Fill array with 1’s using minimum iterations of filling neighbors

Given an array of 0s and 1s, in how many iterations the whole array can be filled with 1s if in a single iteration immediate neighbors of 1s can be filled.

NOTE: If we cannot fill array with 1s, then print “-1” .

Examples :

Input : arr[] = {1, 0, 1, 0, 0, 1, 0, 1,
1, 0, 1, 1, 0, 0, 1}
Output : 1
To convert the whole array into 1s, one iteration
is required. Between indexes i=2 and i=5, the zero
at i=3 would be converted to '1' due to its neighbours
at i=2 similarly the zero at i=4 would be converted
into '1' due to its neighbor at i=5, all this can
be done in a single iteration. Similarly all 0's can
be converted to 1 in single iteration.

Input : arr[] = {0, 0, 1, 1, 0, 0, 1, 1, 0,
1, 1, 1, 1, 0, 0, 0, 1}
Output : 2

Asked in : Amazon

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

It is given that a single 1 can convert both its 0 neighbours to 1. This problem boils down to three cases :

Case 1 : A block of 0s has 1s on both sides

Let count_zero be the count of zeros in the block.

Number of iterations are always equal to :
count_zero/2   if (count_zero is even)
count_zero+1)/2    if(count_zero is odd).

Case 2 : Either single 1 at the end or in
the starting. For example 0 0 0 0 1 and
1 0 0 0 0
In this case the number of iterations required will
always be equal to number of zeros.

Case 3 : There are no 1s (Array has only 0s)
In this case array can't be filled with all 1's.
So print -1.

Algorithm :

1-Start traversing the array.
(a) Traverse until a 0 is found.
while (i < n && a[i] == 1)
{
i++;
flag=true;
}
Flag is set to true just to check at
the last if array contains any 1 or not.

(b) Traverse until a 1 is found and Count
contiguous 0 .
while (i < n && a[i] == 0)
{
count_zero++;
i++;
}

(c) Now check which case is satisfied by
current subarray. And update iterations
using count and update max iterations.

C++

 // C++ program to find number of iterations // to fill with all 1s #include using namespace std;    // Returns count of iterations to fill arr[] // with 1s. int countIterations(int arr[], int n) {     bool oneFound = false;     int res = 0;     // Start traversing the array     for (int i=0; i

Java

 // Java program to find number of iterations // to fill with all 1s    class Test {     // Returns count of iterations to fill arr[]     // with 1s.     static int countIterations(int arr[], int n)     {         boolean oneFound = false;         int res = 0;                    // Start traversing the array         for (int i=0; i

Python3

 # Python3 program to find number  # of iterations to fill with all 1s     # Returns count of iterations  # to fill arr[] with 1s.  def countIterations(arr, n):         oneFound = False;      res = 0;     i = 0;            # Start traversing the array      while (i < n):          if (arr[i] == 1):             oneFound = True;             # Traverse until a 0 is found          while (i < n and arr[i] == 1):              i += 1;             # Count contiguous 0s          count_zero = 0;          while (i < n and arr[i] == 0):             count_zero += 1;              i += 1;             # Condition for Case 3          if (oneFound == False and i == n):              return -1;             # Condition to check          # if Case 1 satisfies:          curr_count = 0;          if (i < n and oneFound == True):                            # If count_zero is even              if ((count_zero & 1) == 0):                  curr_count = count_zero // 2;                 # If count_zero is odd              else:                 curr_count = (count_zero + 1) // 2;                 # Reset count_zero              count_zero = 0;             # Case 2          else:             curr_count = count_zero;              count_zero = 0;             # Update res          res = max(res, curr_count);         return res;     # Driver code  arr = [0, 1, 0, 0, 1, 0, 0,        0, 0, 0, 0, 0, 1, 0];  n = len(arr);  print(countIterations(arr, n));     # This code is contributed by mits

C#

 // C# program to find number of  // iterations to fill with all 1s using System;    class Test {            // Returns count of iterations      // to fill arr[] with 1s.     static int countIterations(int []arr, int n)     {         bool oneFound = false;         int res = 0;                    // Start traversing the array         for (int i = 0; i < n; )         {             if (arr[i] == 1)             oneFound = true;                    // Traverse until a 0 is found             while (i < n && arr[i] == 1)                 i++;                    // Count contiguous 0s             int count_zero = 0;             while (i < n && arr[i] == 0)             {                 count_zero++;                 i++;             }                    // Condition for Case 3             if (oneFound == false && i == n)                 return -1;                    // Condition to check if             // Case 1 satisfies:             int curr_count;             if (i < n && oneFound == true)             {                                    // If count_zero is even                 if ((count_zero & 1) == 0)                     curr_count = count_zero / 2;                        // If count_zero is odd                 else                     curr_count = (count_zero + 1) / 2;                        // Reset count_zero                 count_zero = 0;             }                    // Case 2             else             {                 curr_count = count_zero;                 count_zero = 0;             }                    // Update res             res = Math.Max(res, curr_count);         }                return res;     }            // Driver code     public static void Main()      {         int []arr = {0, 1, 0, 0, 1, 0, 0,                 0, 0, 0, 0, 0, 1, 0};                    Console.Write(countIterations(arr, arr.Length));                } }    // This code is contributed by nitin mittal.

PHP



Output :

4

Time Complexity :
O(n)

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Improved By : nitin mittal, Mithun Kumar

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