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Commit 1bb0219

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solves #1584: Min Cost to Connect All Points in java
1 parent 015cc5f commit 1bb0219

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3 files changed

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-40
lines changed

3 files changed

+135
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lines changed

β€ŽREADME.md

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| 1572 | [Matrix Diagonal Sum](https://leetcode.com/problems/matrix-diagonal-sum) | [![Java](assets/java.png)](src/MatrixDiagonalSum.java) | |
620620
| 1576 | [Replace All ?'s to Avoid Consecutive Repeating Characters](https://leetcode.com/problems/replace-all-s-to-avoid-consecutive-repeating-characters) | [![Java](assets/java.png)](src/ReplaceAllToAvoidConsecutiveRepeatingCharacters.java) | |
621621
| 1582 | [Special Positions in a Binary Matrix](https://leetcode.com/problems/special-positions-in-a-binary-matrix) | [![Java](assets/java.png)](src/SpecialPositionInABinaryMatrix.java) | |
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| 1584 | [Min Cost to Connect All Points](https://leetcode.com/problems/min-cost-to-connect-all-points) | [![Java](assets/java.png)](src/MinCostToConnectAllPoints.java) | |
622623
| 1588 | [Sum of All Odd Length Subarrays](https://leetcode.com/problems/sum-of-all-odd-length-subarrays) | [![Java](assets/java.png)](src/SumOfAllOddLengthSubArrays.java) | |
623624
| 1592 | [Rearrange Spaces Between Words](https://leetcode.com/problems/rearrange-spaces-between-words) | [![Java](assets/java.png)](src/RearrangeSpacesBetweenWords.java) | |
624625
| 1598 | [Crawler Log Folder](https://leetcode.com/problems/crawler-log-folder) | [![Java](assets/java.png)](src/CrawlerLogFolder.java) | |

β€Žsrc/HelloWorld.java

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Original file line numberDiff line numberDiff line change
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// S: O(N)
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import java.security.cert.Certificate;
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import java.util.ArrayList;
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import java.util.LinkedList;
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import java.util.List;
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import java.util.Queue;
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public class HelloWorld {
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public static class Node {
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public int val;
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public Node left;
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public Node right;
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public Node next;
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private static final List<List<Integer>> DIRECTIONS = List.of(
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List.of(1, 0),
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List.of(0, 1),
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List.of(-1, 0),
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List.of(0, -1)
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);
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public Node() {}
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public int orangesRotting(int[][] grid) {
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final Queue<int[]> queue = new LinkedList<>();
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addRottenOrangesToQueue(queue, grid);
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int maxTime = 0;
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publicNode(int_val) {
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val = _val;
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}
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while (!queue.isEmpty()) {
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finalint[] info = queue.poll();
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finalintrow = info[0], column = info[1], time = info[2];
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public Node(int _val, Node _left, Node _right, Node _next) {
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val = _val;
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left = _left;
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right = _right;
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next = _next;
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}
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};
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if (grid[row][column] == 0) {
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continue;
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}
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31-
public Node connect(Node root) {
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if (root == null) {
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return null;
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maxTime = Math.max(time, maxTime);
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grid[row][column] = 0;
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for (int[] neighbour : getNeighbours(grid, row, column)) {
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queue.add(new int[] { neighbour[0], neighbour[1], time + 1});
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}
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}
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36-
final Queue<Node> queue = new LinkedList<>();
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queue.add(root);
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queue.add(null);
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Node current = null;
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if (containsFreshOranges(grid)) {
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return -1;
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}
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return maxTime;
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}
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41-
while (!queue.isEmpty()) {
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final Node node = queue.poll();
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if (node == null) {
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if (!queue.isEmpty()) {
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queue.add(null);
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current = null;
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private static boolean containsFreshOranges(int[][] grid) {
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for (int[] row : grid) {
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for (int orange : row) {
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if (orange == 1) {
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return true;
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}
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continue;
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}
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}
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return false;
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}
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51-
addChildrenToQueue(queue, node);
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if (current != null) {
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current.next = node;
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private static void addRottenOrangesToQueue(Queue<int[]> queue, int[][] grid) {
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for(int row = 0 ; row < grid.length ; row++) {
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for (int column = 0 ; column < grid[0].length ; column++) {
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if (grid[row][column] == 2) {
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queue.add(new int[] { row, column, 0 });
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}
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}
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current = node;
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}
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return root;
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}
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62-
private static void addChildrenToQueue(Queue<Node> queue, Node node) {
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if (node.left != null) queue.add(node.left);
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if (node.right != null) queue.add(node.right);
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private static List<int[]> getNeighbours(int[][] grid, int row, int column) {
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final List<int[]> result = new ArrayList<>();
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for (List<Integer> direction : DIRECTIONS) {
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final int r = row + direction.get(0), c = column + direction.get(1);
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if (r >= 0 && r < grid.length && c >= 0 && c < grid[0].length && grid[r][c] == 1) {
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result.add(new int[] {r, c});
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}
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}
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return result;
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}
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}

