solutions: 2432 - The Employee That Worked on the Longest Task (Easy)

Author: wingkwong

solutions/2400-2499/2432-the-employee-that-worked-on-the-longest-task-easy.md

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+---
+description: 'Author: @heder | https://leetcode.com/problems/the-employee-that-worked-on-the-longest-task/'
+tags: [Array]
+---
+
+# 2432 - The Employee That Worked on the Longest Task (Easy)
+
+## Problem Link
+
+https://leetcode.com/problems/the-employee-that-worked-on-the-longest-task/
+
+## Problem Statement
+
+There are `n` employees, each with a unique id from `0` to `n - 1`.
+
+You are given a 2D integer array `logs` where `logs[i] = [idi, leaveTimei]` where:
+
+- `idi` is the id of the employee that worked on the `ith` task, and
+- `leaveTimei` is the time at which the employee finished the `ith` task. All the values `leaveTimei` are **unique**.
+
+Note that the `ith` task starts the moment right after the `(i - 1)th` task ends, and the `0th` task starts at time `0`.
+
+Return *the id of the employee that worked the task with the longest time.* If there is a tie between two or more employees, return*the **smallest** id among them*.
+
+**Example 1:**
+
+```
+Input: n = 10, logs = [[0,3],[2,5],[0,9],[1,15]]
+Output: 1
+Explanation: 
+Task 0 started at 0 and ended at 3 with 3 units of times.
+Task 1 started at 3 and ended at 5 with 2 units of times.
+Task 2 started at 5 and ended at 9 with 4 units of times.
+Task 3 started at 9 and ended at 15 with 6 units of times.
+The task with the longest time is task 3 and the employee with id 1 is the one that worked on it, so we return 1.
+```
+
+**Example 2:**
+
+```
+Input: n = 26, logs = [[1,1],[3,7],[2,12],[7,17]]
+Output: 3
+Explanation: 
+Task 0 started at 0 and ended at 1 with 1 unit of times.
+Task 1 started at 1 and ended at 7 with 6 units of times.
+Task 2 started at 7 and ended at 12 with 5 units of times.
+Task 3 started at 12 and ended at 17 with 5 units of times.
+The tasks with the longest time is task 1. The employee that worked on it is 3, so we return 3.
+```
+
+**Example 3:**
+
+```
+Input: n = 2, logs = [[0,10],[1,20]]
+Output: 0
+Explanation: 
+Task 0 started at 0 and ended at 10 with 10 units of times.
+Task 1 started at 10 and ended at 20 with 10 units of times.
+The tasks with the longest time are tasks 0 and 1. The employees that worked on them are 0 and 1, so we return the smallest id 0.
+```
+
+**Constraints:**
+
+- `2 <= n <= 500`
+- `1 <= logs.length <= 500`
+- `logs[i].length == 2`
+- `0 <= idi <= n - 1`
+- `1 <= leaveTimei <= 500`
+- `idi != idi+1`
+- `leaveTimei` are sorted in a strictly increasing order.
+
+## Approach 1: Single Pass
+
+This could code be a little bit more compact, but I like unpacking the input and giving it things meaningful names.
+
+- Time complexity: $$O(size(logs))$$ we need to look at all the entries in logs.
+- Space complexity: $$O(1)$$ only a few integers as state.
+
+
+<Tabs>
+<TabItem value="cpp" label="C++">
+<SolutionAuthor name="@heder"/>
+
+```cpp
+static int hardestWorker(int n, const vector<vector<int>>& logs) {
+    int start_time = 0;
+    int worker = n;
+    int longest_task = 0;
+    for (const vector<int>& log : logs) {
+        const int id = log[0];
+        const int end_time = log[1];
+        const int time = end_time - start_time;
+        
+        if (time > longest_task || time == longest_task && id < worker) {
+            worker = id;
+            longest_task = time;
+        }
+
+        start_time = end_time;
+    }
+    return worker;
+}
+```
+
+</TabItem>
+</Tabs>