LeetCode #1235 — HARD

Maximum Profit in Job Scheduling

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

We have n jobs, where every job is scheduled to be done from startTime[i] to endTime[i], obtaining a profit of profit[i].

You're given the startTime, endTime and profit arrays, return the maximum profit you can take such that there are no two jobs in the subset with overlapping time range.

If you choose a job that ends at time X you will be able to start another job that starts at time X.

Example 1:

Input: startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]
Output: 120
Explanation: The subset chosen is the first and fourth job. 
Time range [1-3]+[3-6] , we get profit of 120 = 50 + 70.

Example 2:

Input: startTime = [1,2,3,4,6], endTime = [3,5,10,6,9], profit = [20,20,100,70,60]
Output: 150
Explanation: The subset chosen is the first, fourth and fifth job. 
Profit obtained 150 = 20 + 70 + 60.

Example 3:

Input: startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4]
Output: 6

Constraints:

1 <= startTime.length == endTime.length == profit.length <= 5 * 10⁴
1 <= startTime[i] < endTime[i] <= 10⁹
1 <= profit[i] <= 10⁴

Step 01

Brute Force Baseline

Problem summary: We have n jobs, where every job is scheduled to be done from startTime[i] to endTime[i], obtaining a profit of profit[i]. You're given the startTime, endTime and profit arrays, return the maximum profit you can take such that there are no two jobs in the subset with overlapping time range. If you choose a job that ends at time X you will be able to start another job that starts at time X.

Baseline thinking

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

Pattern signal: Array · Binary Search · Dynamic Programming

Example 1

[1,2,3,3]
[3,4,5,6]
[50,10,40,70]

Example 2

[1,2,3,4,6]
[3,5,10,6,9]
[20,20,100,70,60]

Example 3

[1,1,1]
[2,3,4]
[5,6,4]

Core Insight

What unlocks the optimal approach

Think on DP.
Sort the elements by starting time, then define the dp[i] as the maximum profit taking elements from the suffix starting at i.
Use binarySearch (lower_bound/upper_bound on C++) to get the next index for the DP transition.

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

Largest constraint values

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 #1235: Maximum Profit in Job Scheduling
class Solution {
    private int[][] jobs;
    private int[] f;
    private int n;

    public int jobScheduling(int[] startTime, int[] endTime, int[] profit) {
        n = profit.length;
        jobs = new int[n][3];
        for (int i = 0; i < n; ++i) {
            jobs[i] = new int[] {startTime[i], endTime[i], profit[i]};
        }
        Arrays.sort(jobs, (a, b) -> a[0] - b[0]);
        f = new int[n];
        return dfs(0);
    }

    private int dfs(int i) {
        if (i >= n) {
            return 0;
        }
        if (f[i] != 0) {
            return f[i];
        }
        int e = jobs[i][1], p = jobs[i][2];
        int j = search(jobs, e, i + 1);
        int ans = Math.max(dfs(i + 1), p + dfs(j));
        f[i] = ans;
        return ans;
    }

    private int search(int[][] jobs, int x, int i) {
        int left = i, right = n;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (jobs[mid][0] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

// Accepted solution for LeetCode #1235: Maximum Profit in Job Scheduling
func jobScheduling(startTime []int, endTime []int, profit []int) int {
	n := len(profit)
	type tuple struct{ s, e, p int }
	jobs := make([]tuple, n)
	for i, p := range profit {
		jobs[i] = tuple{startTime[i], endTime[i], p}
	}
	sort.Slice(jobs, func(i, j int) bool { return jobs[i].s < jobs[j].s })
	f := make([]int, n)
	var dfs func(int) int
	dfs = func(i int) int {
		if i >= n {
			return 0
		}
		if f[i] != 0 {
			return f[i]
		}
		j := sort.Search(n, func(j int) bool { return jobs[j].s >= jobs[i].e })
		ans := max(dfs(i+1), jobs[i].p+dfs(j))
		f[i] = ans
		return ans
	}
	return dfs(0)
}

# Accepted solution for LeetCode #1235: Maximum Profit in Job Scheduling
class Solution:
    def jobScheduling(
        self, startTime: List[int], endTime: List[int], profit: List[int]
    ) -> int:
        @cache
        def dfs(i):
            if i >= n:
                return 0
            _, e, p = jobs[i]
            j = bisect_left(jobs, e, lo=i + 1, key=lambda x: x[0])
            return max(dfs(i + 1), p + dfs(j))

        jobs = sorted(zip(startTime, endTime, profit))
        n = len(profit)
        return dfs(0)

// Accepted solution for LeetCode #1235: Maximum Profit in Job Scheduling
struct Solution;

use std::collections::BTreeMap;

impl Solution {
    fn job_scheduling(start_time: Vec<i32>, end_time: Vec<i32>, profit: Vec<i32>) -> i32 {
        let mut jobs = vec![];
        let n = start_time.len();
        for i in 0..n {
            jobs.push((start_time[i], end_time[i], profit[i]));
        }
        jobs.sort_unstable();
        let mut memo: BTreeMap<i32, i32> = BTreeMap::new();
        let mut res = 0;
        for i in (0..n).rev() {
            let mut cur = jobs[i].2;
            if let Some((_, val)) = memo.range(jobs[i].1..).next() {
                cur += val;
            }
            res = res.max(cur);
            memo.insert(jobs[i].0, res);
        }
        res
    }
}

#[test]
fn test() {
    let start_time = vec![1, 2, 3, 3];
    let end_time = vec![3, 4, 5, 6];
    let profit = vec![50, 10, 40, 70];
    let res = 120;
    assert_eq!(Solution::job_scheduling(start_time, end_time, profit), res);
    let start_time = vec![1, 2, 3, 4, 6];
    let end_time = vec![3, 5, 10, 6, 9];
    let profit = vec![20, 20, 100, 70, 60];
    let res = 150;
    assert_eq!(Solution::job_scheduling(start_time, end_time, profit), res);
    let start_time = vec![1, 1, 1];
    let end_time = vec![2, 3, 4];
    let profit = vec![5, 6, 4];
    let res = 6;
    assert_eq!(Solution::job_scheduling(start_time, end_time, profit), res);
}

// Accepted solution for LeetCode #1235: Maximum Profit in Job Scheduling
function jobScheduling(startTime: number[], endTime: number[], profit: number[]): number {
    const n = startTime.length;
    const f = new Array(n).fill(0);
    const idx = new Array(n).fill(0).map((_, i) => i);
    idx.sort((i, j) => startTime[i] - startTime[j]);
    const search = (x: number) => {
        let l = 0;
        let r = n;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (startTime[idx[mid]] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    const dfs = (i: number): number => {
        if (i >= n) {
            return 0;
        }
        if (f[i] !== 0) {
            return f[i];
        }
        const j = search(endTime[idx[i]]);
        return (f[i] = Math.max(dfs(i + 1), dfs(j) + profit[idx[i]]));
    };
    return dfs(0);
}

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(log n)

Space

O(1)

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.

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.