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leetcode_java/src/main/java/LeetCodeJava/BFS/ShortestBridge.java
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| package LeetCodeJava.BFS; | ||
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| // https://leetcode.com/problems/shortest-bridge/description/ | ||
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| import java.util.ArrayList; | ||
| import java.util.List; | ||
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| /** | ||
| * 934. Shortest Bridge | ||
| * Medium | ||
| * Topics | ||
| * Companies | ||
| * You are given an n x n binary matrix grid where 1 represents land and 0 represents water. | ||
| * | ||
| * An island is a 4-directionally connected group of 1's not connected to any other 1's. There are exactly two islands in grid. | ||
| * | ||
| * You may change 0's to 1's to connect the two islands to form one island. | ||
| * | ||
| * Return the smallest number of 0's you must flip to connect the two islands. | ||
| * | ||
| * | ||
| * | ||
| * Example 1: | ||
| * | ||
| * Input: grid = [[0,1],[1,0]] | ||
| * Output: 1 | ||
| * Example 2: | ||
| * | ||
| * Input: grid = [[0,1,0],[0,0,0],[0,0,1]] | ||
| * Output: 2 | ||
| * Example 3: | ||
| * | ||
| * Input: grid = [[1,1,1,1,1],[1,0,0,0,1],[1,0,1,0,1],[1,0,0,0,1],[1,1,1,1,1]] | ||
| * Output: 1 | ||
| * | ||
| * | ||
| * Constraints: | ||
| * | ||
| * n == grid.length == grid[i].length | ||
| * 2 <= n <= 100 | ||
| * grid[i][j] is either 0 or 1. | ||
| * There are exactly two islands in grid. | ||
| */ | ||
| public class ShortestBridge { | ||
| // V0 | ||
| // public int shortestBridge(int[][] grid) { | ||
| // | ||
| // } | ||
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| // V1-1 | ||
| // https://leetcode.com/problems/shortest-bridge/editorial/ | ||
| // IDEA: Depth-First-Search + Breadth-First-Search | ||
| private List<int[]> bfsQueue; | ||
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| // Recursively check the neighboring land cell of current cell grid[x][y] and | ||
| // add all | ||
| // land cells of island A to bfsQueue. | ||
| private void dfs(int[][] grid, int x, int y, int n) { | ||
| grid[x][y] = 2; | ||
| bfsQueue.add(new int[] { x, y }); | ||
| for (int[] pair : new int[][] { { x + 1, y }, { x - 1, y }, { x, y + 1 }, { x, y - 1 } }) { | ||
| int curX = pair[0], curY = pair[1]; | ||
| if (0 <= curX && curX < n && 0 <= curY && curY < n && grid[curX][curY] == 1) { | ||
| dfs(grid, curX, curY, n); | ||
| } | ||
| } | ||
| } | ||
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| // Find any land cell, and we treat it as a cell of island A. | ||
| public int shortestBridge_1_1(int[][] grid) { | ||
| int n = grid.length; | ||
| int firstX = -1, firstY = -1; | ||
| for (int i = 0; i < n; i++) { | ||
| for (int j = 0; j < n; j++) { | ||
| if (grid[i][j] == 1) { | ||
| firstX = i; | ||
| firstY = j; | ||
| break; | ||
| } | ||
| } | ||
| } | ||
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| // Add all land cells of island A to bfsQueue. | ||
| bfsQueue = new ArrayList<>(); | ||
| dfs(grid, firstX, firstY, n); | ||
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| int distance = 0; | ||
| while (!bfsQueue.isEmpty()) { | ||
| List<int[]> newBfs = new ArrayList<>(); | ||
| for (int[] pair : bfsQueue) { | ||
| int x = pair[0], y = pair[1]; | ||
| for (int[] nextPair : new int[][] { { x + 1, y }, { x - 1, y }, { x, y + 1 }, { x, y - 1 } }) { | ||
| int curX = nextPair[0], curY = nextPair[1]; | ||
