Algorithm

1309. Decrypt String from Alphabet to Integer Mapping

https://leetcode.com/problems/decrypt-string-from-alphabet-to-integer-mapping/

Given a string s formed by digits ('0' - '9') and '#' . We want to map s to English lowercase characters as follows:

  • Characters ('a' to 'i') are represented by ('1' to '9') respectively.
  • Characters ('j' to 'z') are represented by ('10#' to '26#') respectively.

Return the string formed after mapping.

It’s guaranteed that a unique mapping will always exist.

Example 1:

Input: s = "10#11#12"
Output: "jkab"
Explanation: "j" -> "10#" , "k" -> "11#" , "a" -> "1" , "b" -> "2".

Example 2:

Input: s = "1326#"
Output: "acz"

Example 3:

Input: s = "25#"
Output: "y"

Example 4:

Input: s = "12345678910#11#12#13#14#15#16#17#18#19#20#21#22#23#24#25#26#"
Output: "abcdefghijklmnopqrstuvwxyz"

Constraints:

  • 1 <= s.length <= 1000
  • s[i] only contains digits letters ('0'-'9') and '#' letter.
  • s will be valid string such that mapping is always possible.

1436. Destination City

https://leetcode.com/problems/destination-city/

You are given the array paths, where paths[i] = [cityAi, cityBi] means there exists a direct path going from cityAi to cityBi. Return the destination city, that is, the city without any path outgoing to another city.

It is guaranteed that the graph of paths forms a line without any loop, therefore, there will be exactly one destination city.

Example 1:

Input: paths = [["London","New York"],["New York","Lima"],["Lima","Sao Paulo"]]
Output: "Sao Paulo"
Explanation: Starting at "London" city you will reach "Sao Paulo" city which is the destination city. Your trip consist of: "London" -> "New York" -> "Lima" -> "Sao Paulo".

Example 2:

Input: paths = [["B","C"],["D","B"],["C","A"]]
Output: "A"
Explanation: All possible trips are:
"D" -> "B" -> "C" -> "A".
"B" -> "C" -> "A".
"C" -> "A".
"A".
Clearly the destination city is "A".

Example 3:

Input: paths = [["A","Z"]]
Output: "Z"

Constraints:

  • 1 <= paths.length <= 100
  • paths[i].length == 2
  • 1 <= cityAi.length, cityBi.length <= 10
  • cityAi != cityBi
  • All strings consist of lowercase and uppercase English letters and the space character.

1450. Number of Students Doing Homework at a Given Time

https://leetcode.com/problems/number-of-students-doing-homework-at-a-given-time/

Given two integer arrays startTime and endTime and given an integer queryTime.

The ith student started doing their homework at the time startTime[i] and finished it at time endTime[i].

Return the number of students doing their homework at time queryTime. More formally, return the number of students where queryTime lays in the interval [startTime[i], endTime[i]] inclusive.

Example 1:

Input: startTime = [1,2,3], endTime = [3,2,7], queryTime = 4
Output: 1
Explanation: We have 3 students where:
The first student started doing homework at time 1 and finished at time 3 and wasn't doing anything at time 4.
The second student started doing homework at time 2 and finished at time 2 and also wasn't doing anything at time 4.
The third student started doing homework at time 3 and finished at time 7 and was the only student doing homework at time 4.

Example 2:

Input: startTime = [4], endTime = [4], queryTime = 4
Output: 1
Explanation: The only student was doing their homework at the queryTime.

Example 3:

Input: startTime = [4], endTime = [4], queryTime = 5
Output: 0

Example 4:

Input: startTime = [1,1,1,1], endTime = [1,3,2,4], queryTime = 7
Output: 0

Example 5:

Input: startTime = [9,8,7,6,5,4,3,2,1], endTime = [10,10,10,10,10,10,10,10,10], queryTime = 5
Output: 5

Constraints:

  • startTime.length == endTime.length
  • 1 <= startTime.length <= 100
  • 1 <= startTime[i] <= endTime[i] <= 1000
  • 1 <= queryTime <= 1000

1381. Design a Stack With Increment Operation

Design a stack which supports the following operations.

Implement the CustomStack class:

  • CustomStack(int maxSize) Initializes the object with maxSize which is the maximum number of elements in the stack or do nothing if the stack reached the maxSize.
  • void push(int x) Adds x to the top of the stack if the stack hasn't reached the maxSize.
  • int pop() Pops and returns the top of stack or -1 if the stack is empty.
  • void inc(int k, int val) Increments the bottom k elements of the stack by val. If there are less than k elements in the stack, just increment all the elements in the stack.

Example 1:

Input
["CustomStack","push","push","pop","push","push","push","increment","increment","pop","pop","pop","pop"]
[[3],[1],[2],[],[2],[3],[4],[5,100],[2,100],[],[],[],[]]
Output
[null,null,null,2,null,null,null,null,null,103,202,201,-1]
Explanation
CustomStack customStack = new CustomStack(3); // Stack is Empty []
customStack.push(1); // stack becomes [1]
customStack.push(2); // stack becomes [1, 2]
customStack.pop(); // return 2 --> Return top of the stack 2, stack becomes [1]
customStack.push(2); // stack becomes [1, 2]
customStack.push(3); // stack becomes [1, 2, 3]
customStack.push(4); // stack still [1, 2, 3], Don't add another elements as size is 4
customStack.increment(5, 100); // stack becomes [101, 102, 103]
customStack.increment(2, 100); // stack becomes [201, 202, 103]
customStack.pop(); // return 103 --> Return top of the stack 103, stack becomes [201, 202]
customStack.pop(); // return 202 --> Return top of the stack 102, stack becomes [201]
customStack.pop(); // return 201 --> Return top of the stack 101, stack becomes []
customStack.pop(); // return -1 --> Stack is empty return -1.

Constraints:

  • 1 <= maxSize <= 1000
  • 1 <= x <= 1000
  • 1 <= k <= 1000
  • 0 <= val <= 100
  • At most 1000 calls will be made to each method of increment, push and pop each separately.

fibonacci

문제

아래와 같이 정의된 피보나치 수열 중 n번째 항의 수를 리턴해야 합니다.

  • 0번째 피보나치 수는 0이고, 1번째 피보나치 수는 1입니다. 그 다음 2번째 피보나치 수부터는 바로 직전의 두 피보나치 수의 합으로 정의합니다.
  • 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, …

입력

인자 1 : n

  • number 타입의 n (n은 0 이상의 정수)

출력

  • number 타입을 리턴해야합니다.

주의사항

  • fibonacci 함수는 재귀함수 형태로 작성해야 합니다.

입출력 예시

let output = fibonacci(0);
console.log(output); // --> 0
output = fibonacci(1);
console.log(output); // --> 1
output = fibonacci(5);
console.log(output); // --> 5
output = fibonacci(9);
console.log(output); // --> 34

Advanced

  • 재귀함수의 형태를 유지한 채, 메모이제이션(memoization)을 적용하여 Advanced 테스트 케이스를 통과해 보세요.

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