What is the time complexity of Min Cost to Connect All Points?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Min Cost to Connect All Points?

This problem uses Array, Union Find, Graph, Minimum Spanning Tree. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Min Cost to Connect All Points?

Insight 1: The problem is essentially asking us to find the Minimum Spanning Tree (MST) of a complete graph where the vertices are the given points and the edges are weighted by the Manhattan distance between the corresponding points. Insight 2: Prim's algorithm is a suitable algorithm for finding the MST, especially since the graph is implicitly complete (every point is connected to every other point). We can adapt Prim's algorithm to work with the given point data without explicitly constructing the graph.

Which companies ask LeetCode 1584. Min Cost to Connect All Points in interviews?

This problem is frequently asked at DE Shaw, Directi, Nutanix.

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Min Cost to Connect All Points - LeetCode 1584 Solution

Min Cost to Connect All Points - Complete Solution Guide

Min Cost to Connect All Points is LeetCode problem 1584, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an array points representing integer coordinates of some points on a 2D-plane, where points[i] = [x i , y i ] . The cost of connecting two points [x i , y i ] and [x j , y j ] is the manhattan distance between them: |x i - x j | + |y i - y j | , 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: Input: points = [[0,0],[2,2],[3,10],[5,2],

Detailed Explanation

The problem asks us to find the minimum cost to connect all points in a 2D plane. The input is an array of points, where each point is represented by its (x, y) coordinates. The cost of connecting two points is defined as the Manhattan distance between them, which is the sum of the absolute differences of their x and y coordinates: |x1 - x2| + |y1 - y2|. The goal is to find the minimum total cost such that there's exactly one simple path between any two points, meaning we need to find a Minimum Spanning Tree (MST).

Solution Approach

The provided code implements Prim's algorithm to find the MST. Instead of explicitly creating an adjacency list or matrix for the graph, it implicitly considers all possible edges between points based on their Manhattan distance. It maintains a `min_cost` array to store the minimum cost to connect each point to the MST being constructed and an `in_mst` array to indicate whether a point is already included in the MST. The algorithm iteratively adds the point with the minimum connection cost to the MST until all points are included.

Step-by-Step Algorithm

Step 1: Initialize `min_cost` array with infinity for all points and `in_mst` array with `False` for all points. Set the `min_cost` of the starting point (point 0) to 0.
Step 2: Initialize `total_cost` to 0 and `mst_size` to 0.
Step 3: While the `mst_size` is less than the total number of points `n`:
Step 4: Find the vertex `u` that is not yet in the MST (`in_mst[u] == False`) and has the minimum `min_cost[u]`.
Step 5: Add the cost to connect `u` to the MST to `total_cost` and mark `u` as being in the MST (`in_mst[u] = True`). Increment `mst_size`.
Step 6: For each vertex `v` that is not yet in the MST:
Step 7: Calculate the Manhattan distance between `u` and `v`.
Step 8: If the calculated distance is less than the current `min_cost[v]`, update `min_cost[v]` with the new distance.
Step 9: Return `total_cost`.

Key Insights

Insight 1: The problem is essentially asking us to find the Minimum Spanning Tree (MST) of a complete graph where the vertices are the given points and the edges are weighted by the Manhattan distance between the corresponding points.
Insight 2: Prim's algorithm is a suitable algorithm for finding the MST, especially since the graph is implicitly complete (every point is connected to every other point). We can adapt Prim's algorithm to work with the given point data without explicitly constructing the graph.
Insight 3: We can optimize the algorithm by using a `min_cost` array to keep track of the minimum distance to connect each point to the growing MST. This avoids redundant distance calculations.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Union Find, Graph, Minimum Spanning Tree.

Companies

Asked at: DE Shaw, Directi, Nutanix.

Min Cost to Connect All Points | LeetCode 1584

Medium

Array Union Find Graph Minimum Spanning Tree

Time: O(n^2)Space: O(n)

Solution Code

Available in:

class Solution {
    public int minCostConnectPoints(int[][] points) {
        int n = points.length;
        if (n <= 1) {
            return 0;
        }

        // min_cost[i] will store the minimum cost to connect point i to the MST being built.
        int[] min_cost = new int[n];
        for (int i = 0; i < n; i++) {
            min_cost[i] = Integer.MAX_VALUE;
        }

        // in_mst[i] is true if point i is already included in the MST.
        boolean[] in_mst = new boolean[n];

        // Start with the first point. The cost to connect it to itself is 0.
        min_cost[0] = 0;

        int total_cost = 0;
        int mst_size = 0;

        while (mst_size < n) {
            // Find the vertex u not yet in MST with the minimum connection cost.
            int min_edge_cost = Integer.MAX_VALUE;
            int u = -1;
            for (int i = 0; i < n; i++) {
                if (!in_mst[i] && min_cost[i] < min_edge_cost) {
                    min_edge_cost = min_cost[i];
                    u = i;
                }
            }

