巴勃罗(Pablo)的方形巧克力盒大小为nxn，最初以“ H”表示存在各种健康巧克力，但他发现某些巧克力变质且不健康，以“ U”表示。在一天之内，烂巧克力会使所有邻近的巧克力变得不健康。这种情况一直持续到巧克力盒中的所有巧克力变得不健康为止。找出整个巧克力盒变得不健康的天数。
(注意：保证至少其中一种巧克力不健康)
例子：
Input : n = 3
H H H
H U H
H H H
Output : 1
Only 1 day is required to turn all the
chocolates unhealthy in the chocolate box.
Input : n = 4
H H H U
H H H H
H U H H
H H H H
Output : 2
Explanation:
In first day chocolate at (0, 0), (0, 1),
(2, 3), (3, 3) will remain healthy and in
the second day all the chocolates will
become unhealthy.
在亚马逊，Accolite，Arcesium中问。
蛮力法：
初始化一个标志=1。使用一个while循环，在其中搜索H(如果我们无法在2D字符数组中找到H，则搜索需要O(n ^ 2)时间复杂度)，在此期间停止递增天数计数器和将标志设置为0可以打破循环。
下面是上述方法的实现：
C++
// CPP program to find number of days before
// all chocolates become unhealthy.
#include
using namespace std;
// Validates out of bounds indexing
bool isValid(int i, int j, int n)
{
if (i < 0 || j < 0 || i >= n || j >= n)
return false;
return true;
}
// function for returning number of days
int numdays(char arr[][4], int n)
{
int numdays = 0;
while (true)
{
// Traverse matrix to look for unhealthy
// chocolates and mark their neighbors.
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (arr[i][j] == 'U')
{
if (isValid(i - 1, j - 1, n) &&
arr[i - 1][j - 1] == 'H')
arr[i - 1][j - 1] = 'V';
if (isValid(i - 1, j, n) &&
arr[i - 1][j] == 'H')
arr[i - 1][j] = 'V';
if (isValid(i - 1, j + 1, n) &&
arr[i - 1][j + 1] == 'H')
arr[i - 1][j + 1] = 'V';
if (isValid(i, j - 1, n) &&
arr[i][j - 1] == 'H')
arr[i][j - 1] = 'V';
if (isValid(i, j + 1, n) &&
arr[i][j + 1] == 'H')
arr[i][j + 1] = 'V';
if (isValid(i + 1, j - 1, n) &&
arr[i + 1][j - 1] == 'H')
arr[i + 1][j - 1] = 'V';
if (isValid(i + 1, j, n) &&
arr[i + 1][j] == 'H')
arr[i + 1][j] = 'V';
if (isValid(i + 1, j + 1, n) &&
arr[i + 1][j + 1] == 'H')
arr[i + 1][j + 1] = 'V';
}
/*Here we are assigning the neighbours of U
with the character V because we don't want
these neighbours to be counted in that
particular day. If we do not do so, in the
next iteration that neighbour will also get
counted which was supposed to be counted in
the next day. */
}
}
// Mark chocolates unhealthy which are made
// unhealthy in current day.
bool Hflag = false;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (arr[i][j] == 'V')
{
arr[i][j] = 'U';
Hflag = true;
}
}
}
// Check if there was any chocoloate
// marked unhealthy in current day
if (Hflag)
numdays++;
else
break;
}
return numdays;
}
// Driver function
int main()
{
int n = 4;
char arr[4][4] = { 'H', 'H', 'H', 'U',
'H', 'H', 'H', 'H',
'H', 'U', 'H', 'H',
'H', 'H', 'H', 'H'
};
int ans = numdays(arr, n);
cout << "number of days taken : "
<< ans << "\n";
return 0;
} C
// C program to find number of days before
// all chocolates become unhealthy.
#include
// Validates out of bounds indexing
int isValid(int i, int j, int n)
{
if (i < 0 || j < 0 || i >= n || j >= n)
return 0;
return 1;
}
// function for returning number of days
int numdays(char arr[][4], int n)
{
int numdays = 0;
while (1)
{
// Traverse matrix to look for unhealthy
// chocolates and mark their neighbors.
