feat(work): add new LeetCode solution files for 1426, 1832, 344, and 15

work/3/Easy/Array/1426.counting-elements.py

@@ -0,0 +1,45 @@
+"""
+1426. Counting Elements
+Difficulty: Easy
+https://leetcode.com/problems/counting-elements/
+
+──────────────────────────────────────────────────
+
+Given an integer array arr, count how many elements x there are, such
+that x + 1 is also in arr. If there are duplicates in arr, count them
+separately.
+
+
+
+Example 1:
+
+Input: arr = [1,2,3]
+Output: 2
+Explanation: 1 and 2 are counted cause 2 and 3 are in arr.
+
+Example 2:
+
+Input: arr = [1,1,3,3,5,5,7,7]
+Output: 0
+Explanation: No numbers are counted, cause there is no 2, 4, 6, or 8
+in arr.
+
+
+
+Constraints:
+
+        • 1 <= arr.length <= 1000
+
+        • 0 <= arr[i] <= 1000
+"""
+
+
+class Solution:
+    def countElements(self, arr: List[int]) -> int:
+        arr_set = set(arr)
+        count = 0
+        for n in arr:
+            sum = n + 1
+            if sum in arr_set:
+                count += 1
+        return count

work/3/Easy/Hash Table/1832.check-if-the-sentence-is-pangram.py

@@ -0,0 +1,40 @@
+"""
+1832. Check if the Sentence Is Pangram
+Difficulty: Easy
+https://leetcode.com/problems/check-if-the-sentence-is-pangram/
+
+──────────────────────────────────────────────────
+
+A pangram is a sentence where every letter of the English alphabet
+appears at least once.
+
+Given a string sentence containing only lowercase English letters,
+return true if sentence is a pangram, or false otherwise.
+
+
+
+Example 1:
+
+Input: sentence = "thequickbrownfoxjumpsoverthelazydog"
+Output: true
+Explanation: sentence contains at least one of every letter of the
+English alphabet.
+
+Example 2:
+
+Input: sentence = "leetcode"
+Output: false
+
+
+
+Constraints:
+
+        • 1 <= sentence.length <= 1000
+
+        • sentence consists of lowercase English letters.
+"""
+
+
+class Solution:
+    def checkIfPangram(self, sentence: str) -> bool:
+        return len(set(sentence.lower())) == 26

work/3/Easy/Two Pointers/344.reverse-string.py

@@ -0,0 +1,47 @@
+"""
+344. Reverse String
+Difficulty: Easy
+https://leetcode.com/problems/reverse-string/
+
+──────────────────────────────────────────────────
+
+Write a function that reverses a string. The input string is given as
+an array of characters s.
+
+You must do this by modifying the input array in-place with O(1)
+extra memory.
+
+
+
+Example 1:
+
+Input: s = ["h","e","l","l","o"]
+Output: ["o","l","l","e","h"]
+
+Example 2:
+
+Input: s = ["H","a","n","n","a","h"]
+Output: ["h","a","n","n","a","H"]
+
+
+
+Constraints:
+
+        • 1 <= s.length <= 10^5
+
+        • s[i] is a printable ascii character.
+"""
+
+
+class Solution:
+    def reverseString(self, s: List[str]) -> None:
+        l = 0
+        r = len(s) - 1
+        while l < r:
+            s[l], s[r] = s[r], s[l]
+            l += 1
+            r -= 1
+
+        """
+        Do not return anything, modify s in-place instead.
+        """

work/3/Medium/Array/15.3sum.py

@@ -0,0 +1,89 @@
+"""
+15. 3Sum
+Difficulty: Medium
+https://leetcode.com/problems/3sum/
+
+──────────────────────────────────────────────────
+
+Given an integer array nums, return all the triplets [nums[i],
+nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] +
+nums[j] + nums[k] == 0.
+
+Notice that the solution set must not contain duplicate triplets.
+
+
+
+Example 1:
+
+Input: nums = [-1,0,1,2,-1,-4]
+Output: [[-1,-1,2],[-1,0,1]]
+Explanation:
+nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
+nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
+nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
+The distinct triplets are [-1,0,1] and [-1,-1,2].
+Notice that the order of the output and the order of the triplets
+does not matter.
+
+Example 2:
+
+Input: nums = [0,1,1]
+Output: []
+Explanation: The only possible triplet does not sum up to 0.
+
+Example 3:
+
+Input: nums = [0,0,0]
+Output: [[0,0,0]]
+Explanation: The only possible triplet sums up to 0.
+
+
+
+Constraints:
+
+        • 3 <= nums.length <= 3000
+
+        • -10^5 <= nums[i] <= 10^5
+"""
+
+
+class Solution:
+    def threeSum(self, nums: list[int]) -> list[list[int]]:
+        nums.sort()
+        result = []
+        n = len(nums)
+
+        for i in range(n):
+            # skip all zero
+            if i > 0 and nums[i] == nums[i - 1]:
+                continue
+
+            # two pointers
+            left = i + 1
+            right = n - 1
+            target = -nums[i]
+
+            while left < right:
+                current = nums[left] + nums[right]
+
+                if current == target:
+                    result.append([nums[i], nums[left], nums[right]])
+
+                    # skip duplicates
+                    while left < right and nums[left] == nums[left + 1]:
+                        left += 1
+                    while left < right and nums[right] == nums[right - 1]:
+                        right -= 1
+
+                    # shift pointers
+                    left += 1
+                    right -= 1
+
+                # since sorted, if current < target, then move left
+                elif current < target:
+                    left += 1
+                # since sorted, if current > target, then move right
+                else:
+                    right -= 1
+
+        return result