β€Žsrc/MinCostToConnectAllPoints.java

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// https://leetcode.com/problems/min-cost-to-connect-all-points
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// P = |points|, E = P^2
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// T: O(E logE + E al(P)) al = inverse Ackerman function
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// S: O(P + E) = S(E)
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import java.util.ArrayList;
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import java.util.Comparator;
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import java.util.List;
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public class MinCostToConnectAllPoints {
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private static class DisjointSet {
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private final int[] roots, rank;
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14+
DisjointSet(int size) {
15+
roots = new int[size];
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rank = new int[size];
17+
for (int i = 0 ; i < size ; i++) {
18+
roots[i] = i;
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rank[i] = i;
20+
}
21+
}
22+
23+
public int find(int num) {
24+
if (num == roots[num]) {
25+
return num;
26+
}
27+
return roots[num] = find(roots[num]);
28+
}
29+
30+
public boolean areConnected(int x, int y) {
31+
return find(x) == find(y);
32+
}
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34+
public void union(int x, int y) {
35+
final int rootX = find(x), rootY = find(y);
36+
if (rootX == rootY) {
37+
return;
38+
}
39+
if (rank[rootX] > rank[rootY]) {
40+
roots[rootY] = rootX;
41+
} else if (rank[rootX] < rank[rootY]) {
42+
roots[rootX] = rootY;
43+
} else {
44+
roots[rootY] = rootX;
45+
rank[rootX]++;
46+
}
47+
}
48+
}
49+
50+
// Kruskal's algorithm
51+
public int minCostConnectPoints(int[][] points) {
52+
final List<int[]> edges = createEdges(points);
53+
final DisjointSet disjointSet = new DisjointSet(points.length);
54+
edges.sort(Comparator.comparingInt(a -> a[2]));
55+
56+
int minCost = 0;
57+
58+
for (int[] edge : edges) {
59+
final int from = edge[0], to = edge[1], weight = edge[2];
60+
if (!disjointSet.areConnected(from, to)) {
61+
disjointSet.union(from, to);
62+
minCost += weight;
63+
}
64+
}
65+
66+
return minCost;
67+
}
68+
69+
private static List<int[]> createEdges(int[][] points) {
70+
final List<int[]> edges = new ArrayList<>();
71+
for (int i = 0 ; i < points.length ; i++) {
72+
for (int j = i + 1 ; j < points.length ; j++) {
73+
edges.add(new int[] { i, j, manhattanDistance(points[i], points[j]) });
74+
}
75+
}
76+
return edges;
77+
}
78+
79+
private static int manhattanDistance(int[] p1, int[] p2) {
80+
return Math.abs(p1[0] - p2[0]) + Math.abs(p1[1] - p2[1]);
81+
}
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}

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