LeetCode #1124 — MEDIUM

Longest Well-Performing Interval

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

The Problem

Problem Statement

We are given hours, a list of the number of hours worked per day for a given employee.

A day is considered to be a tiring day if and only if the number of hours worked is (strictly) greater than 8.

A well-performing interval is an interval of days for which the number of tiring days is strictly larger than the number of non-tiring days.

Return the length of the longest well-performing interval.

Example 1:

Input: hours = [9,9,6,0,6,6,9]
Output: 3
Explanation: The longest well-performing interval is [9,9,6].

Example 2:

Input: hours = [6,6,6]
Output: 0

Constraints:

1 <= hours.length <= 10⁴
0 <= hours[i] <= 16

Step 01

Brute Force Baseline

Problem summary: We are given hours, a list of the number of hours worked per day for a given employee. A day is considered to be a tiring day if and only if the number of hours worked is (strictly) greater than 8. A well-performing interval is an interval of days for which the number of tiring days is strictly larger than the number of non-tiring days. Return the length of the longest well-performing interval.

Baseline thinking

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

Pattern signal: Array · Hash Map · Stack

Example 1

[9,9,6,0,6,6,9]

Example 2

[6,6,6]

Step 02

Core Insight

What unlocks the optimal approach

Make a new array A of +1/-1s corresponding to if hours[i] is > 8 or not. The goal is to find the longest subarray with positive sum.
Using prefix sums (PrefixSum[i+1] = A[0] + A[1] + ... + A[i]), you need to find for each j, the smallest i < j with PrefixSum[i] + 1 == PrefixSum[j].

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 #1124: Longest Well-Performing Interval
class Solution {
    public int longestWPI(int[] hours) {
        int ans = 0, s = 0;
        Map<Integer, Integer> pos = new HashMap<>();
        for (int i = 0; i < hours.length; ++i) {
            s += hours[i] > 8 ? 1 : -1;
            if (s > 0) {
                ans = i + 1;
            } else if (pos.containsKey(s - 1)) {
                ans = Math.max(ans, i - pos.get(s - 1));
            }
            pos.putIfAbsent(s, i);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1124: Longest Well-Performing Interval
func longestWPI(hours []int) (ans int) {
	s := 0
	pos := map[int]int{}
	for i, x := range hours {
		if x > 8 {
			s++
		} else {
			s--
		}
		if s > 0 {
			ans = i + 1
		} else if j, ok := pos[s-1]; ok {
			ans = max(ans, i-j)
		}
		if _, ok := pos[s]; !ok {
			pos[s] = i
		}
	}
	return
}

# Accepted solution for LeetCode #1124: Longest Well-Performing Interval
class Solution:
    def longestWPI(self, hours: List[int]) -> int:
        ans = s = 0
        pos = {}
        for i, x in enumerate(hours):
            s += 1 if x > 8 else -1
            if s > 0:
                ans = i + 1
            elif s - 1 in pos:
                ans = max(ans, i - pos[s - 1])
            if s not in pos:
                pos[s] = i
        return ans

// Accepted solution for LeetCode #1124: Longest Well-Performing Interval
struct Solution;
use std::collections::HashMap;

impl Solution {
    fn longest_wpi(hours: Vec<i32>) -> i32 {
        let n = hours.len();
        let mut hm: HashMap<i32, usize> = HashMap::new();
        let mut score = 0;
        let mut res = 0;
        for i in 0..n {
            score += if hours[i] > 8 { 1 } else { -1 };
            if score > 0 {
                res = i + 1;
            } else {
                hm.entry(score).or_insert(i);
                if let Some(j) = hm.get(&(score - 1)) {
                    res = res.max(i - j);
                }
            }
        }
        res as i32
    }
}

#[test]
fn test() {
    let hours = vec![9, 9, 6, 0, 6, 6, 9];
    let res = 3;
    assert_eq!(Solution::longest_wpi(hours), res);
}

// Accepted solution for LeetCode #1124: Longest Well-Performing Interval
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1124: Longest Well-Performing Interval
// class Solution {
//     public int longestWPI(int[] hours) {
//         int ans = 0, s = 0;
//         Map<Integer, Integer> pos = new HashMap<>();
//         for (int i = 0; i < hours.length; ++i) {
//             s += hours[i] > 8 ? 1 : -1;
//             if (s > 0) {
//                 ans = i + 1;
//             } else if (pos.containsKey(s - 1)) {
//                 ans = Math.max(ans, i - pos.get(s - 1));
//             }
//             pos.putIfAbsent(s, i);
//         }
//         return ans;
//     }
// }

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)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

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.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.