update 200 with more comments

fishercoder1534 · fishercoder1534 · commit c7d4f399eb43 · 2024-06-09T16:54:49.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_200.java b/src/main/java/com/fishercoder/solutions/_200.java
@@ -52,9 +52,10 @@ public UnionFind(char[][] grid) {
                 for (int i = 0; i < m; i++) {
                     for (int j = 0; j < n; j++) {
                         if (grid[i][j] == '1') {
+                            //at initialization, we count each '1' as one island, later, during traversal, we'll union them during which we'll dedup the number of islands.
                             count++;
-                            ids[i * n + j] = i * n + j;
                         }
+                        ids[i * n + j] = i * n + j;
                     }
                 }
             }
@@ -63,9 +64,13 @@ public void union(int i, int j) {
                 int x = find(i);
                 int y = find(j);
                 if (x != y) {
-                    /**note: this is when x != y, only in this case, we should union these two nodes, which makes sense naturally.*/
-                    count--;
+                    /**
+                     * This means when these two nodes should be unioned, however, so far,
+                     * they have not, i.e. they have different ids,
+                     * so we'll have to unify them by assigning one's ids to the other, or vice versa.
+                     * */
                     ids[x] = y;//ids[y] = x; //also works
+                    count--;//since now these two islands are unified/merged, we'll decrement the count by one
                 }
             }
 

Original file line number	Diff line number	Diff line change
`@@ -52,9 +52,10 @@ public UnionFind(char[][] grid) {`
`52`	`52`	`for (int i = 0; i < m; i++) {`
`53`	`53`	`for (int j = 0; j < n; j++) {`
`54`	`54`	`if (grid[i][j] == '1') {`
	`55`	`+ //at initialization, we count each '1' as one island, later, during traversal, we'll union them during which we'll dedup the number of islands.`
`55`	`56`	`count++;`
`56`		`- ids[i * n + j] = i * n + j;`
`57`	`57`	`}`
	`58`	`+ ids[i * n + j] = i * n + j;`
`58`	`59`	`}`
`59`	`60`	`}`
`60`	`61`	`}`
`@@ -63,9 +64,13 @@ public void union(int i, int j) {`
`63`	`64`	`int x = find(i);`
`64`	`65`	`int y = find(j);`
`65`	`66`	`if (x != y) {`
`66`		`- /*note: this is when x != y, only in this case, we should union these two nodes, which makes sense naturally./`
`67`		`- count--;`
	`67`	`+ /**`
	`68`	`+ * This means when these two nodes should be unioned, however, so far,`
	`69`	`+ * they have not, i.e. they have different ids,`
	`70`	`+ * so we'll have to unify them by assigning one's ids to the other, or vice versa.`
	`71`	`+ * */`
`68`	`72`	`ids[x] = y;//ids[y] = x; //also works`
	`73`	`+ count--;//since now these two islands are unified/merged, we'll decrement the count by one`
`69`	`74`	`}`
`70`	`75`	`}`
`71`	`76`