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Posts tagged as “easy”

花花酱 LeetCode 1636. Sort Array by Increasing Frequency

By zxi on October 31, 2020

Given an array of integers nums, sort the array in increasing order based on the frequency of the values. If multiple values have the same frequency, sort them in decreasing order.

Return the sorted array.

Example 1:

Input: nums = [1,1,2,2,2,3]
Output: [3,1,1,2,2,2]
Explanation: '3' has a frequency of 1, '1' has a frequency of 2, and '2' has a frequency of 3.

Example 2:

Input: nums = [2,3,1,3,2]
Output: [1,3,3,2,2]
Explanation: '2' and '3' both have a frequency of 2, so they are sorted in decreasing order.

Example 3:

Input: nums = [-1,1,-6,4,5,-6,1,4,1]
Output: [5,-1,4,4,-6,-6,1,1,1]

Constraints:

  • 1 <= nums.length <= 100
  • -100 <= nums[i] <= 100

Solution: Hashtable + Sorting

Use a hashtable to track the frequency of each number.

Time complexity: O(nlogn)
Space complexity: O(n)

C++

花花酱 LeetCode 1629. Slowest Key

By zxi on October 24, 2020

A newly designed keypad was tested, where a tester pressed a sequence of n keys, one at a time.

You are given a string keysPressed of length n, where keysPressed[i] was the ith key pressed in the testing sequence, and a sorted list releaseTimes, where releaseTimes[i] was the time the ith key was released. Both arrays are 0-indexed. The 0th key was pressed at the time 0, and every subsequent key was pressed at the exact time the previous key was released.

The tester wants to know the key of the keypress that had the longest duration. The ithkeypress had a duration of releaseTimes[i] - releaseTimes[i - 1], and the 0th keypress had a duration of releaseTimes[0].

Note that the same key could have been pressed multiple times during the test, and these multiple presses of the same key may not have had the same duration.

Return the key of the keypress that had the longest duration. If there are multiple such keypresses, return the lexicographically largest key of the keypresses.

Example 1:

Input: releaseTimes = [9,29,49,50], keysPressed = "cbcd"
Output: "c"
Explanation: The keypresses were as follows:
Keypress for 'c' had a duration of 9 (pressed at time 0 and released at time 9).
Keypress for 'b' had a duration of 29 - 9 = 20 (pressed at time 9 right after the release of the previous character and released at time 29).
Keypress for 'c' had a duration of 49 - 29 = 20 (pressed at time 29 right after the release of the previous character and released at time 49).
Keypress for 'd' had a duration of 50 - 49 = 1 (pressed at time 49 right after the release of the previous character and released at time 50).
The longest of these was the keypress for 'b' and the second keypress for 'c', both with duration 20.
'c' is lexicographically larger than 'b', so the answer is 'c'.

Example 2:

Input: releaseTimes = [12,23,36,46,62], keysPressed = "spuda"
Output: "a"
Explanation: The keypresses were as follows:
Keypress for 's' had a duration of 12.
Keypress for 'p' had a duration of 23 - 12 = 11.
Keypress for 'u' had a duration of 36 - 23 = 13.
Keypress for 'd' had a duration of 46 - 36 = 10.
Keypress for 'a' had a duration of 62 - 46 = 16.
The longest of these was the keypress for 'a' with duration 16.

Constraints:

  • releaseTimes.length == n
  • keysPressed.length == n
  • 2 <= n <= 1000
  • 0 <= releaseTimes[i] <= 109
  • releaseTimes[i] < releaseTimes[i+1]
  • keysPressed contains only lowercase English letters.

Solution: Straightforward

Time complexity: O(n)
Space complexity: O(1)

C++

class Solution {
public:
  char slowestKey(vector& releaseTimes, string keysPressed) {
    int l = releaseTimes[0];
    char ans = keysPressed[0];
    
    for (int i = 1; i < releaseTimes.size(); ++i) {
      int t = releaseTimes[i] - releaseTimes[i - 1];
      if (t > l) { 
        ans = keysPressed[i]; 
        l = t;
      } else if (t == l) {
        ans = max(ans, keysPressed[i]);      
      }
    }
    return ans;
  }
};







		

	

	

		

				

						

				花花酱 LeetCode 1624. Largest Substring Between Two Equal Characters
			

			
By zxi on October 18, 2020
		

				

			

Given a string s, return the length of the longest substring between two equal characters, excluding the two characters. If there is no such substring return -1.