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (arr[i][j] == 'U')
{
if (isValid(i - 1, j - 1, n) &&
arr[i - 1][j - 1] == 'H')
arr[i - 1][j - 1] = 'V';
if (isValid(i - 1, j, n) &&
arr[i - 1][j] == 'H')
arr[i - 1][j] = 'V';
if (isValid(i - 1, j + 1, n) &&
arr[i - 1][j + 1] == 'H')
arr[i - 1][j + 1] = 'V';
if (isValid(i, j - 1, n) &&
arr[i][j - 1] == 'H')
arr[i][j - 1] = 'V';
if (isValid(i, j + 1, n) &&
arr[i][j + 1] == 'H')
arr[i][j + 1] = 'V';
if (isValid(i + 1, j - 1, n) &&
arr[i + 1][j - 1] == 'H')
arr[i + 1][j - 1] = 'V';
if (isValid(i + 1, j, n) &&
arr[i + 1][j] == 'H')
arr[i + 1][j] = 'V';
if (isValid(i + 1, j + 1, n) &&
arr[i + 1][j + 1] == 'H')
arr[i + 1][j + 1] = 'V';
}
/*Here we are assigning the neighbours of U
with the character V because we don't want
these neighbours to be counted in that
particular day. If we do not do so, in the
next iteration that neighbour will also get
counted which was supposed to be counted in
the next day. */
}
}
// Mark chocolates unhealthy which are made
// unhealthy in current day.
int Hflag = 0;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (arr[i][j] == 'V')
{
arr[i][j] = 'U';
Hflag = 1;
}
}
}
// Check if there was any chocoloate
// marked unhealthy in current day
if (Hflag)
numdays++;
else
break;
}
return numdays;
}
// Driver Code
int main()
{
int n = 4;
char arr[4][4] = { 'H', 'H', 'H', 'U',
'H', 'H', 'H', 'H',
'H', 'U', 'H', 'H',
'H', 'H', 'H', 'H' };
int ans = numdays(arr, n);
printf("number of days taken : %d\n", ans);
return 0;
}
// This code is contributed by ankush_953 Java
// Java program to find number of days before
// all chocolates become unhealthy.
class GFG
{
// Validates out of bounds indexing
static boolean isValid(int i, int j, int n)
{
if (i < 0 || j < 0 || i >= n || j >= n)
return false;
return true;
}
// function for returning number of days
static int numdays(char [][]arr, int n)
{
int numdays = 0;
while (true)
{
// Traverse matrix to look for unhealthy
// chocolates and mark their neighbors.
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (arr[i][j] == 'U')
{
if (isValid(i - 1, j - 1, n) &&
arr[i - 1][j - 1] == 'H')
arr[i - 1][j - 1] = 'V';
if (isValid(i - 1, j, n) &&
arr[i - 1][j] == 'H')
arr[i - 1][j] = 'V';
if (isValid(i - 1, j + 1, n) &&
arr[i - 1][j + 1] == 'H')
arr[i - 1][j + 1] = 'V';
if (isValid(i, j - 1, n) &&
arr[i][j - 1] == 'H')
arr[i][j - 1] = 'V';
if (isValid(i, j + 1, n) &&
arr[i][j + 1] == 'H')
arr[i][j + 1] = 'V';
if (isValid(i + 1, j - 1, n) &&
arr[i + 1][j - 1] == 'H')
arr[i + 1][j - 1] = 'V';
if (isValid(i + 1, j, n) &&
arr[i + 1][j] == 'H')
arr[i + 1][j] = 'V';
if (isValid(i + 1, j + 1, n) &&
arr[i + 1][j + 1] == 'H')
arr[i + 1][j + 1] = 'V';
}
/*Here we are assigning the neighbours of U
with the character V because we don't want
these neighbours to be counted in that
particular day. If we do not do so, in the
next iteration that neighbour will also get
counted which was supposed to be counted in
the next day. */
}
}
// Mark chocolates unhealthy which are made
// unhealthy in current day.
