#2921HardPremium on LC~50 min

Maximum Profitable Triplets With Increasing Prices II

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

binary search tree prefix sum sorted list binary search stack fenwick tree segment tree

Solutions

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

// prefix sum, bst, binary search, mono stack
class Solution {
public:
    int maxProfit(vector<int>& prices, vector<int>& profits) {
        static const int NEG_INF = numeric_limits<int>::min();

        const auto& query = [](const auto& bst, const auto& k) {
            const auto it = bst.lower_bound(pair(k, 0));
            return it != begin(bst) ? prev(it)->second : NEG_INF;
        };

        const auto& update = [](auto& bst, const auto& k, const auto& v) {
            const auto it = bst.lower_bound(pair(k, 0));
            if (it != end(bst) && it->first == k) {
                if (!(it->second < v)) {
                    return;
                }
                bst.erase(it);
            } else if (!(it == begin(bst) || prev(it)->second < v)) {
                return;
            }
            const auto [jt, _] = bst.emplace(k, v);
            while (next(jt) != end(bst) && next(jt)->second <= jt->second) {
                bst.erase(next(jt));
            }
        };

        int result = NEG_INF;
        set<pair<int, int>> bst1, bst2;
        for (int i = 0; i < size(prices); ++i) {
            const int mx2 = query(bst2, prices[i]);
            if (mx2 != NEG_INF) {
                result = max(result, mx2 + profits[i]);
            }
            update(bst1, prices[i], profits[i]);
            const int mx1 = query(bst1, prices[i]);
            if (mx1 != NEG_INF) {
                update(bst2, prices[i], mx1 + profits[i]);
            }
        }
        return result != NEG_INF ? result : -1;
    }
};

// Time:  O(nlogn)
// Space: O(n)
// prefix sum, bst, binary search, mono stack
class Solution2 {
public:
    int maxProfit(vector<int>& prices, vector<int>& profits) {
        static const int NEG_INF = numeric_limits<int>::min();

        vector<int> right(size(prices), NEG_INF);
        set<pair<int, int>> bst;
        for (int i = size(prices) - 1; i >= 0; --i) {
            const auto it = bst.lower_bound(pair(-prices[i], 0));
            if (it != begin(bst)) {
                right[i] = prev(it)->second;
            }
            if (!(it == begin(bst) || prev(it)->second < profits[i])) {
                continue;
            }
            const auto [jt, _] = bst.emplace(-prices[i], profits[i]);
            while (next(jt) != end(bst) && next(jt)->second <= jt->second) {
                bst.erase(next(jt));
            }
        }
        int result = NEG_INF;
        bst.clear();
        for (int i = 0; i < size(prices); ++i) {
            const auto it = bst.lower_bound(pair(prices[i], 0));
            if (it != begin(bst)) {
                if (prev(it)->second != NEG_INF && right[i] != NEG_INF) {
                    result = max(result, prev(it)->second + profits[i] + right[i]);
                }
            }
            if (!(it == begin(bst) || prev(it)->second < profits[i])) {
                continue;
            }
            const auto [jt, _] = bst.emplace(prices[i], profits[i]);
            while (next(jt) != end(bst) && next(jt)->second <= jt->second) {
                bst.erase(next(jt));
            }
        }
        return result != NEG_INF ? result : -1;
    }
};

// Time:  O(nlogn)
// Space: O(n)
// prefix sum, bit, fenwick tree
class Solution3 {
public:
    int maxProfit(vector<int>& prices, vector<int>& profits) {
        static const int NEG_INF = numeric_limits<int>::min();

        unordered_set<int> prices_set(cbegin(prices), cend(prices));
        vector<int> sorted_prices(cbegin(prices_set), cend(prices_set));
        sort(begin(sorted_prices), end(sorted_prices));
        unordered_map<int, int> price_to_idx;
        for (int i = 0; i < size(sorted_prices); ++i) {
            price_to_idx[sorted_prices[i]] = i;
        }
        int result = NEG_INF;
        const auto& fn = [](int a, int b) {
            return max(a, b);
        };
        BIT bit1(size(price_to_idx), NEG_INF, fn), bit2(size(price_to_idx), NEG_INF, fn);
        for (int i = 0; i < size(prices); ++i) {
            const int mx2 = bit2.query(price_to_idx[prices[i]] - 1);
            if (mx2 != NEG_INF) {
                result = max(result, mx2 + profits[i]);
            }
            bit1.update(price_to_idx[prices[i]], profits[i]);
            const int mx1 = bit1.query(price_to_idx[prices[i]] - 1);
            if (mx1 != NEG_INF) {
                bit2.update(price_to_idx[prices[i]], mx1 + profits[i]);
            }
        }
        return result != NEG_INF ? result : -1;
    }

