Complete the function: use Java and add comments to the program static int[] cou
ID: 3736784 • Letter: C
Question
Complete the function: use Java and add comments to the program
static int[] countGroups(int[][] m, int[] t)
{
}
For this question, you will be tinding groups ot adjacent cells in a two-dimensional array and determining how many groups of each specified size exist. As part of the input, you will be provided with a two-dimensional array where each cell has a value of either 0 or 1. Cells are considered to be in the same group if they both have a value of 1 and they are adjacent an adjacent cell is directly to the left, to the nght, up or down; diagonal is not considered Qdjacent lo define it strictly: any two cells M y) and 2 y2 fall into the same group f X, X2 y, y 7 and both cells have a value of 1 The group-size of a group is the number of cells in tht roup. Complete the countGroups function in your editor. It has 2 parameters: 1. An n x n two-dimensional array ot integers, m, where the value of each element miy (where jExplanation / Answer
CODE:
class FindGroups
{
static boolean is_visited[][];
static int count = 0;
public static void main(String args[])
{
int row, col, no_of_t;
int m[][], t[];
Scanner in = new Scanner(System.in);
//Get no of rows
System.out.println("Enter no of rows : ");
row = in.nextInt();
System.out.println("Enter no of cols : ");
col = in.nextInt();
m = new int[row][col];
is_visited = new boolean[row][col];
System.out.println("Enter elements : ");
for(int i=0; i<row; i++) {
for(int j=0; j<col; j++) {
m[i][j] = in.nextInt();
is_visited[i][j] = false;
}
}
System.out.println("Enter no of t : ");
no_of_t = in.nextInt();
t = new int[no_of_t];
System.out.println("Enter no of groups : ");
for(int i=0; i<no_of_t; i++) {
t[i] = in.nextInt();
}
System.out.println("-------------------------");
System.out.println("Input :: ");
for(int i=0; i<row; i++) {
for(int j=0; j<col; j++) {
System.out.print(m[i][j] + " ");
}
System.out.println("");
}
int[] count_groups = countGroups(m, t);
System.out.println(" -------------------------");
System.out.println("Output :: ");
for(int k=0; k<no_of_t; k++) {
System.out.println(t[k] + " : " + count_groups[k]);
}
}
static void isConnected( int numb, int m[][], int y, int x, boolean is_visited[][]) {
if(y != 0) {
if ( m[y-1][x] == numb && !is_visited[y-1][x]) {
is_visited[y-1][x] = true;
count++;
isConnected( numb, m, y-1, x, is_visited);
}
}
if(y < (m.length - 1)) {
if ( m[y+1][x] == numb && !is_visited[y+1][x]) {
is_visited[y+1][x] = true;
count++;
isConnected( numb, m, y+1, x, is_visited);
}
}
if(x != 0) {
if ( m[y][x-1] == numb && !is_visited[y][x-1]) {
is_visited[y][x-1] = true;
count++;
isConnected( numb, m, y, x-1, is_visited);
}
}
if(x < (m.length - 1)) {
if ( m[y][x+1] == numb && !is_visited[y][x+1]) {
is_visited[y][x+1] = true;
count++;
isConnected( numb, m, y, x+1, is_visited);
}
}
}
static int[] countGroups(int[][] m, int[] t) {
int res_len = t.length;
int ret[] = new int[res_len];
for(int i=0; i<m.length; i++) {
for(int j=0; j<m[0].length; j++) {
if(!is_visited[i][j]) {
count = 0;
isConnected( m[i][j], m, i, j, is_visited);
if(count == 0)
count = 1;
for(int k=0; k<res_len; k++) {
if(t[k] == count)
ret[k]++;
}
}
}
}
return ret;
}
}
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