[alphaorderly] WEEK 03 Solutions (#2707)

* week 1

* [alphaorderly] WEEK 02 Solutions

* fix: description에 맞게 코드 수정

* fix: 좀 더 간결하게 수정

* [alphaorderly] WEEK 03 Solutions

* fix: 불필요한 코드 삭제

* fix: 줄바꿈 린트 문제 수정

* [alphaorderly] WEEK 03 Solutions

* fix: 린트 오류 수정

* fix: 파이썬 내장함수 사용 코드 추가

* fix: valid-palindrome 문제에 투포인터 구현 답안 추가

* fix: 로직 가독성 수정

Author: alphaorderly

combination-sum/alphaorderly.py

@@ -0,0 +1,51 @@
+
+"""
+Time Complexity: O(N^target)
+Space Complexity: O(target)
+
+Classic backtracking approach.
+- Use a helper function to backtrack and generate all possible combinations.
+"""
+# class Solution:
+#     def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
+#         N = len(candidates)
+#         candidates.sort()
+        
+#         ans = []
+#         app = []
+
+#         def backtracking(last: int, left: int) -> None:
+#             if left == 0:
+#                 ans.append(app[:])
+#                 return
+
+#             for i in range(last, N):
+#                 if candidates[i] > left:
+#                     continue
+
+#                 app.append(candidates[i])
+#                 backtracking(i, left - candidates[i])
+#                 app.pop()
+
+#         backtracking(0, target)
+
+#         return ans
+
+"""
+Time Complexity: O(N * target)
+Space Complexity: O(target)
+
+- Use a dynamic programming approach to store the combinations that sum to each target.
+- dp[j] is a list of lists that sum to j.
+"""
+class Solution:
+    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
+        dp = [list() for _ in range(target + 1)]
+        dp[0] = [[]]
+
+        for c in candidates:
+            for j in range(c, target + 1):
+                for partial in dp[j - c]:
+                    dp[j].append(partial + [c])
+
+        return dp[-1]

decode-ways/alphaorderly.py

@@ -0,0 +1,31 @@
+"""
+Time Complexity: O(N)
+Space Complexity: O(1)
+
+prev_2 : number of ways to decode the string ending with the previous two characters
+prev_1 : number of ways to decode the string ending with the previous character
+new_prev : number of ways to decode the string ending with the current character
+"""
+class Solution:
+    def numDecodings(self, s: str) -> int:
+        N = len(s)
+
+        if s[0] == "0":
+            return 0
+
+        if N == 1:
+            return 1
+
+        prev_2 = 1
+        prev_1 = int(1 <= int(s[0] + s[1]) <= 26) + int(s[1] != "0")
+
+        for i in range(2, N):
+            new_prev = 0
+            if 1 <= int(s[i - 1] + s[i]) <= 26 and s[i - 1] != "0":
+                new_prev += prev_2
+            if s[i] != "0":
+                new_prev += prev_1
+
+            prev_2, prev_1 = prev_1, new_prev
+
+        return prev_1

maximum-subarray/alphaorderly.py

@@ -0,0 +1,71 @@
+
+
+"""
+Time Complexity: O(NLogN)
+Space Complexity: O(N) ( Reason: Recursive stack pointer for every element )
+
+Divide and Conquer Approach
+
+1. Divide the array into two halves
+2. Find the maximum subarray sum in the left half
+3. Find the maximum subarray sum in the right half
+4. Find the maximum subarray sum that crosses the midpoint
+5. Return the maximum of the three sums
+"""
+# class Solution:
+#     def maxSubArray(self, nums: List[int]) -> int:
+
+#         def divide(start: int, end: int) -> int:
+#             if start >= end:
+#                 return nums[start]
+
+#             mid = (start + end) // 2
+
+#             left = divide(start, mid)
+#             right = divide(mid + 1, end)
+
+#             left_largest = -float('inf')
+#             right_largest = -float('inf')
+
+#             left_acc = 0
+#             right_acc = 0
+
+#             left_index = mid
+#             right_index = mid + 1
+
+#             while left_index >= start:
+#                 left_acc += nums[left_index]
+#                 left_largest = max(left_largest, left_acc)
+#                 left_index -= 1
+
+#             while right_index <= end:
+#                 right_acc += nums[right_index]
+#                 right_largest = max(right_largest, right_acc)
+#                 right_index += 1
+
+#             return max(left_largest + right_largest, left, right)
+
+#         return divide(0, len(nums) - 1)
+
+
+
+
+"""
+Time Complexity: O(N)
+Space Complexity: O(1)
+
+prev : maximum sum of the subarray ending with the previous element
+ans : maximum sum of the subarray
+curr : maximum sum of the subarray ending with the current element
+"""
+class Solution:
+    def maxSubArray(self, nums: List[int]) -> int:
+        N = len(nums)
+        prev = nums[0]
+        ans = nums[0]
+
+        for i in range(1, N):
+            prev = max(nums[i], prev + nums[i])
+            ans = max(ans, prev)
+
+        return ans

number-of-1-bits/alphaorderly.py

@@ -0,0 +1,11 @@
+class Solution:
+    def hammingWeight(self, n: int) -> int:
+        ans = 0
+        while n:
+            ans += n & 1
+            n >>= 1
+        return ans
+
+# class Solution:
+#     def hammingWeight(self, n: int) -> int:
+#         return bin(n).count('1')

valid-palindrome/alphaorderly.py

@@ -0,0 +1,42 @@
+"""
+Time Complexity: O(N)
+Space Complexity: O(N)
+
+- Process string with lower() and isalnum() to remove non-alphanumeric characters
+- Check if the processed string is a palindrome by comparing it to its reverse
+"""
+# class Solution:
+#     def isPalindrome(self, s: str) -> bool:
+#         processed = "".join(ch.lower() for ch in s if ch.isalnum())
+
+#         return processed == processed[::-1]
+
+"""
+Time Complexity: O(N)
+Space Complexity: O(1)
+
+- Use two pointers to check if the string is a palindrome
+- Skip non-alphanumeric characters
+- Compare characters from both ends towards the center
+"""
+class Solution:
+    def isPalindrome(self, s: str) -> bool:
+        N = len(s)
+
+        left, right = 0, N - 1
+
+        while True:
+            while not s[left].isalnum() and left < right:
+                left += 1
+            while not s[right].isalnum() and left < right:
+                right -= 1
+
+            if left >= right:
+                break
+
+            if s[left].lower() != s[right].lower():
+                return False
+
+            left, right = left + 1, right - 1
+
+        return True