[Helena] Week 9

find-minimum-in-rotated-sorted-array/yolophg.py

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+# Time Complexity: O(log n) - using binary search, so cut the search space in half each time.
+# Space Complexity: O(1) - only use a few variables (low, high, mid), no extra space.
+
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        low = 0
+        # start with the full range of the array
+        high = len(nums) - 1  
+
+        # find the middle index
+        while low < high:
+            mid = low + (high - low) // 2  
+
+            # if mid is greater than the last element, the min must be on the right
+            if nums[mid] > nums[high]:
+                # move the low pointer to the right
+                low = mid + 1  
+            else:
+                # min could be mid or in the left part
+                high = mid  
+        # low and high converge to the minimum element
+        return nums[low]

linked-list-cycle/yolophg.py

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+# Time Complexity: O(n) - traverse the linked list at most once.
+# Space Complexity: O(1) - only use two pointers, no extra memory.
+
+class Solution:
+    def hasCycle(self, head: Optional[ListNode]) -> bool:
+        # two pointers for fast moves twice as fast as slow.
+        fast = head
+        slow = head
+
+        # loop the list while fast and fast.next exist.
+        while fast and fast.next:
+            # move fast pointer two steps.
+            fast = fast.next.next
+            # move slow pointer one step.
+            slow = slow.next
+
+            # if they meet, there's a cycle.
+            if fast == slow:
+                return True
+        # if they don't meet, there's no cycle.
+        return False

maximum-product-subarray/yolophg.py

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+# Time Complexity: O(N) - just one pass through the array, so it's linear time.
+# Space Complexity: O(1) - no extra arrays, just a few variables.
+
+class Solution:
+    def maxProduct(self, nums: List[int]) -> int:
+        # tracking max product from both ends
+        prefix_product, suffix_product = 1, 1  
+        # start with the biggest single number
+        max_product = max(nums)  
+
+        for i in range(len(nums)):
+            # move forward, multiplying
+            prefix_product *= nums[i]  
+            # move backward, multiplying
+            suffix_product *= nums[len(nums) - i - 1]  
+            # update max product
+            max_product = max(max_product, prefix_product, suffix_product)
+
+            # if hit zero, reset to 1 (zero kills the product chain)
+            if prefix_product == 0:
+                prefix_product = 1
+            if suffix_product == 0:
+                suffix_product = 1
+
+        return max_product

minimum-window-substring/yolophg.py

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+# Time Complexity: O(N) - go through the string with two pointers, so it's basically O(N).
+# Space Complexity: O(1) - only storing character frequencies (max 52 keys for a-z & A-Z), so it's effectively constant space.
+
+class Solution:
+    def minWindow(self, s: str, t: str) -> str:
+        # store character counts for t
+        target_count = Counter(t)
+        # window character count
+        window_count = defaultdict(int)
+
+        # start of the window
+        left = 0  
+        # min substring
+        min_substring = ""
+        # tracks how many characters match the required count
+        matched_chars = 0  
+        # unique characters needed
+        required_chars = len(target_count)  
+
+        for right, char in enumerate(s):
+            # expand window by adding the rightmost character
+            if char in target_count:
+                window_count[char] += 1
+                if window_count[char] == target_count[char]:
+                    matched_chars += 1  
+
+            # try shrinking the window if all required characters are present
+            while matched_chars == required_chars:
+                # update min substring if this one is shorter
+                if min_substring == "" or (right - left + 1) < len(min_substring):
+                    min_substring = s[left:right + 1]
+
+                # remove leftmost character and move left pointer
+                if s[left] in window_count:
+                    window_count[s[left]] -= 1
+                    if window_count[s[left]] < target_count[s[left]]:
+                        matched_chars -= 1
+                left += 1  
+
+        return min_substring

pacific-atlantic-water-flow/yolophg.py

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+# Time Complexity: O(m * n) - running DFS from each border, so worst case, we visit each cell twice.
+# Space Complexity: O(m * n) - using two sets to track which cells can reach each ocean and the recursion stack.
+
+class Solution:
+    def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:
+
+        rows = len(heights)
+        cols = len(heights[0])
+        result = []  
+
+        # tracking which cells can reach each ocean
+        pac, atl = set(), set() 
+
+        def dfs(r, c, visited, preHeight):
+            # if out of bounds, already visited, or can't flow from prev height → just dip
+            if (r < 0 or c < 0 or r == rows or c == cols or 
+                (r, c) in visited or heights[r][c] < preHeight):
+                return
+
+            # mark as visited
+            visited.add((r, c))  
+
+            # go in all 4 directions
+            dfs(r + 1, c, visited, heights[r][c])  # down
+            dfs(r - 1, c, visited, heights[r][c])  # up
+            dfs(r, c + 1, visited, heights[r][c])  # right
+            dfs(r, c - 1, visited, heights[r][c])  # left
+
+        # hit up all border cells for both oceans
+        for c in range(cols):
+            dfs(0, c, pac, heights[0][c])  # top row (pacific)
+            dfs(rows - 1, c, atl, heights[rows - 1][c])  # bottom row (atlantic)
+
+        for r in range(rows):
+            dfs(r, 0, pac, heights[r][0])  # leftmost col (pacific)
+            dfs(r, cols - 1, atl, heights[r][cols - 1])  # rightmost col (atlantic)
+
+        # now just check which cells are in both sets
+        for r in range(rows):
+            for c in range(cols):
+                if (r, c) in pac and (r, c) in atl:
+                    result.append([r, c])
+
+        return result