            // Add the found vertex to the MST.
            total_cost += min_edge_cost;
            in_mst[u] = true;
            mst_size += 1;

            // Update the min_cost for all adjacent vertices of u.
            // In this problem, all points are adjacent to all other points.
            int p1_x = points[u][0];
            int p1_y = points[u][1];
            for (int v = 0; v < n; v++) {
                if (!in_mst[v]) {
                    int p2_x = points[v][0];
                    int p2_y = points[v][1];
                    int distance = Math.abs(p1_x - p2_x) + Math.abs(p1_y - p2_y);
                    // If we found a cheaper way to connect v, update its cost.
                    if (distance < min_cost[v]) {
                        min_cost[v] = distance;
                    }
                }
            }
        }

        return total_cost;
    }
}

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Problem Info

Description

Problem 1584: Min Cost to Connect All Points - Detailed explanation and solution approaches.

Topics

Array Union Find Graph Minimum Spanning Tree

Companies

DE ShawDirectiNutanix

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:1584

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Initialize `min_cost` array with infinity for all points and `in_mst` array with `False` for all points. Set the `min_cost` of the starting point (point 0) to 0.
2
Step 2: Initialize `total_cost` to 0 and `mst_size` to 0.
3
Step 3: While the `mst_size` is less than the total number of points `n`:
4
Step 4: Find the vertex `u` that is not yet in the MST (`in_mst[u] == False`) and has the minimum `min_cost[u]`.
5
Step 5: Add the cost to connect `u` to the MST to `total_cost` and mark `u` as being in the MST (`in_mst[u] = True`). Increment `mst_size`.
6
Step 6: For each vertex `v` that is not yet in the MST:
7
Step 7: Calculate the Manhattan distance between `u` and `v`.
8
Step 8: If the calculated distance is less than the current `min_cost[v]`, update `min_cost[v]` with the new distance.
9
Step 9: Return `total_cost`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The outer loop of Prim's algorithm runs `n` times (once for each point). Inside the outer loop, we iterate through all the remaining points to find the minimum cost to connect to the MST (O(n)). Also, within the outer loop is another loop iterating over all the vertices to update their minimum costs, which is another O(n) operation. Therefore, the total time complexity is O(n * n) = O(n^2). The `min_cost` and `in_mst` arrays both require O(n) space, leading to a total space complexity of O(n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Kruskal's Algorithm

Kruskal's algorithm sorts all edges by weight and then iteratively adds the smallest edges to the MST, as long as they don't create a cycle. Requires Union-Find data structure to detect cycles efficiently.

Time: O(E log E) where E is the number of edges. In this case, E = n(n-1)/2, so O(n^2 log n).Space: O(n^2) to store all edges or O(n) if edges are generated on the fly + O(n) for Union-Find.

Trade-offs: Kruskal's Algorithm might be slightly faster than Prim's algorithm for sparse graphs. However, for this problem, the graph is implicitly complete, making Prim's algorithm generally more efficient due to its avoidance of the sorting step needed in Kruskal's.

✨

Key Insights

1Insight 1: The problem is essentially asking us to find the Minimum Spanning Tree (MST) of a complete graph where the vertices are the given points and the edges are weighted by the Manhattan distance between the corresponding points.
2Insight 2: Prim's algorithm is a suitable algorithm for finding the MST, especially since the graph is implicitly complete (every point is connected to every other point). We can adapt Prim's algorithm to work with the given point data without explicitly constructing the graph.
3Insight 3: We can optimize the algorithm by using a `min_cost` array to keep track of the minimum distance to connect each point to the growing MST. This avoids redundant distance calculations.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: `points` array has only one point. The algorithm correctly returns 0 because there's nothing to connect.
Edge case 2: All points are located at the same location. The Manhattan distance between all points will be 0, and the MST cost will correctly be computed as 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to initialize the `min_cost` array with infinity (or a sufficiently large value) can lead to incorrect results, as it might inadvertently consider disconnected points with costs of 0.

✗Common mistake 2: Not correctly implementing the Manhattan distance calculation can also lead to incorrect MST cost. It is important to use absolute values.

✗Common mistake 3: Premature optimization. Attempting to reduce the n^2 complexity could lead to very complicated code. The n^2 complexity is usually acceptable given the constraints of the problem.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Network design: Finding the minimum cost to connect a set of computers or servers in a network, where the cost is the length of cable or the latency between them.

🏦Real-world scenario 2: Transportation: Finding the minimum cost to build roads or railways connecting a set of cities, where the cost is the distance or construction cost.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 23, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Min Cost to Connect All Points"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether union find techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?