From 950abb6b37df8b2b532844df9f01824a70efaf44 Mon Sep 17 00:00:00 2001
From: Sangeeth Sudheer <sangeeth96@users.noreply.github.com>
Date: Wed, 27 Apr 2022 02:47:42 +0530
Subject: [PATCH] min-cost-to-connect-all-points py3

---
 1706_min-cost-to-connect-all-points/README.md | 38 ++++++++++
 .../python3/solution.py                       | 69 +++++++++++++++++++
 2 files changed, 107 insertions(+)
 create mode 100644 1706_min-cost-to-connect-all-points/README.md
 create mode 100644 1706_min-cost-to-connect-all-points/python3/solution.py

diff --git a/1706_min-cost-to-connect-all-points/README.md b/1706_min-cost-to-connect-all-points/README.md
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+---
+source:
+  platform: leetcode
+  problemId: 1584
+tags: graphs, prims-algorithm
+---
+
+You are given an array `points` representing integer coordinates of some points on a 2D-plane, where `points[i] = [xi, yi]`.
+
+The cost of connecting two points `[xi, yi]` and `[xj, yj]` is the **manhattan distance** between them: `|xi - xj| + |yi - yj|`, where `|val|` denotes the absolute value of `val`.
+
+Return _the minimum cost to make all points connected._ All points are connected if there is **exactly one** simple path between any two points.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2020/08/26/d.png)
+
+    Input: points = [[0,0],[2,2],[3,10],[5,2],[7,0]]
+    Output: 20
+    Explanation: 
+    
+    We can connect the points as shown above to get the minimum cost of 20.
+    Notice that there is a unique path between every pair of points.
+    
+
+**Example 2:**
+
+    Input: points = [[3,12],[-2,5],[-4,1]]
+    Output: 18
+    
+
+**Constraints:**
+
+*   `1 <= points.length <= 1000`
+*   `-106 <= xi, yi <= 106`
+*   All pairs `(xi, yi)` are distinct.
+
+https://leetcode.com/problems/min-cost-to-connect-all-points/
diff --git a/1706_min-cost-to-connect-all-points/python3/solution.py b/1706_min-cost-to-connect-all-points/python3/solution.py
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+# Time: O(N^2·logN)
+# Space: O(N^2) ; we could be pushing N·N(N - 1) / 2 edges into the heap
+from collections import defaultdict
+from heapq import heappush, heappop
+
+class Solution:
+    def minCostConnectPoints(self, points: List[List[int]]) -> int:
+        '''
+        BFS + Prim's algorithm
+        '''
+        N = len(points)
+        neighbors = defaultdict(list)
+        
+        # Build adjacency lists by connecting each point to every other
+        # point since we are building a new graph here
+        for i in range(N):
+            x1, y1 = points[i]
+            
+            for j in range(i + 1, N):
+                x2, y2 = points[j]
+                
+                # We are considering the Manhattan Distance as stated in
+                # the problem
+                distance = abs(x1 - x2) + abs(y1 - y2)
+                
+                # We are going to add both the neighbor of given point as
+                # well as the distance so that we can later use it to create
+                # Prim's min heap
+                neighbors[i].append((distance, j))
+                neighbors[j].append((distance, i))
+        
+        
+        # Perform Prim's algorithm
+        
+        # To avoid cycles
+        visited = set()
+        
+        # We start at the first point, distance would be 0 since it's the first
+        # point
+        min_heap = [(0, 0)]
+        
+        # We need to find distances from one node to all other nodes (as long as
+        # its not already visited), keep track of it in the min heap and once
+        # all neighbors of a node is visited, we should have found the next node
+        # to pop from the heap since it'd have the least distance.
+        #
+        # We need to keep doing this until we have visited all the nodes
+        total = 0
+        
+        while len(visited) < N:
+            curr_dist, min_point = heappop(min_heap)
+            
+            # It's possible we've already visited the node (let's say the very first node we
+            # initialized the heap with) since it could end up being added to the heap again
+            # while visiting other nodes. If this is the case, we skip this iteration of the
+            # loop
+            if min_point in visited:
+                continue
+            
+            total += curr_dist
+            visited.add(min_point)
+            
+            for neighbor in neighbors[min_point]:
+                if neighbor[1] not in visited:
+                    heappush(min_heap, neighbor)
+        
+        return total
+        
+        
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