(players), end(players));
        set<int> sorted_ages(cbegin(ages), cend(ages));
        unordered_map<int, int> dp;
        int result = 0;
        for (const auto& [score, age] : players) {
            int max_score = 0;
            for (const auto& a : sorted_ages) {
                if (a > age) {
                    break;
                }
                max_score = max(max_score, dp[a]);
            }
            dp[age] = max_score + score;
            result = max(result, dp[age]);
        }
        return result;
    }
};

// Time:  O(n * s)
// Space: O(n)
// optimized from Solution6
class Solution4 {
public:
    int bestTeamScore(vector<int>& scores, vector<int>& ages) {
        vector<pair<int, int>> players;
        for (int i = 0; i < size(scores); ++i) {
            players.emplace_back(ages[i], scores[i]);
        }
        sort(begin(players), end(players));
        set<int> sorted_scores(cbegin(scores), cend(scores));
        unordered_map<int, int> dp;
        int result = 0;
        for (const auto& [age, score] : players) {
            int max_score = 0;
            for (const auto& s : sorted_scores) {
                if (s > score) {
                    break;
                }
                max_score = max(max_score, dp[s]);
            }
            dp[score] = max_score + score;
            result = max(result, dp[score]);
        }
        return result;
    }
};

// Time:  O(n^2)
// Space: O(n)
// longest_increasing_subsequence like dp solution
class Solution5 {
public:
    int bestTeamScore(vector<int>& scores, vector<int>& ages) {
        vector<pair<int, int>> players;
        for (size_t i = 0; i < scores.size(); ++i) {
            players.emplace_back(scores[i], ages[i]);
        }
        sort(begin(players), end(players));
        vector<int> dp(size(scores));
        int result = 0;
        for (int i = 0; i < size(players); ++i) {
            dp[i] = players[i].first;
            for (int j = 0; j < i; ++j) {
                if (players[j].second <= players[i].second) {
                    dp[i] = max(dp[i], dp[j] + players[i].first);
                }
            }
            result = max(result, dp[i]);
        }
        return result;
    }
};

// Time:  O(n^2)
// Space: O(n)
// longest_increasing_subsequence like dp solution
class Solution6 {
public:
    int bestTeamScore(vector<int>& scores, vector<int>& ages) {
        vector<pair<int, int>> players;
        for (size_t i = 0; i < scores.size(); ++i) {
            players.emplace_back(ages[i], scores[i]);
        }
        sort(begin(players), end(players));
        vector<int> dp(size(scores));
        int result = 0;
        for (int i = 0; i < size(players); ++i) {
            dp[i] = players[i].second;
            for (int j = 0; j < i; ++j) {
                if (players[j].second <= players[i].second) {
                    dp[i] = max(dp[i], dp[j] + players[i].second);
                }
            }
            result = max(result, dp[i]);
        }
        return result;
    }
};

Beginner Explanation

What is Best Team With No Conflicts?

Best Team With No Conflicts (LeetCode #1626) is a Medium problem that primarily trains dynamic programming.

How to think about it

Restate the goal in your own words before coding.
Work a tiny example by hand so the invariant becomes obvious.
Identify the pattern — this problem aligns with sort, dynamic programming, and segment tree.
Only then translate the idea into code.

Why this problem matters

It sits in the sweet spot of interview difficulty: multiple valid approaches, clear trade-offs. Official solution notes mention: Sort, DP, Segment Tree.

AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.

Interview Walkthrough

Interview approach for Best Team With No Conflicts

Opening (30–60 seconds)

Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
State a brute force so the interviewer knows you can solve it naively.
Propose the optimal direction tied to sort, dynamic programming, and segment tree.

Core solution narrative

Define the state you track (pointers, DP cell, set membership, stack top, etc.).
Explain the transition when you process the next element.
Call out time (O(nloga)) and space (O(n)) before coding.
Code cleanly; narrate variable names.

