Question

1124. Longest Well-Performing Interval

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Description

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

Solutions

Solution 1: Prefix Sum + Hash Table

We can use the idea of prefix sum, maintaining a variable $s$, which represents the difference between the number of "tiring days" and "non-tiring days" from index $0$ to the current index. If $s$ is greater than $0$, it means that the segment from index $0$ to the current index is a "well-performing time period". In addition, we use a hash table $pos$ to record the first occurrence index of each $s$.

Next, we traverse the hours array, for each index $i$:

• If $hours[i] > 8$, we increment $s$ by $1$, otherwise we decrement $s$ by $1$.
• If $s > 0$, it means that the segment from index $0$ to the current index $i$ is a "well-performing time period", we update the result $ans = i + 1$. Otherwise, if $s - 1$ is in the hash table $pos$, let $j = pos[s - 1]$, it means that the segment from index $j + 1$ to the current index $i$ is a "well-performing time period", we update the result $ans = \max(ans, i - j)$.
• Then, if $s$ is not in the hash table $pos$, we record $pos[s] = i$.

After the traversal, return the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the hours array.

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

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;
  }
}

class Solution {
public:
  int longestWPI(vector<int>& hours) {
    int ans = 0, s = 0;
    unordered_map<int, int> pos;
    for (int i = 0; i < hours.size(); ++i) {
      s += hours[i] > 8 ? 1 : -1;
      if (s > 0) {
        ans = i + 1;
      } else if (pos.count(s - 1)) {
        ans = max(ans, i - pos[s - 1]);
      }
      if (!pos.count(s)) {
        pos[s] = i;
      }
    }
    return ans;
  }
};

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
}