private:
    class BIT {
    public:
        BIT(int n, int val, const function<int (int, int)> fn)
          : bit_(n + 1, val),
            fn_(fn) {  // 0-indexed
        }
        
        void update(int i, int val) {
            ++i;
            for (; i < size(bit_); i += lower_bit(i)) {
                bit_[i] = fn_(bit_[i], val);
            }
        }

        int query(int i) const {
            ++i;
            int total = bit_[0];
            for (; i > 0; i -= lower_bit(i)) {
                total = fn_(total, bit_[i]);
            }
            return total;
        }
    
    private:
        int lower_bit(int i) const {
            return i & -i;
        }
        
        vector<int> bit_;
        const function<int (int, int)> fn_;
    };
};

// Time:  O(nlogn)
// Space: O(n)
// prefix sum, segment tree
class Solution4 {
public:
    int maxProfit(vector<int>& prices, vector<int>& profits) {
        static const int NEG_INF = numeric_limits<int>::min();

        unordered_set<int> prices_set(cbegin(prices), cend(prices));
        vector<int> sorted_prices(cbegin(prices_set), cend(prices_set));
        sort(begin(sorted_prices), end(sorted_prices));
        unordered_map<int, int> price_to_idx;
        for (int i = 0; i < size(sorted_prices); ++i) {
            price_to_idx[sorted_prices[i]] = i;
        }
        int result = NEG_INF;
        SegmentTree st1(size(price_to_idx)), st2(size(price_to_idx));
        for (int i = 0; i < size(prices); ++i) {
            const int mx2 = st2.query(0, price_to_idx[prices[i]] - 1);
            if (mx2 != NEG_INF) {
                result = max(result, mx2 + profits[i]);
            }
            st1.update(price_to_idx[prices[i]], profits[i]);
            const int mx1 = st1.query(0, price_to_idx[prices[i]] - 1);
            if (mx1 != NEG_INF) {
                st2.update(price_to_idx[prices[i]], mx1 + profits[i]);
            }
        }
        return result != NEG_INF ? result : -1;
    }

private:
    class SegmentTree {
    private:
        const int NEG_INF = numeric_limits<int>::min();
      
    public:
        explicit SegmentTree(int N)
          : tree(N > 1 ? 1 << (__lg(N - 1) + 2) : 2, NEG_INF),
            base(N > 1 ? 1 << (__lg(N - 1) + 1) : 1) {

        }

        void update(int i, int h) {
            int x = base + i;
            tree[x] = max(tree[x], h);
            while (x > 1) {
                x /= 2;
                tree[x] = max(tree[x * 2], tree[x * 2 + 1]);
            }
        }

        int query(int L, int R) {
            int result = NEG_INF;
            if (L > R) {
                return result;
            }
            L += base;
            R += base;
            for (; L <= R; L /= 2, R /= 2) {
                if (L & 1) {
                    result = max(result, tree[L]);
                    ++L;
                }
                if ((R & 1) == 0) {
                    result = max(tree[R], result);
                    --R;
                }
            }
            return result;
        }

        vector<int> tree;
        int base;
    };
};

// Time:  O(nlogn)
// Space: O(n)
// prefix sum, segment tree
class Solution5 {
public:
    int maxProfit(vector<int>& prices, vector<int>& profits) {
        static const int NEG_INF = numeric_limits<int>::min();

        unordered_set<int> prices_set(cbegin(prices), cend(prices));
        vector<int> sorted_prices(cbegin(prices_set), cend(prices_set));
        sort(begin(sorted_prices), end(sorted_prices));
        unordered_map<int, int> price_to_idx;
        for (int i = 0; i < size(sorted_prices); ++i) {
            price_to_idx[sorted_prices[i]] = i;
        }
        vector<int> right(size(prices), NEG_INF);
        SegmentTree st2(size(price_to_idx));
        for (int i = size(prices) - 1; i >= 0; --i) {
            right[i] = st2.query(price_to_idx[prices[i]] + 1, size(price_to_idx) - 1);
            st2.update(price_to_idx[prices[i]], profits[i]);
        }
        int result = NEG_INF;
        vector<int> left(size(prices), NEG_INF);
        SegmentTree st1(size(price_to_idx));
        for (int i = 0; i < size(prices); ++i) {
            const int left = st1.query(0, price_to_idx[prices[i]] - 1);
            if (left != NEG_INF && right[i] != NEG_INF) {
                result = max(result, left + profits[i] + right[i]);
            }
            st1.update(price_to_idx[prices[i]], profits[i]);
        }
        return result != NEG_INF ? result : -1;
    }

private:
    class SegmentTree {
    private:
        const int NEG_INF = numeric_limits<int>::min();
      
    public:
        explicit SegmentTree(int N)
          : tree(N > 1 ? 1 << (__lg(N - 1) + 2) : 2, NEG_INF),
            base(N > 1 ? 1 << (__lg(N - 1) + 1) : 1) {