| if (0 <= curX && curX < n && 0 <= curY && curY < n) { | ||
| if (grid[curX][curY] == 1) { | ||
| return distance; | ||
| } else if (grid[curX][curY] == 0) { | ||
| newBfs.add(nextPair); | ||
| grid[curX][curY] = -1; | ||
| } | ||
| } | ||
| } | ||
| } | ||
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| // Once we finish one round without finding land cells of island B, we will | ||
| // start the next round on all water cells that are 1 cell further away from | ||
| // island A and increment the distance by 1. | ||
| bfsQueue = newBfs; | ||
| distance++; | ||
| } | ||
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| return distance; | ||
| } | ||
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| // V1-2 | ||
| // https://leetcode.com/problems/shortest-bridge/editorial/ | ||
| // IDEA: BFS | ||
| public int shortestBridge_1_2(int[][] grid) { | ||
| int n = grid.length; | ||
| int firstX = -1, firstY = -1; | ||
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| // Find any land cell, and we treat it as a cell of island A. | ||
| for (int i = 0; i < n; i++) { | ||
| for (int j = 0; j < n; j++) { | ||
| if (grid[i][j] == 1) { | ||
| firstX = i; | ||
| firstY = j; | ||
| break; | ||
| } | ||
| } | ||
| } | ||
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| // bfsQueue for BFS on land cells of island A; secondBfsQueue for BFS on water | ||
| // cells. | ||
| List<int[]> bfsQueue = new ArrayList<>(); | ||
| List<int[]> secondBfsQueue = new ArrayList<>(); | ||
| bfsQueue.add(new int[] { firstX, firstY }); | ||
| secondBfsQueue.add(new int[] { firstX, firstY }); | ||
| grid[firstX][firstY] = 2; | ||
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| // BFS for all land cells of island A and add them to secondBfsQueue. | ||
| while (!bfsQueue.isEmpty()) { | ||
| List<int[]> newBfs = new ArrayList<>(); | ||
| for (int[] cell : bfsQueue) { | ||
| int x = cell[0]; | ||
| int y = cell[1]; | ||
| for (int[] next : new int[][] { { x + 1, y }, { x - 1, y }, { x, y + 1 }, { x, y - 1 } }) { | ||
| int curX = next[0]; | ||
| int curY = next[1]; | ||
| if (curX >= 0 && curX < n && curY >= 0 && curY < n && grid[curX][curY] == 1) { | ||
| newBfs.add(new int[] { curX, curY }); | ||
| secondBfsQueue.add(new int[] { curX, curY }); | ||
| grid[curX][curY] = 2; | ||
| } | ||
| } | ||
| } | ||
| bfsQueue = newBfs; | ||
| } | ||
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| int distance = 0; | ||
| while (!secondBfsQueue.isEmpty()) { | ||
| List<int[]> newBfs = new ArrayList<>(); | ||
| for (int[] cell : secondBfsQueue) { | ||
| int x = cell[0]; | ||
| int y = cell[1]; | ||
| for (int[] next : new int[][] { { x + 1, y }, { x - 1, y }, { x, y + 1 }, { x, y - 1 } }) { | ||
| int curX = next[0]; | ||
| int curY = next[1]; | ||
| if (curX >= 0 && curX < n && curY >= 0 && curY < n) { | ||
| if (grid[curX][curY] == 1) { | ||
| return distance; | ||
| } else if (grid[curX][curY] == 0) { | ||
| newBfs.add(new int[] { curX, curY }); | ||
| grid[curX][curY] = -1; | ||
| } | ||
| } | ||
| } | ||
| } | ||
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| // Once we finish one round without finding land cells of island B, we will | ||
| // start the next round on all water cells that are 1 cell further away from | ||
| // island A and increment the distance by 1. | ||
| secondBfsQueue = newBfs; | ||
| distance++; | ||
| } | ||
| return distance; | ||
| } | ||
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| // V2 | ||
| } |
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