[Swift]LeetCode1124. 表现良好的最长时间段 | Longest Well-Performing Interval

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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]. 

Constraints:

• 1 <= hours.length <= 10000
• 0 <= hours[i] <= 16

---

给你一份工作时间表 hours，上面记录着某一位员工每天的工作小时数。

我们认为当员工一天中的工作小时数大于 8 小时的时候，那么这一天就是「劳累的一天」。

所谓「表现良好的时间段」，意味在这段时间内，「劳累的天数」是严格 大于「不劳累的天数」。

请你返回「表现良好时间段」的最大长度。 

示例 1：

输入：hours = [9,9,6,0,6,6,9]
输出：3
解释：最长的表现良好时间段是 [9,9,6]。 

提示：

• 1 <= hours.length <= 10000
• 0 <= hours[i] <= 16

---

140ms

class Solution {
    func longestWPI(_ hours: [Int]) -> Int {
        var maxInterval = 0
        var wellDays = 0
        var dict = [Int: Int]()
        
        for i in 0..<hours.count {
            wellDays += hours[i] > 8 ? 1 : -1
            if(wellDays >= 1) {
                maxInterval = max(maxInterval, i + 1);
            }

            if dict[wellDays] != nil {
                guard let maxSum = dict[wellDays-1] else { continue }
                maxInterval = max(maxInterval, i-maxSum)
            } else {
                dict[wellDays] = i
            }
        }
        if maxInterval == 0 {
            maxInterval = hours.filter{$0 > 8}.count >= 1 ? 1 : 0
        }
        return maxInterval
    }
}

---

148ms

class Solution {
    func longestWPI(_ hours: [Int]) -> Int {
        var result = 0
        var score = 0 
        var n = hours.count 
        var seen = [Int: Int]()
        
        for i in 0..<n {
            score += hours[i] > 8 ? 1 : -1
            if score > 0 {
                result = i + 1
            } else {
                if seen[score] == nil {
                    seen[score] = i 
                }
                if let idx = seen[score - 1] {
                    result = max(result, i - idx)
                }
            }
        }
        return result
    }
}

---

164ms

class Solution {
    func longestWPI(_ hours: [Int]) -> Int {
    var memo : [Int:Int] = [0:-1]
    var maxLength = 0
    var countAboveTiringDays = 0
    for i in 0..<hours.count {
        if hours[i] > 8 {
            countAboveTiringDays += 1
        } else {
            countAboveTiringDays -= 1
        }

        // if the count is positive, then it is just the number of days
        if countAboveTiringDays > 0 {
            maxLength = i + 1
        } else {
        
            if memo[countAboveTiringDays - 1] != nil {
                maxLength = max(maxLength, i - memo[countAboveTiringDays - 1]!)
            }
            
            if memo[countAboveTiringDays] == nil
            {
                memo[countAboveTiringDays] = i
            }

        }
    }
    return maxLength
    }
}

---

1900ms

class Solution {
    func longestWPI(_ hours: [Int]) -> Int {
        var ans = 0
        let size  = hours.count
        let hoursCopy = hours.map { (val) -> Int in
            val > 8 ? 1 : -1
        }
        
        var sum = Array<Int>(repeating: 0, count: size + 1)
        for i in 0..<size {
            sum[i + 1] = sum[i] + hoursCopy[i]
        }
        for i in 0..<size {
            var j = size
            while j > i && j - i > ans  {
                if  sum[i] < sum[j] {
                    ans = j - i
                }
                j -= 1
            }
        }
        return ans
    }
}

---

Runtime: 3896 ms

Memory Usage: 21.2 MB

class Solution {
    func longestWPI(_ hours: [Int]) -> Int {
        var hours = hours
        for i in 0..<hours.count
        {
            if hours[i] > 8 {hours[i] = 1}
            else {hours[i] = -1}
        }
        var ans:Int = 0
        for i in 0..<hours.count
        {
            var bal:Int = 0
            var cur:Int = -1
            for j in i..<hours.count
            {
                bal += hours[j]
                if bal > 0
                {
                    cur = j
                }
            }
            ans = max(ans, cur - i + 1)
        }
        return ans        
    }
}