BPS-2-1

EmployeeImportance.py

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+# Time Complexity : O(n)
+# Space Complexity : O(n)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach : Map every employee ID to their object for easy access.
+# Do BFS starting from the given employee, adding their importance.
+# Keep exploring subordinates until all levels are visited and summed up.
+
+"""
+# Definition for Employee.
+class Employee:
+    def __init__(self, id: int, importance: int, subordinates: List[int]):
+        self.id = id
+        self.importance = importance
+        self.subordinates = subordinates
+"""
+
+class Solution:
+    def getImportance(self, employees: List['Employee'], id: int) -> int:
+        id_emp_map = {}
+
+        for employee in employees:
+            id_emp_map[employee.id] = employee
+
+        q = deque()
+        q.append(id)
+        res = 0
+
+        while q:
+            currId = q.popleft() 
+            currEmp = id_emp_map[currId]
+            res += currEmp.importance
+
+            for subId in currEmp.subordinates:
+                q.append(subId)
+
+        return res       
+
+

RottenOranges.py

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+# Time Complexity : O(m*n)
+# Space Complexity : O(m*n)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach : Collect all the rotten oranges and count how many fresh ones are there.
+# Then, every minute, rot the nearby fresh ones using BFS until none are left.
+# If we finish rotting all, return the time; else, return -1 since some can't be reached.
+
+class Solution:
+    def orangesRotting(self, grid: List[List[int]]) -> int:
+        rows, cols, fresh = len(grid), len(grid[0]), 0
+        dirs = [[-1,0],[1,0],[0,1],[0,-1]]
+
+        q = deque()
+        for i in range(rows):
+            for j in range(cols):
+                if grid[i][j] == 2:
+                    q.append((i,j))
+                if grid[i][j] == 1:
+                    fresh += 1
+
+        time = 0
+        if fresh == 0:
+            return time
+
+        while q:
+            size = len(q)
+            time += 1
+            for i in range(size):
+                curr = q.popleft()
+                for dx, dy in dirs:
+                    r = dx + curr[0]
+                    c = dy + curr[1]
+
+                    if r >= 0 and c >= 0 and r < rows and c < cols and grid[r][c] == 1:
+                            grid[r][c] = 2
+                            q.append((r, c))
+                            fresh -= 1
+                            if fresh == 0:
+                                return time
+
+        return -1
+
+
+
+
+