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Number of days until all chocolates become unhealthy

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Pablo has a 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 unhealthy denoted by ‘U’. In one day the rotten chocolates make all its neighboring chocolates unhealthy. This goes on and on until all chocolates present in the chocolate box become Unhealthy to eat. Find out the number of days in which the whole chocolate box becomes Unhealthy. 

(Note: It is guaranteed that at least 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.

Asked in Amazon, Accolite, and Arcesium.

Brute Force Approach: 

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

Below is the implementation of the above approach: 

C++




// CPP program to find number of days before
// all chocolates become unhealthy.
#include <bits/stdc++.h>
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 <stdio.h>
 
// 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


Javascript




<script>
 
// Python3 program to find number of days before
// all chocolates become unhealthy.
 
// Validates out of bounds indexing
function isValid(i, j, n){
    if (i < 0 || j < 0 || i >= n || j >= n)
        return false
    return true
}
 
// function for returning number of days
function numdays(arr, n){
    let numdays = 0
    while (true)
    {
         
        // Traverse matrix to look for unhealthy
        // chocolates and mark their neighbors.
        for(let i = 0; i < n; i++)
        {
            for(let 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.
        let Hflag = false
        for(let i=0;i<n;i++){
            for(let 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
let n = 4
let arr = [['H', 'H', 'H', 'U'],
    ['H', 'H', 'H', 'H'],
    ['H', 'U', 'H', 'H'],
    ['H', 'H', 'H', 'H']]
let ans = numdays(arr, n)
document.write("number of days taken : ", ans)
 
// This code is contributed by shinjanpatra
 
</script>


Output

number of days taken : 2

Time Complexity: O(n2)
Auxiliary Space: O(1)

Efficient Approach (Uses BFS):

In this approach, declare a queue which inputs pairs that 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 the root node and increment numdays instead of the 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 encountered two consecutive (-1, -1) pairs.

Below is the implementation of the above approach:

C++




// CPP program using Efficient approach
// to find number of days
#include <bits/stdc++.h>
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<pair<int, int> > 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<int, int> 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;
}


Java




// Java program using Efficient approach
// to find number of days
import java.util.*;
 
class gfg2 {
  static class pair {
    int first, second;
    pair(int f, int s)
    {
      first = f;
      second = s;
    }
  }
  // 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;
    int i, j;
    Queue<pair> q = new ArrayDeque<>();
 
    // 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.add(new pair(i, j));
      }
 
    q.add(new 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.isEmpty()) {
      temp = q.peek();
      i = temp.first;
      j = temp.second;
      q.remove();
      if (i == -1 && j == -1) {
        flag++;
        q.add(new 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.add(new pair(i - 1, j - 1));
          arr[i - 1][j - 1] = 'U';
        }
 
        if (isValid(i - 1, j, n)
            && arr[i - 1][j] == 'H') {
          q.add(new pair(i - 1, j));
          arr[i - 1][j] = 'U';
        }
 
        if (isValid(i - 1, j + 1, n)
            && arr[i - 1][j + 1] == 'H') {
          q.add(new pair(i - 1, j + 1));
          arr[i - 1][j + 1] = 'U';
        }
 
        if (isValid(i, j - 1, n)
            && arr[i][j - 1] == 'H') {
          q.add(new pair(i, j - 1));
          arr[i][j - 1] = 'U';
        }
 
        if (isValid(i, j + 1, n)
            && arr[i][j + 1] == 'H') {
          q.add(new pair(i, j + 1));
          arr[i][j + 1] = 'U';
        }
 
        if (isValid(i + 1, j - 1, n)
            && arr[i + 1][j - 1] == 'H') {
          q.add(new pair(i + 1, j - 1));
          arr[i + 1][j - 1] = 'U';
        }
 
        if (isValid(i + 1, j, n)
            && arr[i + 1][j] == 'H') {
          q.add(new pair(i + 1, j));
          arr[i + 1][j] = 'U';
        }
 
        if (isValid(i + 1, j + 1, n)
            && arr[i + 1][j + 1] == 'H') {
          q.add(new pair(i + 1, j + 1));
          arr[i + 1][j + 1] = 'U';
        }
      }
    }
 
    return numdays - 1;
  }
 
  // Driver function
  public static void main(String[] args)
  {
    int n = 4;
    char arr[][] = { { 'H', 'H', 'H', 'U' },
                    { 'H', 'H', 'H', 'H' },
                    { 'H', 'H', 'U', 'H' },
                    { 'H', 'H', 'H', 'H' } };
    int ans = numdays(arr, n);
    System.out.println("number of days taken : " + ans);
  }
}
 
// This code is contributed by karandeep1234


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


C#




// C# program using Efficient approach
// to find number of days
using System;
using System.Collections.Generic;
 
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;
    int i, j;
    List< int[]> q = new List< int[]>();
 
    // 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.Add(new int[] {i, j});
      }
 
    q.Add(new int[] {-1, -1});
 
    // (-1, -1) is used as a checkpoint
    // to count the number of days
    int[] temp = {-1, -1};
 
    // temporary pair to store the indexes
    int flag = 0;
    while (q.Count > 0){
      temp = q[0];
      i = temp[0];
      j = temp[1];
      q.RemoveAt(0);
      if (i == -1 && j == -1) {
        flag++;
        q.Add(new int[] {-1, -1});
 