        }

        void update(int i, int h) {
            int x = base + i;
            tree[x] = max(tree[x], h);
            while (x > 1) {
                x /= 2;
                tree[x] = max(tree[x * 2], tree[x * 2 + 1]);
            }
        }

        int query(int L, int R) {
            int result = NEG_INF;
            if (L > R) {
                return result;
            }
            L += base;
            R += base;
            for (; L <= R; L /= 2, R /= 2) {
                if (L & 1) {
                    result = max(result, tree[L]);
                    ++L;
                }
                if ((R & 1) == 0) {
                    result = max(tree[R], result);
                    --R;
                }
            }
            return result;
        }

        vector<int> tree;
        int base;
    };
};

Beginner Explanation

What is Maximum Profitable Triplets With Increasing Prices II?

Maximum Profitable Triplets With Increasing Prices II (LeetCode #2921) is a Hard problem that primarily trains binary search tree.

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 prefix sum, sorted list, binary search, stack, fenwick tree, and segment tree.
Only then translate the idea into code.

Why this problem matters

Hard problems force you to combine patterns and prove complexity carefully — interview gold. Official solution notes mention: Prefix Sum, Sorted List, BST, Binary Search.

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

Interview Walkthrough

Interview approach for Maximum Profitable Triplets With Increasing Prices II

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 prefix sum, sorted list, binary search, stack, fenwick tree, 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(nlogn)) 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(nlogn) time and O(n) space.

Pattern focus: prefix sum, sorted list, binary search, stack, fenwick tree, and segment tree

Use the pattern as a checklist:

prefix sum — confirm the invariant holds after each step
sorted list — confirm the invariant holds after each step
binary search — confirm the invariant holds after each step
stack — confirm the invariant holds after each step
fenwick tree — 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(nlogn)
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 Maximum Profitable Triplets With Increasing Prices II

Skipping edge cases — empty collections, single-element inputs, max constraints.
Wrong invariant for prefix sum, sorted list, binary search, stack, fenwick tree, 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 prefix sum, sorted list, binary search, stack, fenwick tree, 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: binary search tree.

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 Maximum Profitable Triplets With Increasing Prices II in a second language (cpp, python).
Drill 3–5 more problems tagged binary search tree.
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 Maximum Profitable Triplets With Increasing Prices II: show input structure (binary search tree), highlight the moving parts of the prefix sum, sorted list, binary search, stack, fenwick tree, 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 prefix sum, sorted list, binary search, stack, fenwick tree, 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

Maximum Profitable Triplets With Increasing Prices II (#2921) — Hard. Pattern: prefix sum, sorted list, binary search, stack, fenwick tree, and segment tree. Complexity: O(nlogn) time / O(n) space. Re-derive the invariant before coding.

FAQs

What is the time complexity of Maximum Profitable Triplets With Increasing Prices II?+

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

What pattern does Maximum Profitable Triplets With Increasing Prices II use?+

It primarily maps to prefix sum, sorted list, binary search, stack, fenwick tree, and segment tree, within the broader topic of binary search tree.

Is Maximum Profitable Triplets With Increasing Prices II good for interviews?+

Yes — as a Hard 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/maximum-profitable-triplets-with-increasing-prices-ii/

#	Title	Difficulty	Topics
2907	Maximum Profitable Triplets With Increasing Prices I	Medium	binary search tree
2926	Maximum Balanced Subsequence Sum	Hard	binary search tree
3161	Block Placement Queries	Hard	binary search tree
2426	Number of Pairs Satisfying Inequality	Hard	binary search tree
3930	Power Update After K-th Largest Insertion II	Hard	binary search tree
2817	Minimum Absolute Difference Between Elements With Constraint	Medium	binary search tree
3073	Maximum Increasing Triplet Value	Medium	binary search tree
3072	Distribute Elements Into Two Arrays II	Hard	binary search tree