Complete Graph-3

Exercise_1.py

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+# S30 Problem #216 Optimize Water Distribution in a Village
+#LeetCode #1168 https://leetcode.com/problems/optimize-water-distribution-in-a-village/
+
+# Author : Akaash Trivedi
+# Did this code successfully run on Leetcode :  Yes #
+# Any problem you faced while coding this : No
+
+# union find technique
+# connect the village node to a dummy node 0 with cost of wells as edge
+# add pipe to edges and sort them according to cost
+# use union find to connect each nodes until all nodes are connected and return result
+# TC: O(E log E) Here E = edges
+# SC: O(E + V)
+class Solution:
+    def minCostToSupplyWater(self, n: int, wells: List[int], pipes: List[List[int]]) -> int:
+        edges = []
+        uf = []  # union find
+        for i in range(n+1):
+            uf.append(i)
+        # make a dummy node 0 and draw edges from it to each node and give well cost
+        for i in range(1,n+1):
+            # edge from 0 to, node, cost of well
+            edges.append((0, i, wells[i-1]))
+
+        # add pipes cost to edges
+        edges.extend(pipes)
+        edges.sort(key=lambda x:x[2])
+        print(edges)
+
+        res = 0
+        for edge in edges:
+            x, y = edge[0], edge[1]
+            px = self.find(uf, x)  # parent x
+            py = self.find(uf, y)  # parent y
+
+            if px != py:
+                res += edge[2]
+                uf[py] = px
+
+        return res
+
+    # function to find ulimate parent
+    def find(self, uf, x):
+        # base
+        if x != uf[x]:
+            uf[x] = self.find(uf, uf[x])
+        return uf[x]

Exercise_2.py

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+# S30 Problem #194 find celebrity 
+#LeetCode #277 https://leetcode.com/problems/find-the-celebrity/description/
+
+# Author : Akaash Trivedi
+# Did this code successfully run on Leetcode :  Yes #
+# Any problem you faced while coding this : No
+
+# The knows API is already defined for you.
+# return a bool, whether a knows b
+# def knows(a: int, b: int) -> bool:
+
+# topological sort 
+# indegrees array: (a,b) a knows b
+# increase the indegree of person b and degreees indegree of person a
+# Person whos indegree is n-1, meaning that everyone knows that person => celebrity
+# TC: O(n^2)
+# SC: O(n)
+class Solution:
+    def findCelebrity(self, n: int) -> int:
+        indegree = [0] * n
+
+        for i in range(n):
+            for j in range(n):
+                if i == j: continue
+                if knows(i, j):
+                    # increase the j's indegree and decrease i's
+                    indegree[j] += 1
+                    indegree[i] -= 1
+
+        # anyone has indegree n-1, means everyone knows that person 
+        for i in range(n):
+            if indegree[i] == n - 1:
+                return i
+
+        return -1
+
+
+# linear 3n solution
+# Assume one of them is a celeb, and update the celeb when they knows someone
+# after 1st pass, we found some celeb: check if everyone knows celeb; celeb knows no one
+# return the celeb, if not return -1
+# TC: O(n)
+# SC: O(1)
+
+class Solution:
+    def findCelebrity(self, n: int) -> int:
+        celeb = 0  # assume starting with 0 is celebrity
+        for i in range(1, n):
+            if knows(celeb, i):
+                # celebrity cant know anyone, move the celebrity to i
+                celeb = i
+
+        # celeberity should not know anyone and everyone should know celeb
+        for i in range(n):
+            if i == celeb:continue
+            if not knows(i, celeb) or knows(celeb, i):
+                return -1
+
+        return celeb
+
+
+# linear 3n solution - little faster
+# Assume one of them is a celeb, and update the celeb when they knows someone
+# after 1st pass, we found some celeb: check if everyone knows celeb; celeb knows no one
+# return the celeb, if not return -1
+# TC: O(n)
+# SC: O(1)
+
+class Solution:
+    def findCelebrity(self, n: int) -> int:
+        celeb = 0  # assume starting with 0 is celebrity
+        for i in range(1, n):
+            if knows(celeb, i):
+                # celebrity cant know anyone, move the celebrity to i
+                celeb = i
+
+        # celeberity should not know anyone and everyone should know celeb
+        for i in range(n):
+            if i == celeb:continue
+            if not knows(i, celeb):
+                return -1
+            if i < celeb and knows(celeb, i):
+                return -1
+        return celeb