        // Adding 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.Add(new int[] {i - 1, j - 1});
          arr[i - 1, j - 1] = 'U';
        }
 
        if (isValid(i - 1, j, n) &&
            arr[i - 1, j] == 'H') {
          q.Add(new int[] {i - 1, j});
          arr[i - 1, j] = 'U';
        }
 
        if (isValid(i - 1, j + 1, n) &&
            arr[i - 1, j + 1] == 'H') {
          q.Add(new int[] {i - 1, j + 1});
          arr[i - 1, j + 1] = 'U';
        }
 
        if (isValid(i, j - 1, n) &&
            arr[i, j - 1] == 'H') {
          q.Add(new int[] {i, j - 1});
          arr[i, j - 1] = 'U';
        }
 
        if (isValid(i, j + 1, n) &&
            arr[i, j + 1] == 'H') {
          q.Add(new int[] {i, j + 1});
          arr[i, j + 1] = 'U';
        }
 
        if (isValid(i + 1, j - 1, n) &&
            arr[i + 1, j - 1] == 'H') {
          q.Add(new int[] {i + 1, j - 1});
          arr[i + 1, j - 1] = 'U';
        }
 
        if (isValid(i + 1, j, n) &&
            arr[i + 1, j] == 'H') {
          q.Add(new int[] {i + 1, j});
          arr[i + 1, j] = 'U';
        }
 
        if (isValid(i + 1, j + 1, n) &&
            arr[i + 1, j + 1] == 'H') {
          q.Add(new int[] {i + 1, j + 1});
          arr[i + 1, j + 1] = 'U';
        }
      }
    }
 
    return numdays - 1;
  }
 
  // Driver function
  public static void Main(string[] args)
  {
    int n = 4;
    char[,] arr = { {'H', 'H', 'H', 'U'},
                   {'H', 'H', 'H', 'H'},
                   {'H', 'H', 'U', 'H'},
                   {'H', 'H', 'H', 'H' }};
    int ans = numdays(arr, n);
    Console.Write("number of days taken : " + ans);
  }
}
 
// This code is contributed by phasing17.


Javascript




<script>
 
// JavaScript program using Efficient approach
// to find number of days
 
// Validates out of bounds indexing
function isValid(i, j, n){
    if (i < 0 || j < 0 || i >= n || j >= n)
        return false
 
    return true
}
 
// function for returning number of days
function numdays(arr, n){
     
    let numdays = 0
    let i = 0
    let j = 0
    let q = []
     
    // Initializing queue with initial
    // positions of unhealthy chocolates
    for(let i=0;i<n;i++){
        for(let j=0;j<n;j++){
            if (arr[i][j] == 'U'){
                q.push([i, j])
            }
        }
    }
     
    q.push([-1, -1])
     
    // (-1, -1) is used as a checkpo
    // to count the number of days
    let temp = []
     
    // temporary pair to store the indexes
    let flag = 0
    while (q.length){
        let temp = q.shift()
        let i = temp[0]
        let j = temp[1]
 
        if (i == -1 && j == -1){
            flag += 1
            q.push([-1, -1])
             
            // pushing the respective
            // checkpo
            if (flag == 2)
                break
            numdays += 1
        }
        else{
             
            flag = 0
            if (isValid(i - 1, j - 1, n) && arr[i - 1][j - 1] == 'H'){
                q.push([i - 1, j - 1])
                arr[i - 1][j - 1] = 'U'
            }
             
            if (isValid(i - 1, j, n) && arr[i - 1][j] == 'H'){
                q.push([i - 1, j])
                arr[i - 1][j] = 'U'
            }
             
            if (isValid(i - 1, j + 1, n) && arr[i - 1][j + 1] == 'H'){
                q.push([i - 1, j + 1])
                arr[i - 1][j + 1] = 'U'
            }
             
            if (isValid(i, j - 1, n) && arr[i][j - 1] == 'H'){
                q.push([i, j - 1])
                arr[i][j - 1] = 'U'
            }
             
            if (isValid(i, j + 1, n) && arr[i][j + 1] == 'H'){
                q.push([i, j + 1])
                arr[i][j + 1] = 'U'
            }
                 
            if (isValid(i + 1, j - 1, n) && arr[i + 1][j - 1] == 'H'){
                q.push([i + 1, j - 1])
                arr[i + 1][j - 1] = 'U'
            }
                 
            if (isValid(i + 1, j, n) && arr[i + 1][j] == 'H'){
                q.push([i + 1, j])
                arr[i + 1][j] = 'U'
            }
             
            if (isValid(i + 1, j + 1, n) && arr[i + 1][j + 1] == 'H'){
                q.push([i + 1, j + 1])
                arr[i + 1][j + 1] = 'U'
            }
        }
    }
                 
    return numdays - 1
}
 
 
// Driver function
let n = 4
let arr = [['H', 'H', 'H', 'U'],['H', 'H', 'H', 'H'],
        ['H', 'H', 'U', 'H'],['H', 'H', 'H', 'H']]
let ans = numdays(arr, n)
document.write("number of days taken :",ans,"</br>")
 
// This code is contributed by shinjanpatra
 
</script>


Output

number of days taken : 2

Time Complexity: O(n2)
Auxiliary Space: O(n2)



Last Updated : 14 Dec, 2022
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