boolean Hflag = false;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (arr[i][j] == 'V')
{
arr[i][j] = 'U';
Hflag = true;
}
}
}
// Check if there was any chocoloate
// marked unhealthy in current day
if (Hflag)
numdays++;
else
break;
}
return numdays;
}
// Driver Code
public static void main(String []args)
{
int n = 4;
char [][]arr = {{'H', 'H', 'H', 'U'},
{'H', 'H', 'H', 'H'},
{'H', 'U', 'H', 'H'},
{'H', 'H', 'H', 'H'}};
int ans = numdays(arr, n);
System.out.println("number of days taken : " + ans);
}
}
// This code is contributed by ankush_953Python3
# Python3 program to find number of days before
# all chocolates become unhealthy.
# Validates out of bounds indexing
def isValid(i, j, n):
if (i < 0 or j < 0 or i >= n or j >= n):
return False
return True
# function for returning number of days
def numdays(arr, n):
numdays = 0
while (True):
# Traverse matrix to look for unhealthy
# chocolates and mark their neighbors.
for i in range(n):
for j in range(n):
if (arr[i][j] == 'U'):
if (isValid(i - 1, j - 1, n) and\
arr[i - 1][j - 1] == 'H'):
arr[i - 1][j - 1] = 'V'
if (isValid(i - 1, j, n) and\
arr[i - 1][j] == 'H'):
arr[i - 1][j] = 'V'
if (isValid(i - 1, j + 1, n) and\
arr[i - 1][j + 1] == 'H'):
arr[i - 1][j + 1] = 'V'
if (isValid(i, j - 1, n) and\
arr[i][j - 1] == 'H'):
arr[i][j - 1] = 'V'
if (isValid(i, j + 1, n) and\
arr[i][j + 1] == 'H'):
arr[i][j + 1] = 'V'
if (isValid(i + 1, j - 1, n) and\
arr[i + 1][j - 1] == 'H'):
arr[i + 1][j - 1] = 'V'
if (isValid(i + 1, j, n) and\
arr[i + 1][j] == 'H'):
arr[i + 1][j] = 'V'
if (isValid(i + 1, j + 1, n) and\
arr[i + 1][j + 1] == 'H'):
arr[i + 1][j + 1] = 'V'
# Here we are assigning the neighbours of U
# with the character V because we don't want
# these neighbours to be counted in that
# particular day. If we do not do so, in the
# next iteration that neighbour will also get
# counted which was supposed to be counted in
# the next day.
# Mark chocolates unhealthy which are made
# unhealthy in current day.
Hflag = False
for i in range(n):
for j in range(n):
if (arr[i][j] == 'V'):
arr[i][j] = 'U'
Hflag = True
# Check if there was any chocoloate
# marked unhealthy in current day
if (Hflag):
numdays += 1
else:
break
return numdays
# Driver Code
n = 4
arr = [['H', 'H', 'H', 'U'],
['H', 'H', 'H', 'H'],
['H', 'U', 'H', 'H'],
['H', 'H', 'H', 'H']]
ans = numdays(arr, n)
print("number of days taken :", ans)
# This code is contributed by ankush_953C#
// C# program to find number of days before
// all chocolates become unhealthy.
using System;
class GFG{
// Validates out of bounds indexing
static bool isValid(int i, int j, int n)
{
if (i < 0 || j < 0 || i >= n || j >= n)
return false;
return true;
}
// function for returning number of days
static int numdays(char[][] arr, int n)
{
int numdays = 0;
while (true)
{
// Traverse matrix to look for unhealthy
// chocolates and mark their neighbors.
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (arr[i][j] == 'U')
{
if (isValid(i - 1, j - 1, n) &&
arr[i - 1][j - 1] == 'H')
arr[i - 1][j - 1] = 'V';
if (isValid(i - 1, j, n) &&
arr[i - 1][j] == 'H')
arr[i - 1][j] = 'V';
if (isValid(i - 1, j + 1, n) &&
arr[i - 1][j + 1] == 'H')
arr[i - 1][j + 1] = 'V';
if (isValid(i, j - 1, n) &&
arr[i][j - 1] == 'H')
arr[i][j - 1] = 'V';
if (isValid(i, j + 1, n) &&
arr[i][j + 1] == 'H')
arr[i][j + 1] = 'V';
if (isValid(i + 1, j - 1, n) &&
arr[i + 1][j - 1] == 'H')
arr[i + 1][j - 1] = 'V';
if (isValid(i + 1, j, n) &&
arr[i + 1][j] == 'H')
arr[i + 1][j] = 'V';
if (isValid(i + 1, j + 1, n) &&
arr[i + 1][j + 1] == 'H')
arr[i + 1][j + 1] = 'V';
}
/*Here we are assigning the neighbours of U
with the character V because we don't want
these neighbours to be counted in that
particular day. If we do not do so, in the
next iteration that neighbour will also get
counted which was supposed to be counted in
the next day. */
}
}
// Mark chocolates unhealthy which are made
// unhealthy in current day.
