python & js練習

README.md

@@ -4,6 +4,8 @@
 2. 題目來源:
 - LeetCode 
 - CodeWars
+- HackerRank
 
 雖是JavaScript，但實際上使用Node.js，因此不需要打開瀏覽器便可執行JS的環境
-CMD打上node app.js
+CMD打上node {檔案名稱.js}，EG.node index.js <br>
+而Python，則是打上 python3 {檔案名稱.py} EG.python3 index.py

javascript/LeetCode/Array/3683.js

@@ -0,0 +1,19 @@
+/**
+ * 3683. Earliest Time to Finish One Task
+ * 
+ * task = [start time,finsh time]
+ * 回傳task最早完成的時間
+ * @param {number[][]} tasks
+ * @return {number}
+ */
+var earliestTime = function(tasks) {
+  let ans = Infinity;
+  for(let i = 0;i < tasks.length;i++) {
+    ans = Math.min(ans,tasks[i][0] + tasks[i][1]);
+  }
+  return ans;
+};
+let tasks = [[1,6],[2,3]];
+// 5
+// The first task starts at time t = 1 and finishes at time 1 + 6 = 7. The second task finishes at time 2 + 3 = 5. You can finish one task at time 5.
+console.log(earliestTime(tasks));

javascript/LeetCode/math/3658.js

@@ -0,0 +1,37 @@
+/**
+ * 3658. GCD of Odd and Even Sums
+ * 
+ * GCD 是 最大公因數（Greatest Common Divisor）的縮寫，指的是能夠整除兩個或以上非零整數的最大正整數。 例如，8和12的最大公因數是4，因為4是8和12的公因數中最大的那個
+ *
+ * sumOdd = 從1開始，n個奇數的總和
+ * sumEven = 從1開始，n個偶數總和
+ * gcd(sumOdd,sumEven) = answer
+ * 
+ * @param {number} n
+ * @return {number}
+ */
+var gcdOfOddEvenSums = function(n) {
+  // ans 能夠整除sumOdd & sumEven
+  // let sumOdd = [],sumEven = [];
+  // for(let i = 1;i <= n*2;i++) {
+  //   if(i % 2 === 0 && sumOdd.length <= n){
+  //     sumEven.push(i);
+  //   }else{
+  //     sumOdd.push(i);
+  //   }
+  // }
+  // let odd = sumOdd.reduce((a,b)=>a+b,0);
+  // let even = sumEven.reduce((a,b)=>a+b,0);
+  // return Math.abs(odd-even);
+
+  // solution 2
+  let sumEven = n * (n + 1);
+  let sumOdd = n * n;
+  return Math.abs(sumOdd - sumEven);
+};
+let n = 4;
+// 4
+// Sum of the first 4 odd numbers sumOdd = 1 + 3 + 5 + 7 = 16
+// Sum of the first 4 even numbers sumEven = 2 + 4 + 6 + 8 = 20
+// Hence, GCD(sumOdd, sumEven) = GCD(16, 20) = 4.
+console.log(gcdOfOddEvenSums(n))

javascript/index.js

@@ -1024,8 +1024,7 @@ var closetPair = function(arr1,arr2,x) {
   }
 
 }
-// let arr1 = [1,4,5,7],arr2 = [10,20,30,40],x = 32;
+// let arr1 = [1,4,5,7],arr2 = [10,20,30,40],x = 32; 
 // [1,30];
 // console.log(closetPair(arr1,arr2,x));
 
-

python/index.py

@@ -10,6 +10,7 @@
 from math import sqrt
 from operator import le
 from typing import List
+import math
 
 
 # from numpy import diff
@@ -22,38 +23,29 @@
 
 # oop
 # from file name(module) import the class.
+from oop.Animals import Animals
+from oop.Tortoises import Tortoises
+from oop.Lion import Lion
+
+greek = Tortoises("福氣","7個月大","veggie","地中海型陸龜")
+greek.eat()
+greek.environment()
+
+lion = Lion("獅子","未知","肉食")
+lion.environment()
+
 # from oop.Fruit import Fruit
 # from oop.Melon import Melon
 # fruits = Fruit("Apple",3,5)
-# print(fruits.calculate())
+# print("價格：",f"{fruits.calculate()}")
 
-# fruits = Fruit()
 # fruits.make_watering()
 