What interviewers listen for

Correctness on edge cases
Complexity honesty
Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)

Follow-up questions they may ask

Can you solve it with less memory?
What if the input stream is infinite / doesn't fit in RAM?
How would tests look for adversarial inputs?

Optimized Approach

Optimized solution notes

The reference solutions on AlgoForge target O(nloga) time and O(n) space.

Pattern focus: sort, dynamic programming, and segment tree

Use the pattern as a checklist:

sort — confirm the invariant holds after each step
dynamic programming — confirm the invariant holds after each step
segment tree — confirm the invariant holds after each step

Multiple methods appear in the source solutions — compare them and explain when each is preferable.

Implementation tips

Prefer readable names over micro-optimizations in interviews.
Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
After AC-level logic, re-scan for off-by-one and null checks.

Complexity Analysis

Complexity

Measure	Bound
Time	O(nloga)
Space	O(n)

How to justify this in an interview

Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
Space: include hash maps, recursion stack, and output allocation when the problem asks for it.

If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.

Common Mistakes

Common mistakes on Best Team With No Conflicts

Skipping edge cases — empty collections, single-element inputs, max constraints.
Wrong invariant for sort, dynamic programming, and segment tree — updating state too early or too late.
Mutating input unexpectedly when the problem forbids it.
Off-by-one in windows, ranges, or binary search bounds.
Ignoring overflow / precision for integer arithmetic problems.
Overengineering — jumping to an advanced structure when a simpler approach works.

Alternative Approaches

Alternatives

The source file includes more than one method. Compare:

Primary optimized path — best complexity for typical interviews.
Secondary approach — often brute force, sorting-based, or space-optimized variant.

Practice articulating when you would pick each (constraints, readability, follow-ups).

Edge Cases

Edge cases checklist

Minimum input size
Maximum input size / time limits
Duplicates and already-sorted input
Negative numbers / zeros (if applicable)
Disconnected structures (graphs/trees)
Single path vs branching recursion depth

Pattern Recognition

Spotting this pattern

Signal phrases that point to sort, dynamic programming, and segment tree:

Sorted input or ability to sort without changing the answer class
Need for contiguous subarray / substring → consider sliding window
Need for O(1) membership → hash set/map
Optimal substructure + overlapping subproblems → DP
Connectivity / components → graph DFS/BFS or Union-Find

Primary topics: dynamic programming.

Follow-up Interview Questions

Follow-ups

How does the solution change if the input is a stream?
Can you solve it in-place?
What if duplicates must be handled differently?
How would you parallelize the approach?
Design tests that would break a buggy implementation.

Practice Recommendations

What to practice next

Re-solve Best Team With No Conflicts in a second language (cpp, python).
Drill 3–5 more problems tagged dynamic programming.
Teach the solution out loud in under 5 minutes.
Add this problem to your revision calendar in 3 days and 14 days.

Visualization

Conceptual diagram for Best Team With No Conflicts: show input structure (dynamic programming), highlight the moving parts of the sort, dynamic programming, and segment tree approach, and annotate each step with the maintained invariant and complexity.

Study checklist

Read the official problem statement on LeetCode
Solve on paper / whiteboard first
Implement the sort, dynamic programming, and segment tree approach
Verify edge cases from the checklist
State time and space complexity aloud
Compare with the AlgoForge reference solution
Schedule a revision session

Revision notes

Best Team With No Conflicts (#1626) — Medium. Pattern: sort, dynamic programming, and segment tree. Complexity: O(nloga) time / O(n) space. Re-derive the invariant before coding.

FAQs

What is the time complexity of Best Team With No Conflicts?+

The reference solutions aim for O(nloga) time and O(n) space. Always re-derive complexity from the code you write in the interview.

What pattern does Best Team With No Conflicts use?+

It primarily maps to sort, dynamic programming, and segment tree, within the broader topic of dynamic programming.

Is Best Team With No Conflicts good for interviews?+

Yes — as a Medium problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.

Where can I read the official statement?+

Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/best-team-with-no-conflicts/

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