bool Hflag = false;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (arr[i][j] == 'V')
{
arr[i][j] = 'U';
Hflag = true;
}
}
}
// Check if there was any chocoloate
// marked unhealthy in current day
if (Hflag)
numdays++;
else
break;
}
return numdays;
}
// Driver function
static public void Main ()
{
int n = 4;
char[][] arr = new char[][]{"HHHU".ToCharArray(),
"HHHH".ToCharArray(),
"HUHH".ToCharArray(),
"HHHH".ToCharArray()};
int ans = numdays(arr, n);
Console.WriteLine("number of days taken : "+ ans);
}
}
// This code is contributed by shubhamsingh10C++
// CPP program using Efficient approach
// to find number of days
#include
using namespace std;
// Validates out of bounds indexing
bool isValid(int i, int j, int n)
{
if (i < 0 || j < 0 || i >= n || j >= n)
return false;
return true;
}
// function for returning number of days
int numdays(char arr[][4], int n)
{
int numdays = 0;
int i, j;
queue > q;
// Initializing queue with initial
// positions of unhealthy chocolates
for (i = 0; i < n; i++)
for (j = 0; j < n; j++) {
if (arr[i][j] == 'U')
q.push(make_pair(i, j));
}
q.push(make_pair(-1, -1));
// (-1, -1) is used as a checkpoint
// to count the number of days
pair temp;
// temporary pair to store the indexes
int flag = 0;
while (!q.empty()) {
temp = q.front();
i = temp.first;
j = temp.second;
q.pop();
if (i == -1 && j == -1) {
flag++;
q.push(make_pair(-1, -1));
// pushing the respective
// checkpoint
if (flag == 2)
break;
numdays++;
}
else {
flag = 0;
if (isValid(i - 1, j - 1, n) &&
arr[i - 1][j - 1] == 'H') {
q.push(make_pair(i - 1, j - 1));
arr[i - 1][j - 1] = 'U';
}
if (isValid(i - 1, j, n) &&
arr[i - 1][j] == 'H') {
q.push(make_pair(i - 1, j));
arr[i - 1][j] = 'U';
}
if (isValid(i - 1, j + 1, n) &&
arr[i - 1][j + 1] == 'H') {
q.push(make_pair(i - 1, j + 1));
arr[i - 1][j + 1] = 'U';
}
if (isValid(i, j - 1, n) &&
arr[i][j - 1] == 'H') {
q.push(make_pair(i, j - 1));
arr[i][j - 1] = 'U';
}
if (isValid(i, j + 1, n) &&
arr[i][j + 1] == 'H') {
q.push(make_pair(i, j + 1));
arr[i][j + 1] = 'U';
}
if (isValid(i + 1, j - 1, n) &&
arr[i + 1][j - 1] == 'H') {
q.push(make_pair(i + 1, j - 1));
arr[i + 1][j - 1] = 'U';
}
if (isValid(i + 1, j, n) &&
arr[i + 1][j] == 'H') {
q.push(make_pair(i + 1, j));
arr[i + 1][j] = 'U';
}
if (isValid(i + 1, j + 1, n) &&
arr[i + 1][j + 1] == 'H') {
q.push(make_pair(i + 1, j + 1));
arr[i + 1][j + 1] = 'U';
}
}
}
return numdays - 1;
}
// Driver function
int main()
{
int n = 4;
char arr[4][4] = { 'H', 'H', 'H', 'U',
'H', 'H', 'H', 'H',
'H', 'H', 'U', 'H',
'H', 'H', 'H', 'H' };
int ans = numdays(arr, n);
cout << "number of days taken : "
<< ans << "\n";
return 0;
} Python3
# Python3 program using Efficient approach
# to find number of days
# Validates out of bounds indexing
def isValid(i, j, n):
if (i < 0 or j < 0 or i >= n or j >= n):
return False
return True
# function for returning number of days
def numdays(arr, n):
numdays = 0
