Longest Well-Performing Interval - Practice Coding Problems

Longest Well-Performing Interval - Problem

Array Hash Table Stack Monotonic Stack Prefix Sum Medium

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.

Input & Output

Example 1 — Basic Well-Performing Interval

$ Input: hours = [9,9,6,0,6,6,9]

› Output: 3

💡 Note: The longest well-performing interval is [9,9,6] from index 0 to 2. It has 2 tiring days (9>8, 9>8) and 1 non-tiring day (6≤8), so tiring > non-tiring.

Example 2 — No Well-Performing Interval

$ Input: hours = [6,6,6]

› Output: 0

💡 Note: All days have ≤8 hours, so no interval can have more tiring days than non-tiring days. The longest well-performing interval has length 0.

Example 3 — Entire Array is Well-Performing

$ Input: hours = [9,9,9]

› Output: 3

💡 Note: All days are tiring (>8 hours), so the entire array [9,9,9] is well-performing with 3 tiring days and 0 non-tiring days.

Constraints

1 ≤ hours.length ≤ 10⁴
0 ≤ hours[i] ≤ 16

Visualization

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Understanding the Visualization

Input Analysis

Classify each day as tiring (>8 hours) or non-tiring (≤8 hours)

Interval Search

Find intervals where tiring days > non-tiring days

Maximum Length

Return length of longest well-performing interval

Key Takeaway

🎯 Key Insight: Transform to +1/-1 scoring system and find longest positive-sum subarray

Asked in

G Google 25 a Amazon 18 M Microsoft 15 f Facebook 12

The key insight is to transform this into a subarray sum problem: convert tiring days to +1 and non-tiring to -1, then find the longest subarray with positive sum. Best approach uses prefix sum with hash map in O(n) time and O(n) space.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Check All Intervals	O(n³)	O(1)	Check every possible interval and count tiring vs non-tiring days
Prefix Sum with Hash Map	O(n)	O(n)	Transform to +1/-1 problem, use prefix sums to find longest positive subarray
Monotonic Stack	O(n)	O(n)	Use monotonic stack to efficiently find longest positive-sum subarray

Brute Force - Check All Intervals — Algorithm Steps

For each starting day i
For each ending day j from i onwards
Count tiring days (>8 hours) and non-tiring days in interval
If tiring > non-tiring, update maximum length

Visualization

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Step-by-Step Walkthrough

Start Position

Try each starting position i

Extend Interval

For each start, try all ending positions j

Count Days

Count tiring (>8) vs non-tiring (≤8) in interval

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* hours, int n) {
    int maxLength = 0;
    
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            int tiringDays = 0;
            int nonTiringDays = 0;
            
            for (int k = i; k <= j; k++) {
                if (hours[k] > 8) {
                    tiringDays++;
                } else {
                    nonTiringDays++;
                }
            }
            
            if (tiringDays > nonTiringDays) {
                int length = j - i + 1;
                if (length > maxLength) {
                    maxLength = length;
                }
            }
        }
    }
    
    return maxLength;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int hours[1000];
    int n = 0;
    
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != '\n') {
            hours[n++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(hours, n);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Triple nested loops: n starting positions × n ending positions × n counting operations

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables for counting

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Intervals — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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