-# v = Melon()
+# v = Melon("西瓜",65,10,30)
 # v.make_seedling()
 # v.palnt()
 
 
-
-def diagonalDifference(arr):
-  '''
-  Diagonal Difference
-  Complete the 'diagonalDifference' function below.
-
-  The function is expected to return an INTEGER.
-  The function accepts 2D_INTEGER_ARRAY arr as parameter.
-  '''
-  left = 0
-  right = 0
-  for i in range(len(arr)):
-    left += arr[i][i]
-    right += arr[i][len(arr) - 1 - i]
-  return abs(left - right)
-
-
-
-
 '''
 1415. The k-th Lexicographical String of All Happy Strings of Length n
 
@@ -221,53 +213,6 @@ def getFinalState(self, nums: List[int], k: int, multiplier: int) -> List[int]:
 # print(a.getFinalState(nums,k,multiplier))
 
 
-
-'''
-3407. Substring Matching Pattern
-
-You are given a string s and a pattern string p, where p contains exactly one '*' character.
-The '*' in p can be replaced with any sequence of zero or more characters.
-Return true if p can be made a substring of s, and false otherwise.
-
-Hints:
-1. Divide the pattern in two strings and search in the string.
-
-Example 1:
-Input: s = "leetcode", p = "ee*e"
-Output: true
-Explanation:
-By replacing the '*' with "tcod", the substring "eetcode" matches the pattern.
-
-Example 2:
-Input: s = "car", p = "c*v"
-Output: false
-Explanation:
-There is no substring matching the pattern.
-
-Example 3:
-Input: s = "luck", p = "u*"
-Output: true
-Explanation:
-The substrings "u", "uc", and "uck" match the pattern.
-
-Constraints:
-1 <= s.length <= 50
-1 <= p.length <= 50 
-s contains only lowercase English letters.
-p contains only lowercase English letters and exactly one '*'
-
-參數為一個字串s和字串p，p內有一個"*"符號，而該符號可被替換成任一或多個字母
-若p的*號在替換成字母後可變成s的子字串，則回傳true
-否則false
-'''
-def hasMatch(s: str, p: str) -> bool:
-  #  將字串拆成兩部分再搜尋
-  return False
-# s = "leetcode"
-# p = "ee*e"
-# True
-# print(hasMatch(s,p))
-
 '''
 2523. Closest Prime Numbers in Range
 
@@ -296,8 +241,6 @@ def hasMatch(s: str, p: str) -> bool:
 
 Constraints:
 1 <= left <= right <= 106
-
-
 '''
 def closestPrimes(left: int, right: int) -> List[int]:
   '''
@@ -353,48 +296,3 @@ def isPrime(element:int):
 # print(closestPrimes(left,right))
 
 
-'''
-2873. Maximum Value of an Ordered Triplet I
-
-You are given a 0-indexed integer array nums.
-Return the maximum value over all triplets of indices (i, j, k) such that i < j < k. If all such triplets have a negative value, return 0.
-The value of a triplet of indices (i, j, k) is equal to (nums[i] - nums[j]) * nums[k].
-
-Hints:
-1.Use three nested loops to find all the triplets.
-
-Example 1:
-Input: nums = [12,6,1,2,7]
-Output: 77
-Explanation: The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
-It can be shown that there are no ordered triplets of indices with a value greater than 77. 
-
-Example 2:
-Input: nums = [1,10,3,4,19]
-Output: 133
-Explanation: The value of the triplet (1, 2, 4) is (nums[1] - nums[2]) * nums[4] = 133.
-It can be shown that there are no ordered triplets of indices with a value greater than 133.
-Example 3:
-
-Input: nums = [1,2,3]
-Output: 0
-Explanation: The only ordered triplet of indices (0, 1, 2) has a negative value of (nums[0] - nums[1]) * nums[2] = -3. Hence, the answer would be 0.
-
-
-Constraints:
-3 <= nums.length <= 100
-1 <= nums[i] <= 106
-
-'''
-def maximumTripletValue(nums: List[int]) -> int:
-  # i < j < k
-  # (nums[i] - nums[j]) * nums[k]
-  ans = 0
-
-  return ans
-
-
-nums = [12,6,1,2,7]
-# 77
-# (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
-print(maximumTripletValue(nums))

