Minimum no. of operations required to make all Array Elements Zero

Given an array of N elements and each element is either 1 or 0. You need to make all the elements of the array equal to 0 by performing below operations:

  • If an element is 1, You can change it’s value equal to 0 then,
    • if the next consecutive element is 1, it will automatically get converted to 0.
    • if the next consecutive element is already 0, nothing will happen.

Now, the task is to find the minimum number of operations required to make all elements equal to 0.

Examples:

Input : arr[] = {1, 1, 0, 0, 1, 1, 1, 0, 0, 1} 
Output : Minimum changes: 3 

Input : arr[] = {1, 1, 1, 1}
Output : Minimum changes: 1 

Approach: To find the minimum number of changes required, iterate the array from left to right and check if the current element is 1 or not. If the current element is 1, then change it to 0 and increment the count by 1 and search for the 0 for the next operation as all consecutive 1’s will be automatically converted to 0.

Below is the implementation of above approach:

C++

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// CPP program to find  minimum number of 
// operations required to make all 
// array elements zero
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find  minimum number of 
// operations required to make all 
// array elements zero
int minimumChanges(int arr[], int n)
{
    int i;
      
    // It will maintain total changes required
    int changes = 0; 
      
    for (i = 0; i < n; i++) 
    {   
        // Check for the first 1
        if (arr[i] == 1) 
        {   
            int j;
              
            // Check for number of 
            // consecutive 1's
            for(j = i+1; j<n; j++)
            {
                if(arr[j]==0)
                    break;
            }
              
            // Increment i to the position of 
            // last consecutive 1
            i = j-1;
              
            changes++;
        }
    }
      
    return changes;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 1, 0, 0, 0, 1, 0, 1, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
      
    cout << "Minimum operations: " << minimumChanges(arr, n);
      
    return 0;
}

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Java

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// Java program to find minimum 
// number of operations required 
// to make all array elements zero 
  
class GFG 
{
  
// Function to find minimum number 
// of operations required to make  
// all array elements zero 
static int minimumChanges(int arr[], 
                          int n) 
    int i; 
      
    // It will maintain total
    // changes required 
    int changes = 0
      
    for (i = 0; i < n; i++) 
    
        // Check for the first 1 
        if (arr[i] == 1
        
            int j; 
              
            // Check for number of 
            // consecutive 1's 
            for(j = i + 1; j < n; j++) 
            
                if(arr[j] == 0
                    break
            
              
            // Increment i to the position  
            // of last consecutive 1 
            i = j - 1
              
            changes++; 
        
    
      
    return changes; 
  
// Driver code 
public static void main (String args[])
{
    int arr[] = { 1, 1, 0, 0, 0
                     1, 0, 1, 1 }; 
    int n = arr.length ;
      
    System.out.println("Minimum operations: "
                        minimumChanges(arr, n)); 
      
}
}
  
// This code is contributed by ANKITRAI1

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Python 3

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# Python 3 program to find 
# minimum number of operations 
# required to make all array
# elements zero 
  
# Function to find minimum number 
# of operations required to make 
# all array elements zero 
def minimumChanges(arr, n) :
  
    # It will maintain total
    # changes required 
    changes = 0
      
    i = 0
      
    while i < n :
  
        # Check for the first 1
        if arr[i] == 1 :
  
            j = i + 1
  
            # Check for number of 
            # consecutive 1's 
            while j < n:
  
                if arr[j] == 0 :
                    break
  
                j += 1
  
            # Increment i to the position 
            # of last consecutive 1 
            i = j - 1
              
            changes += 1
  
        i += 1
          
    return changes
  
# Driver code     
if __name__ == "__main__" :
  
    arr = [ 1, 1, 0, 0, 0, 1, 0, 1, 1]
    n = len(arr)
  
    print("Minimum operations:"
           minimumChanges(arr, n))
  
# This code is contributed by ANKITRAI1

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C#

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// C# program to find minimum 
// number of operations required 
// to make all array elements zero 
class GFG 
{
  
// Function to find minimum number 
// of operations required to make 
// all array elements zero 
static int minimumChanges(int[] arr, 
                          int n) 
    int i; 
      
    // It will maintain total
    // changes required 
    int changes = 0; 
      
    for (i = 0; i < n; i++) 
    
        // Check for the first 1 
        if (arr[i] == 1) 
        
            int j; 
              
            // Check for number of 
            // consecutive 1's 
            for(j = i + 1; j < n; j++) 
            
                if(arr[j] == 0) 
                    break
            
              
            // Increment i to the position 
            // of last consecutive 1 
            i = j - 1; 
              
            changes++; 
        
    
      
    return changes; 
  
// Driver code 
static void Main()
{
    int[] arr = new int[]{ 1, 1, 0, 0, 0, 
                           1, 0, 1, 1 }; 
    int n = arr.Length ;
      
    System.Console.WriteLine("Minimum operations: " +
                             minimumChanges(arr, n)); 
}
}
  
// This code is contributed by mits

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PHP

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<?php
// PHP program to find minimum number  
// of operations required to make all 
// array elements zero
  
// Function to find minimum number  
// of operations required to make  
// all array elements zero
function minimumChanges($arr, $n)
{
    $i;
      
    // It will maintain total
    // changes required
    $changes = 0; 
      
    for ($i = 0; $i < $n; $i++) 
    
        // Check for the first 1
        if ($arr[$i] == 1) 
        
            $j;
              
            // Check for number of 
            // consecutive 1's
            for($j = $i + 1; $j < $n; $j++)
            {
                if($arr[$j] == 0)
                    break;
            }
              
            // Increment i to the position
            // of last consecutive 1
            $i = $j - 1;
              
            $changes++;
        }
    }
      
    return $changes;
}
  
// Driver code
$arr = array( 1, 1, 0, 0, 0, 
                 1, 0, 1, 1 );
$n = sizeof($arr);
      
echo "Minimum operations: "
    minimumChanges($arr, $n);
  
// This code is contributed 
// by Akanksha Rai(Abby_akku)

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Output:

Minimum operations: 3


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