Fill array with 1's using minimum iterations of filling neighbors - GeeksforGeeks

28 Apr, 2021
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 
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++
Java
Python3
C#
PHP
Javascript
// C++ program to find number of iterations// to fill with all 1s#include<bits/stdc++.h>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<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 = max(res, curr_count);    }     return res;} // Driver codeint main(){    int arr[] = {0, 1, 0, 0, 1, 0, 0,                 0, 0, 0, 0, 0, 1, 0};    int n = sizeof(arr) / sizeof(arr[0]);    cout << countIterations(arr, n);    return 0;}
// 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<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 method    public static void main(String[] args)    {        int arr[] = {0, 1, 0, 0, 1, 0, 0,                0, 0, 0, 0, 0, 1, 0};                 System.out.println(countIterations(arr, arr.length));             }}
# 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 codearr = [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# program to find number of// iterations to fill with all 1susing 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// PHP program to find number// of iterations to fill with all 1s // Returns count of iterations// to fill arr[] with 1s.function countIterations($arr, $n){    $oneFound = false;    $res = 0;    // Start traversing the array    for ( $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        $count_zero = 0;        while ($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:        $curr_count;        if ($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        $res = max($res, $curr_count);    }     return $res;} // Driver code$arr = array(0, 1, 0, 0, 1, 0, 0,            0, 0, 0, 0, 0, 1, 0);$n = sizeof($arr) / sizeof($arr[0]);echo countIterations($arr, $n); // This code is contributed by nitin mittal.?>
<script> // Javascript program to find number of// iterations to fill with all 1s // Returns count of iterations to fill arr[]// with 1s.function countIterations(arr, n){    var oneFound = false;    var res = 0;         // Start traversing the array    for(var 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        var 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:        var 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 Codevar arr = [ 0, 1, 0, 0, 1, 0, 0,            0, 0, 0, 0, 0, 1, 0 ];         document.write(countIterations(arr, arr.length));         // This code is contributed by Ankita saini </script>
Output :  
4
Time Complexity : O(n)This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
nitin mittal
Mithun Kumar
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