A substring is a contiguous sequence of characters within a string.





Example 1:





Input: s = "aa"
Output: 0
Explanation: The optimal substring here is an empty substring between the two 'a's.


Example 2:


Input: s = "abca"
Output: 2
Explanation: The optimal substring here is "bc".



Example 3:


Input: s = "cbzxy"
Output: -1
Explanation: There are no characters that appear twice in s.



Example 4:


Input: s = "cabbac"
Output: 4
Explanation: The optimal substring here is "abba". Other non-optimal substrings include "bb" and "".



Constraints:


  • 1 <= s.length <= 300
  • s contains only lowercase English letters.


Solution: Hashtable


Remember the first position each letter occurs.


Time complexity: O(n)
Space complexity: O(26) 




C++











花花酱 LeetCode 1619. Mean of Array After Removing Some Elements



By zxi on October 17, 2020


Given an integer array arr, return the mean of the remaining integers after removing the smallest 5% and the largest 5% of the elements.


Answers within 10-5 of the actual answer will be considered accepted.


Example 1:


Input: arr = [1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3]
Output: 2.00000
Explanation: After erasing the minimum and the maximum values of this array, all elements are equal to 2, so the mean is 2.



Example 2:


Input: arr = [6,2,7,5,1,2,0,3,10,2,5,0,5,5,0,8,7,6,8,0]
Output: 4.00000



Example 3:


Input: arr = [6,0,7,0,7,5,7,8,3,4,0,7,8,1,6,8,1,1,2,4,8,1,9,5,4,3,8,5,10,8,6,6,1,0,6,10,8,2,3,4]
Output: 4.77778



Example 4:


Input: arr = [9,7,8,7,7,8,4,4,6,8,8,7,6,8,8,9,2,6,0,0,1,10,8,6,3,3,5,1,10,9,0,7,10,0,10,4,1,10,6,9,3,6,0,0,2,7,0,6,7,2,9,7,7,3,0,1,6,1,10,3]
Output: 5.27778



Example 5:


Input: arr = [4,8,4,10,0,7,1,3,7,8,8,3,4,1,6,2,1,1,8,0,9,8,0,3,9,10,3,10,1,10,7,3,2,1,4,9,10,7,6,4,0,8,5,1,2,1,6,2,5,0,7,10,9,10,3,7,10,5,8,5,7,6,7,6,10,9,5,10,5,5,7,2,10,7,7,8,2,0,1,1]
Output: 5.29167



Constraints:


  • 20 <= arr.length <= 1000
  • arr.lengthis a multiple of 20.
  • 0 <= arr[i] <= 105


Solution: Sorting


Time complexity: O(nlogn)
Space complexity: O(1)




C++











花花酱 LeetCode 1614. Maximum Nesting Depth of the Parentheses



By zxi on October 10, 2020


A string is a valid parentheses string (denoted VPS) if it meets one of the following:


  • It is an empty string "", or a single character not equal to "(" or ")",
  • It can be written as AB (A concatenated with B), where A and B are VPS‘s, or
  • It can be written as (A), where A is a VPS.


We can similarly define the nesting depth depth(S) of any VPS S as follows:


  • depth("") = 0
  • depth(A + B) = max(depth(A), depth(B)), where A and B are VPS‘s
  • depth("(" + A + ")") = 1 + depth(A), where A is a VPS.


For example, "", "()()", and "()(()())" are VPS‘s (with nesting depths 0, 1, and 2), and ")(" and "(()" are not VPS‘s.


Given a VPS represented as string s, return the nesting depth of s.


Example 1:


Input: s = "(1+(2*3)+((8)/4))+1"
Output: 3
Explanation: Digit 8 is inside of 3 nested parentheses in the string.



Example 2:


Input: s = "(1)+((2))+(((3)))"
Output: 3



Example 3:


Input: s = "1+(2*3)/(2-1)"
Output: 1



Example 4:


Input: s = "1"
Output: 0



Constraints:


  • 1 <= s.length <= 100
  • s consists of digits 0-9 and characters '+', '-', '*', '/', '(', and ')'.
  • It is guaranteed that parentheses expression s is a VPS.


Solution: Stack


We only need to deal with ‘(‘ and ‘)’


Time complexity: O(n)
Space complexity: O(1)




C++