Chengze Li

Code Once, Think Twice

2 minute read

From today I will try to participate the Leetcode weekly contest and share my solution here

2933 High-Access Employees

Difficulty: Medium

You are given a 2D 0-indexed array of strings, access_times, with size n. For each i where 0 <= i <= n - 1, access_times[i][0] represents the name of an employee, and access_times[i][1] represents the access time of that employee. All entries in access_times are within the same day.

The access time is represented as four digits using a 24-hour time format, for example, “0800” or “2250”.

An employee is said to be high-access if he has accessed the system three or more times within a one-hour period.

Times with exactly one hour of difference are not considered part of the same one-hour period. For example, “0815” and “0915” are not part of the same one-hour period.

Access times at the start and end of the day are not counted within the same one-hour period. For example, “0005” and “2350” are not part of the same one-hour period.

Return a list that contains the names of high-access employees with any order you want.

Example

Input: access_times = [["a","0549"],["b","0457"],["a","0532"],["a","0621"],["b","0540"]]
Output: ["a"]
Explanation: "a" has three access times in the one-hour period of [05:32, 06:31] which are 05:32, 05:49, and 06:21.
But "b" does not have more than two access times at all.
So the answer is ["a"].

Solution

my solution

Introspect

  • I was blocked at function has3TimesIn1Hour
    • edge case understanding is not clear
    • moving right boundary and updating the windows state is not aligned perfectly
  • the ways to traverse map
    • Map.Entry<String, List<String>> entry: accessTimeMap.entrySet()
  • convert list to array
    • String[] arr = access_times.toArray(new String[0]);

2934 Minimum Operations to Maximize Last Elements in Arrays

You are given two 0-indexed integer arrays, nums1 and nums2, both having length n.

You are allowed to perform a series of operations (possibly none).

In an operation, you select an index i in the range [0, n - 1] and swap the values of nums1[i] and nums2[i].

Your task is to find the minimum number of operations required to satisfy the following conditions:

nums1[n - 1] is equal to the maximum value among all elements of nums1, i.e., nums1[n - 1] = max(nums1[0], nums1[1], …, nums1[n - 1]). nums2[n - 1] is equal to the maximum value among all elements of nums2, i.e., nums2[n - 1] = max(nums2[0], nums2[1], …, nums2[n - 1]). Return an integer denoting the minimum number of operations needed to meet both conditions, or -1 if it is impossible to satisfy both conditions.

Example

Input: nums1 = [1,2,7], nums2 = [4,5,3]
Output: 1
Explanation: In this example, an operation can be performed using index i = 2.
When nums1[2] and nums2[2] are swapped, nums1 becomes [1,2,3] and nums2 becomes [4,5,7].
Both conditions are now satisfied.
It can be shown that the minimum number of operations needed to be performed is 1.
So, the answer is 1.

Solution

the solution I come up by myself

seems it’s good enough haha, don’t find better solution

Introspect

  • don’t try to come up with perfect solution, figure out the most basic and correct version
  • when encountering the logic operations, try both way, either remove the invalid cases first, or either confirm valid case first.
  • when make assumption, and add another assumption, have to figure out whether the new assumption is valid under previous assumption

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