Keeping track of parents and unioning only when 1 is present

Author: Arjiit

UnionFind/numIslands.java

@@ -0,0 +1,73 @@
+class Solution {
+    /*
+    This problem reduces to number of connected componenets with 1's as the components.
+    */
+    int[][] distance = {{1,0}, {-1,0}, {0,1}, {0,-1}};
+    public int numIslands(char[][] grid) {
+        
+        if (grid == null || grid.length == 0 || grid[0].length == 0) {
+            return 0;
+        }
+        
+        UnionFind uf = new UnionFind(grid);
+        int row = grid.length;
+        int col = grid[0].length;
+        
+        for (int i=0; i<row; i++) {
+            for (int j=0; j< col; j++) {
+                if (grid[i][j] == '1') {
+                    for (int[] d: distance) {
+                        int x = i + d[0];
+                        int y = j + d[1];
+                        if (x >= 0 && x < row && y >= 0 && y < col && grid[x][y] == '1') {
+                            int id1 = i*col + j;
+                            int id2 = x*col + y;
+                            uf.union(id1, id2);
+                        }
+                    }
+                }
+            }
+        }
+        
+        
+        return uf.count;
+    }
+}
+
+class UnionFind{
+    int parent[];
+    int m, n;
+    int count;
+    public UnionFind(char[][] grid) {
+        m = grid.length;
+        n = grid[0].length;
+        parent = new int[m*n];
+        
+        for (int i=0; i<m; i++) {
+            for (int j=0; j <n; j++) {
+                if (grid[i][j] == '1') {
+                    int id = i*n + j; //this forms the 'id' in sequential format only for cells that are 1
+                    parent[id] = id;
+                    count++;
+                    
+                }
+            }
+        }
+    }
+    
+    public int find(int p) {
+        while(parent[p] != p) {
+            p = parent[p];
+        }
+        return p;
+    }
+    
+    public void union(int p, int q) {
+        int pParent = find(p);
+        int qParent = find(q);
+        if (pParent != qParent) {
+            parent[qParent] = pParent;
+            count--;
+        }
+    }
+}