i = 0
j = 0
q = []
# Initializing queue with initial
# positions of unhealthy chocolates
for i in range(n):
for j in range(n):
if (arr[i][j] == 'U'):
q.append([i, j])
q.append([-1, -1])
# (-1, -1) is used as a checkpo
# to count the number of days
temp = []
# temporary pair to store the indexes
flag = 0
while (len(q)):
temp = q[0]
i = temp[0]
j = temp[1]
q.pop(0)
if (i == -1 and j == -1):
flag += 1
q.append([-1, -1])
# appending the respective
# checkpo
if (flag == 2):
break
numdays += 1
else:
flag = 0
if (isValid(i - 1, j - 1, n) and arr[i - 1][j - 1] == 'H'):
q.append([i - 1, j - 1])
arr[i - 1][j - 1] = 'U'
if (isValid(i - 1, j, n) and arr[i - 1][j] == 'H'):
q.append([i - 1, j])
arr[i - 1][j] = 'U'
if (isValid(i - 1, j + 1, n) and arr[i - 1][j + 1] == 'H'):
q.append([i - 1, j + 1])
arr[i - 1][j + 1] = 'U'
if (isValid(i, j - 1, n) and arr[i][j - 1] == 'H'):
q.append([i, j - 1])
arr[i][j - 1] = 'U'
if (isValid(i, j + 1, n) and arr[i][j + 1] == 'H'):
q.append([i, j + 1])
arr[i][j + 1] = 'U'
if (isValid(i + 1, j - 1, n) and arr[i + 1][j - 1] == 'H'):
q.append([i + 1, j - 1])
arr[i + 1][j - 1] = 'U'
if (isValid(i + 1, j, n) and arr[i + 1][j] == 'H'):
q.append([i + 1, j])
arr[i + 1][j] = 'U'
if (isValid(i + 1, j + 1, n) and arr[i + 1][j + 1] == 'H'):
q.append([i + 1, j + 1])
arr[i + 1][j + 1] = 'U'
return numdays - 1
# Driver function
n = 4
arr = [['H', 'H', 'H', 'U'],['H', 'H', 'H', 'H'],
['H', 'H', 'U', 'H'],['H', 'H', 'H', 'H']]
ans = numdays(arr, n)
print("number of days taken :",ans)
# This code is contributed by shubhamsingh10
输出：
number of days taken : 2
高效方法(使用BFS)
在这种方法中，声明一个队列，该队列输入与不健康的巧克力的索引相对应的对，然后一旦达到索引(-1，-1)，我们便增加numdays计数器。此解决方案基本上基于计算二叉树的级别顺序遍历(迭代版本)中的级别，在该级别中，我们将不健康的巧克力的初始索引(而不是根节点)推入，并以数字(numdays)而不是级别计数器作为索引(-1) ，-1)而不是NULL。当标志的计数器达到2时，我们就中断循环，表示队列已经占用了两个连续的(-1，-1)对。
下面是上述方法的实现：
C++
// CPP program using Efficient approach
// to find number of days
#include
using namespace std;
// Validates out of bounds indexing
bool isValid(int i, int j, int n)
{
if (i < 0 || j < 0 || i >= n || j >= n)
return false;
return true;
}
// function for returning number of days
int numdays(char arr[][4], int n)
{
int numdays = 0;
int i, j;
queue > q;
// Initializing queue with initial
// positions of unhealthy chocolates
for (i = 0; i < n; i++)
for (j = 0; j < n; j++) {
if (arr[i][j] == 'U')
q.push(make_pair(i, j));
}
q.push(make_pair(-1, -1));
// (-1, -1) is used as a checkpoint
// to count the number of days
pair temp;
// temporary pair to store the indexes
int flag = 0;
while (!q.empty()) {
temp = q.front();
i = temp.first;
j = temp.second;
q.pop();
if (i == -1 && j == -1) {
flag++;
q.push(make_pair(-1, -1));
// pushing the respective
// checkpoint
if (flag == 2)
break;
numdays++;
}
else {
flag = 0;
if (isValid(i - 1, j - 1, n) &&
arr[i - 1][j - 1] == 'H') {
q.push(make_pair(i - 1, j - 1));
arr[i - 1][j - 1] = 'U';
}
if (isValid(i - 1, j, n) &&
