[Helena] Week 7

longest-substring-without-repeating-characters/yolophg.py

@@ -0,0 +1,24 @@
+# Time Complexity: O(n): both pointers (left and right) traverse the string once.
+# Space Complexity: O(n): worst case, all characters are unique, so the set will store all characters.
+
+class Solution:
+    def lengthOfLongestSubstring(self, s: str) -> int:
+        # to store characters in the current window
+        window_set = set()  
+        # to store the max length of a substring
+        max_length = 0  
+        # left pointer for the sliding window
+        left = 0  
+
+        # iterate through each char in the string using the right pointer
+        for right in range(len(s)):
+            # if the char is already in the window, shrink the window
+            while s[right] in window_set:
+                window_set.remove(s[left])  # remove the leftmost char
+                left += 1  # move the left pointer to the right
+            # add the new char
+            window_set.add(s[right])
+            # update the max length if the current window is longer
+            max_length = max(max_length, right - left + 1)
+
+        return max_length

number-of-islands/yolophg.py

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+# Time Complexity: O(N * M), where N is the number of rows and M is the number of columns.
+# Space Complexity: O(N * M), in the worst case if the grid is entirely land.
+
+class Solution:
+    def numIslands(self, grid: List[List[str]]) -> int:
+        # get the number of rows and columns in the grid
+        rows, cols = len(grid), len(grid[0])
+
+        # define a dfs function to mark visited cells
+        def dfs(i, j):
+            # if out of bounds or the cell is water, stop here
+            if i < 0 or i >= rows or j < 0 or j >= cols or grid[i][j] == "0":
+                return True
+
+            # mark the current cell as visited by changing "1" to "0"
+            grid[i][j] = "0"
+
+            # visit recursively all neighboring cells
+            return True if (dfs(i - 1, j) and dfs(i + 1, j) and dfs(i, j - 1) and dfs(i, j + 1)) else False
+
+        count = 0
+        # loop through the entire grid to find islands
+        for i in range(rows):
+            for j in range(cols):
+                # if we find land ("1"), start a dfs
+                if grid[i][j] == "1":
+                    # increment count if dfs confirms a new island
+                    if dfs(i, j):
+                        count += 1
+        return count

reverse-linked-list/yolophg.py

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+# Time Complexity: O(n)
+# Space Complexity: O(1)
+
+class Solution:
+    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        prev = None
+        # to traverse the original list, starting from head
+        current = head
+        while current:
+            # reverse the link and move to the next node
+            prev, prev.next, current = current, prev, current.next
+
+        # prev is now the head of the reversed list
+        return prev

set-matrix-zeroes/yolophg.py

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+# Time Complexity: O(m * n) - iterate through the matrix multiple times.
+# Space Complexity: O(1) - no extra space is used apart from variables.
+
+class Solution:
+    def setZeroes(self, matrix: List[List[int]]) -> None:
+        m, n = len(matrix), len(matrix[0])
+
+        # check if the first row contains any zero
+        first_row_zero = any(matrix[0][j] == 0 for j in range(n))
+        # check if the first column contains any zero
+        first_col_zero = any(matrix[i][0] == 0 for i in range(m))
+
+        # use the first row and column to mark zero
+        for i in range(1, m):  
+            for j in range(1, n): 
+                if matrix[i][j] == 0:
+                    matrix[i][0] = 0  
+                    matrix[0][j] = 0 
+
+        # update the matrix using the marks from the first row and column
+        for i in range(1, m):
+            for j in range(1, n):
+                if matrix[i][0] == 0 or matrix[0][j] == 0:
+                    matrix[i][j] = 0 
+
+        # handle the first row separately if it initially had any zero
+        if first_row_zero:
+            for j in range(n):
+                matrix[0][j] = 0 
+
+        # handle the first column separately if it initially had any zero
+        if first_col_zero:
+            for i in range(m):
+                matrix[i][0] = 0 

unique-paths/yolophg.py

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+# Time Complexity: O(N * M) : iterate through a grid of size m x n once.
+# Space Complexity: O(N * M) : use a DP table of size (m+1) x (n+1) to store the number of paths.
+
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        # create a dp table and add extra rows and columns for easier indexing
+        table = [[0 for x in range(n+1)] for y in range(m+1)]
+
+        # set the starting point, one way to be at the start
+        table[1][1] = 1
+
+        # iterate the grid to calculate paths
+        for i in range(1, m+1):
+            for j in range(1, n+1):
+                # add paths going down
+                if i+1 <= m:
+                    table[i+1][j] += table[i][j] 
+                # add paths going right
+                if j+1 <= n:
+                    table[i][j+1] += table[i][j]
+
+        return table[m][n]