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

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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
Explanation: 
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 Practice

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.    

Implementation:

C++




// 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 code
int 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




// 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




# 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




<?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.
?>


Javascript




<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 Code
var 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)
Auxiliary Space: O(1)

 



Last Updated : 14 Sep, 2023
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