LeetCode #2830 — MEDIUM

Maximize the Profit as the Salesman

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

Solve on LeetCode

The Problem

Problem Statement

You are given an integer n representing the number of houses on a number line, numbered from 0 to n - 1.

Additionally, you are given a 2D integer array offers where offers[i] = [start_i, end_i, gold_i], indicating that i^th buyer wants to buy all the houses from start_i to end_i for gold_i amount of gold.

As a salesman, your goal is to maximize your earnings by strategically selecting and selling houses to buyers.

Return the maximum amount of gold you can earn.

Note that different buyers can't buy the same house, and some houses may remain unsold.

Example 1:

Input: n = 5, offers = [[0,0,1],[0,2,2],[1,3,2]]
Output: 3
Explanation: There are 5 houses numbered from 0 to 4 and there are 3 purchase offers.
We sell houses in the range [0,0] to 1^st buyer for 1 gold and houses in the range [1,3] to 3^rd buyer for 2 golds.
It can be proven that 3 is the maximum amount of gold we can achieve.

Example 2:

Input: n = 5, offers = [[0,0,1],[0,2,10],[1,3,2]]
Output: 10
Explanation: There are 5 houses numbered from 0 to 4 and there are 3 purchase offers.
We sell houses in the range [0,2] to 2^nd buyer for 10 golds.
It can be proven that 10 is the maximum amount of gold we can achieve.

Constraints:

1 <= n <= 10⁵
1 <= offers.length <= 10⁵
offers[i].length == 3
0 <= start_i <= end_i <= n - 1
1 <= gold_i <= 10³

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: You are given an integer n representing the number of houses on a number line, numbered from 0 to n - 1. Additionally, you are given a 2D integer array offers where offers[i] = [starti, endi, goldi], indicating that ith buyer wants to buy all the houses from starti to endi for goldi amount of gold. As a salesman, your goal is to maximize your earnings by strategically selecting and selling houses to buyers. Return the maximum amount of gold you can earn. Note that different buyers can't buy the same house, and some houses may remain unsold.

Baseline thinking

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

Pattern signal: Array · Hash Map · Binary Search · Dynamic Programming

Example 1

5
[[0,0,1],[0,2,2],[1,3,2]]

Example 2

5
[[0,0,1],[0,2,10],[1,3,2]]

Step 02

Core Insight

What unlocks the optimal approach

<div class="_1l1MA">The intended solution uses a dynamic programming approach to solve the problem.</div>
<div class="_1l1MA">Sort the array offers by <code>start<sub>i</sub></code>.</div>
<div class="_1l1MA">Let <code>dp[i]</code> = { the maximum amount of gold if the sold houses are in the range <code>[0 … i]</code> }.</div>

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 #2830: Maximize the Profit as the Salesman
class Solution {
    public int maximizeTheProfit(int n, List<List<Integer>> offers) {
        offers.sort((a, b) -> a.get(1) - b.get(1));
        n = offers.size();
        int[] f = new int[n + 1];
        int[] g = new int[n];
        for (int i = 0; i < n; ++i) {
            g[i] = offers.get(i).get(1);
        }
        for (int i = 1; i <= n; ++i) {
            var o = offers.get(i - 1);
            int j = search(g, o.get(0));
            f[i] = Math.max(f[i - 1], f[j] + o.get(2));
        }
        return f[n];
    }

    private int search(int[] nums, int x) {
        int l = 0, r = nums.length;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

// Accepted solution for LeetCode #2830: Maximize the Profit as the Salesman
func maximizeTheProfit(n int, offers [][]int) int {
	sort.Slice(offers, func(i, j int) bool { return offers[i][1] < offers[j][1] })
	n = len(offers)
	f := make([]int, n+1)
	g := []int{}
	for _, o := range offers {
		g = append(g, o[1])
	}
	for i := 1; i <= n; i++ {
		j := sort.SearchInts(g, offers[i-1][0])
		f[i] = max(f[i-1], f[j]+offers[i-1][2])
	}
	return f[n]
}

# Accepted solution for LeetCode #2830: Maximize the Profit as the Salesman
class Solution:
    def maximizeTheProfit(self, n: int, offers: List[List[int]]) -> int:
        offers.sort(key=lambda x: x[1])
        f = [0] * (len(offers) + 1)
        g = [x[1] for x in offers]
        for i, (s, _, v) in enumerate(offers, 1):
            j = bisect_left(g, s)
            f[i] = max(f[i - 1], f[j] + v)
        return f[-1]

// Accepted solution for LeetCode #2830: Maximize the Profit as the Salesman
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2830: Maximize the Profit as the Salesman
// class Solution {
//     public int maximizeTheProfit(int n, List<List<Integer>> offers) {
//         offers.sort((a, b) -> a.get(1) - b.get(1));
//         n = offers.size();
//         int[] f = new int[n + 1];
//         int[] g = new int[n];
//         for (int i = 0; i < n; ++i) {
//             g[i] = offers.get(i).get(1);
//         }
//         for (int i = 1; i <= n; ++i) {
//             var o = offers.get(i - 1);
//             int j = search(g, o.get(0));
//             f[i] = Math.max(f[i - 1], f[j] + o.get(2));
//         }
//         return f[n];
//     }
// 
//     private int search(int[] nums, int x) {
//         int l = 0, r = nums.length;
//         while (l < r) {
//             int mid = (l + r) >> 1;
//             if (nums[mid] >= x) {
//                 r = mid;
//             } else {
//                 l = mid + 1;
//             }
//         }
//         return l;
//     }
// }

// Accepted solution for LeetCode #2830: Maximize the Profit as the Salesman
function maximizeTheProfit(n: number, offers: number[][]): number {
    offers.sort((a, b) => a[1] - b[1]);
    n = offers.length;
    const f: number[] = Array(n + 1).fill(0);
    const g = offers.map(x => x[1]);
    const search = (x: number) => {
        let l = 0;
        let r = n;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (g[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    for (let i = 1; i <= n; ++i) {
        const j = search(offers[i - 1][0]);
        f[i] = Math.max(f[i - 1], f[j] + offers[i - 1][2]);
    }
    return f[n];
}

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 × log n)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

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.