[ayosecu] Week 07 Solutions

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

@@ -0,0 +1,34 @@
+class Solution:
+    """
+        - Time Complexity: O(n), n = len(s)
+        - Space Complexity: O(n)
+    """
+    def lengthOfLongestSubstring(self, s: str) -> int:
+        check_set = set()
+
+        longest_length, length = 0, 0
+        l, r = 0, 0
+        while r < len(s):
+            # check each character (s[r]) is duplicated, and expand or narrow the length
+            if s[r] not in check_set:
+                check_set.add(s[r])
+                length += 1
+                longest_length = max(longest_length, length)
+                r += 1
+            else:
+                check_set.remove(s[l])
+                length -= 1
+                l += 1
+
+        return longest_length
+
+tc = [
+        ("abcabcbb", 3),
+        ("bbbbb", 1),
+        ("pwwkew", 3)
+]
+
+sol = Solution()
+for i, (s, e) in enumerate(tc, 1):
+    r = sol.lengthOfLongestSubstring(s)
+    print(f"TC {i} is Passed!" if r == e else f"TC {i} is Failed! - Expected: {e}, Result: {r}")

number-of-islands/ayosecu.py

@@ -0,0 +1,47 @@
+from typing import List
+
+class Solution:
+    """
+        - Time Complexity: O(N), N = The number of items = len(grid) * len(grid[0])
+        - Space Complexity: O(N)
+            - In worst case (all "1"s), The depth of dfs is N => O(N)
+    """
+    def numIslands(self, grid: List[List[str]]) -> int:
+        m, n = len(grid), len(grid[0])
+
+        def dfs(i, j):
+            if i < 0 or i >= m or j < 0 or j >= n or grid[i][j] != "1":
+                return
+
+            grid[i][j] = "#"    # Visited
+            for dx, dy in [(1, 0), (0, 1), (0, -1), (-1, 0)]:
+                dfs(i + dx, j + dy)               
+
+        count = 0            
+        for i in range(m):
+            for j in range(n):
+                if grid[i][j] == "1":
+                    count += 1
+                    dfs(i, j)
+
+        return count
+
+tc = [
+        ([
+            ["1","1","1","1","0"],
+            ["1","1","0","1","0"],
+            ["1","1","0","0","0"],
+            ["0","0","0","0","0"]
+        ], 1),
+        ([
+            ["1","1","0","0","0"],
+            ["1","1","0","0","0"],
+            ["0","0","1","0","0"],
+            ["0","0","0","1","1"]
+        ], 3)
+]
+
+sol = Solution()
+for i, (grid, e) in enumerate(tc, 1):
+    r = sol.numIslands(grid)
+    print(f"TC {i} is Passed!" if r == e else f"TC {i} is Failed! - Expected: {e}, Result: {r}")

reverse-linked-list/ayosecu.py

@@ -0,0 +1,50 @@
+from typing import Optional
+
+# Definition for singly-linked list.
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+class Solution:
+    """
+        - Time Complexity: O(n), n = The number of nodes.
+        - Space Complexity: O(1)
+    """
+    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        pre = None
+        while head:
+            nxt = head.next
+            head.next = pre
+            pre = head
+            head = nxt
+
+        return pre
+
+def toLinkedList(lists):
+    dummy = ListNode(-1)
+    ptr = dummy
+    for item in lists:
+        ptr.next = ListNode(item)
+        ptr = ptr.next
+    return dummy.next
+
+def toList(head):
+    result = []
+
+    while head:
+        result.append(head.val)
+        head = head.next
+
+    return result
+
+tc = [
+        ([1, 2, 3, 4, 5], [5, 4, 3, 2, 1]),
+        ([1, 2], [2, 1]),
+        ([], [])
+]
+
+sol = Solution()
+for i, (l, e) in enumerate(tc, 1):
+    r = toList(sol.reverseList(toLinkedList(l)))
+    print(f"TC {i} is Passed!" if r == e else f"TC {i} is Failed! - Expected: {e}, Result: {r}")

set-matrix-zeroes/ayosecu.py

@@ -0,0 +1,44 @@
+from typing import List
+
+class Solution:
+    """
+        - Time Complexity: O(mn), m = len(matrix), n = len(matrix[0])
+        - Space Complexity: O(1)
+    """
+    def setZeroes(self, matrix: List[List[int]]) -> None:
+        """
+        Do not return anything, modify matrix in-place instead.
+        """
+        m = len(matrix)
+        n = len(matrix[0])
+
+        # Update row and column with a flag (infinite) if the value is not zero
+        def updateRowCol(i, j):
+            for k in range(m):
+                if matrix[k][j] != 0:
+                    matrix[k][j] = float("inf") 
+            for k in range(n):
+                if matrix[i][k] != 0:
+                    matrix[i][k] = float("inf") 
+
+        # Visit all and update row and column if the value is zero
+        for i in range(m):
+            for j in range(n):
+                if matrix[i][j] == 0:
+                    updateRowCol(i, j)
+
+        # Update flagged data to zero
+        for i in range(m):
+            for j in range(n):
+                if matrix[i][j] == float("inf"):
+                    matrix[i][j] = 0
+
+tc = [
+        ([[1,1,1],[1,0,1],[1,1,1]], [[1,0,1],[0,0,0],[1,0,1]]),
+        ([[0,1,2,0],[3,4,5,2],[1,3,1,5]], [[0,0,0,0],[0,4,5,0],[0,3,1,0]])
+]
+
+sol = Solution()
+for i, (m, e) in enumerate(tc, 1):
+    sol.setZeroes(m)
+    print(f"TC {i} is Passed!" if m == e else f"TC {i} is Failed! - Expected: {e}, Result: {m}")

unique-paths/ayosecu.py

@@ -0,0 +1,24 @@
+class Solution:
+    """
+        - Time Complexity: O(mn)
+        - Space Complexity: O(mn), the "dp" variable
+    """
+    def uniquePaths(self, m: int, n: int) -> int:
+        dp = [ [1] * n for _ in range(m) ]
+
+        # DP Approach
+        for i in range(1, m):
+            for j in range(1, n):
+                dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
+
+        return dp[-1][-1]
+
+tc = [
+        (3, 7, 28),
+        (3, 2, 3)
+]
+
+sol = Solution()
+for i, (m, n, e) in enumerate(tc, 1):
+    r = sol.uniquePaths(m, n)
+    print(f"TC {i} is Passed!" if r == e else f"TC {i} is Failed! - Expected: {e}, Result: {r}")