arr[i - 1][j] == 'H') {
q.push(make_pair(i - 1, j));
arr[i - 1][j] = 'U';
}
if (isValid(i - 1, j + 1, n) &&
arr[i - 1][j + 1] == 'H') {
q.push(make_pair(i - 1, j + 1));
arr[i - 1][j + 1] = 'U';
}
if (isValid(i, j - 1, n) &&
arr[i][j - 1] == 'H') {
q.push(make_pair(i, j - 1));
arr[i][j - 1] = 'U';
}
if (isValid(i, j + 1, n) &&
arr[i][j + 1] == 'H') {
q.push(make_pair(i, j + 1));
arr[i][j + 1] = 'U';
}
if (isValid(i + 1, j - 1, n) &&
arr[i + 1][j - 1] == 'H') {
q.push(make_pair(i + 1, j - 1));
arr[i + 1][j - 1] = 'U';
}
if (isValid(i + 1, j, n) &&
arr[i + 1][j] == 'H') {
q.push(make_pair(i + 1, j));
arr[i + 1][j] = 'U';
}
if (isValid(i + 1, j + 1, n) &&
arr[i + 1][j + 1] == 'H') {
q.push(make_pair(i + 1, j + 1));
arr[i + 1][j + 1] = 'U';
}
}
}
return numdays - 1;
}
// Driver function
int main()
{
int n = 4;
char arr[4][4] = { 'H', 'H', 'H', 'U',
'H', 'H', 'H', 'H',
'H', 'H', 'U', 'H',
'H', 'H', 'H', 'H' };
int ans = numdays(arr, n);
cout << "number of days taken : "
<< ans << "\n";
return 0;
}
Python3
# Python3 program using Efficient approach
# to find number of days
# Validates out of bounds indexing
def isValid(i, j, n):
if (i < 0 or j < 0 or i >= n or j >= n):
return False
return True
# function for returning number of days
def numdays(arr, n):
numdays = 0
i = 0
j = 0
q = []
# Initializing queue with initial
# positions of unhealthy chocolates
for i in range(n):
for j in range(n):
if (arr[i][j] == 'U'):
q.append([i, j])
q.append([-1, -1])
# (-1, -1) is used as a checkpo
# to count the number of days
temp = []
# temporary pair to store the indexes
flag = 0
while (len(q)):
temp = q[0]
i = temp[0]
j = temp[1]
q.pop(0)
if (i == -1 and j == -1):
flag += 1
q.append([-1, -1])
# appending the respective
# checkpo
if (flag == 2):
break
numdays += 1
else:
flag = 0
if (isValid(i - 1, j - 1, n) and arr[i - 1][j - 1] == 'H'):
q.append([i - 1, j - 1])
arr[i - 1][j - 1] = 'U'
if (isValid(i - 1, j, n) and arr[i - 1][j] == 'H'):
q.append([i - 1, j])
arr[i - 1][j] = 'U'
if (isValid(i - 1, j + 1, n) and arr[i - 1][j + 1] == 'H'):
q.append([i - 1, j + 1])
arr[i - 1][j + 1] = 'U'
if (isValid(i, j - 1, n) and arr[i][j - 1] == 'H'):
q.append([i, j - 1])
arr[i][j - 1] = 'U'
if (isValid(i, j + 1, n) and arr[i][j + 1] == 'H'):
q.append([i, j + 1])
arr[i][j + 1] = 'U'
if (isValid(i + 1, j - 1, n) and arr[i + 1][j - 1] == 'H'):
q.append([i + 1, j - 1])
arr[i + 1][j - 1] = 'U'
if (isValid(i + 1, j, n) and arr[i + 1][j] == 'H'):
q.append([i + 1, j])
arr[i + 1][j] = 'U'
if (isValid(i + 1, j + 1, n) and arr[i + 1][j + 1] == 'H'):
q.append([i + 1, j + 1])
arr[i + 1][j + 1] = 'U'
return numdays - 1
# Driver function
n = 4
arr = [['H', 'H', 'H', 'U'],['H', 'H', 'H', 'H'],
['H', 'H', 'U', 'H'],['H', 'H', 'H', 'H']]
ans = numdays(arr, n)
print("number of days taken :",ans)
# This code is contributed by shubhamsingh10
输出：
number of days taken : 2
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