python/leetcode/list/2873.py

@@ -0,0 +1,51 @@
+from typing import List
+'''
+2873. Maximum Value of an Ordered Triplet I
+
+You are given a 0-indexed integer array nums.
+Return the maximum value over all triplets of indices (i, j, k) such that i < j < k. If all such triplets have a negative value, return 0.
+The value of a triplet of indices (i, j, k) is equal to (nums[i] - nums[j]) * nums[k].
+
+Hints:
+1.Use three nested loops to find all the triplets.
+
+Example 1:
+Input: nums = [12,6,1,2,7]
+Output: 77
+Explanation: The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
+It can be shown that there are no ordered triplets of indices with a value greater than 77. 
+
+Example 2:
+Input: nums = [1,10,3,4,19]
+Output: 133
+Explanation: The value of the triplet (1, 2, 4) is (nums[1] - nums[2]) * nums[4] = 133.
+It can be shown that there are no ordered triplets of indices with a value greater than 133.
+Example 3:
+
+Input: nums = [1,2,3]
+Output: 0
+Explanation: The only ordered triplet of indices (0, 1, 2) has a negative value of (nums[0] - nums[1]) * nums[2] = -3. Hence, the answer would be 0.
+
+
+Constraints:
+3 <= nums.length <= 100
+1 <= nums[i] <= 106
+
+'''
+def maximumTripletValue(nums: List[int]) -> int:
+  # i < j < k
+  # (nums[i] - nums[j]) * nums[k]
+  sortedSum = (nums[0] - nums[1]) * nums[2]
+  for i in range(len(nums)):
+    for j in range(i+1,len(nums)):
+      for k in range(j+1,len(nums)):
+          total = (nums[i] - nums[j]) * nums[k]
+          if(total > sortedSum):
+            sortedSum = total
+  return sortedSum if sortedSum > 0 else 0
+
+
+nums = [12,6,1,2,7]
+# 77
+# (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
+print(maximumTripletValue(nums))

python/leetcode/math/3683.py

@@ -0,0 +1,19 @@
+from typing import List
+import math
+
+'''
+3683. Earliest Time to Finish One Task
+
+task = [start time,finsh time]
+回傳task最早完成的時間
+'''
+def earliestTime(tasks: List[List[int]]) -> int:
+  ans = math.inf
+  for i in range(len(tasks)):
+    ans = min(ans,tasks[i][0] + tasks[i][1])
+
+  return ans
+
+tasks = [[1,6],[2,3]]
+# 5
+print(earliestTime(tasks))

python/leetcode/string/3407.py

@@ -0,0 +1,53 @@
+'''
+3407. Substring Matching Pattern
+
+You are given a string s and a pattern string p, where p contains exactly one '*' character.
+The '*' in p can be replaced with any sequence of zero or more characters.
+Return true if p can be made a substring of s, and false otherwise.
+
+Hints:
+1. Divide the pattern in two strings and search in the string.
+
+Example 1:
+Input: s = "leetcode", p = "ee*e"
+Output: true
+Explanation:
+By replacing the '*' with "tcod", the substring "eetcode" matches the pattern.
+
+Example 2:
+Input: s = "car", p = "c*v"
+Output: false
+Explanation:
+There is no substring matching the pattern.
+
+Example 3:
+Input: s = "luck", p = "u*"
+Output: true
+Explanation:
+The substrings "u", "uc", and "uck" match the pattern.
+
+Constraints:
+1 <= s.length <= 50
+1 <= p.length <= 50 
+s contains only lowercase English letters.
+p contains only lowercase English letters and exactly one '*'
+
+參數為一個字串s和字串p，p內有一個"*"符號，而該符號可被替換成任一或多個字母
+若p的*號在替換成字母後可變成s的子字串，則回傳true
+否則false
+'''
+def hasMatch(s: str, p: str) -> bool:
+  #  將字串拆成兩部分再搜尋
+  left,right = p.split("*")
+  left_part = s.find(left)
+  right_part = s.rfind(right)
+
+  if left_part == -1 or right_part == -1:
+    return False
+  return left_part + len(left) <= right_part  
+
+s = "luck"
+p = "u*"
+# True
+# 將s的tcod塞進p的*內，p等同是s的子字串
+print(hasMatch(s,p))