# Number of days until all chocolates become unhealthy

Pablo has square chocolate Box of size n x n in which a variety of healthy chocolates are present denoted by ‘H’ initially but he finds out that some of the chocolates are rotten and are unhealthy denoted by ‘U’. In one day the rotten chocolates make all its neighbouring chocolates as unhealthy. This goes on and on until all chocolates present in the chocolate box becomes Unhealthy to eat. Find out the number of days in which the whole chocolate box becomes Unhealthy.
(Note : It is guaranteed that atleast one of the chocolate is Unhealthy)

Examples:

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Brute Force Approach:
Initialize a flag = 1. Use a while loop, inside that while search for an H (searching requires O(n^2) time complexity if we are unable to find a H in the 2-D character array stop incrementing the day counter and set flag as 0 to break the loop.

Below is the implementation of above approach:

## 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_953

## Python3

 # 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_953

## C#

 // 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 shubhamsingh10

Output:

number of days taken : 2

Efficient Approach (Uses BFS)
In this approach, declare a queue which inputs pairs which corresponds to the index of the unhealthy chocolates and then as soon as the index (-1, -1) is reached we increment the numdays counter. This Solution is basically based on calculating levels in level order traversal (Iterative version) of a binary tree in which we push the initial indexes of the unhealthy chocolates instead of root node and incrementing numdays instead of level counter as soon as the index (-1, -1) is reached instead of NULL. As soon as the counter of the flag reaches 2 we break the loop denoting that queue has encoutered two consecutive (-1, -1) pair.

Below is the implementation of above approach:

## 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

Output:

number of days taken : 2

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