Minimum number of 1’s to be replaced in a binary array

Given a binary array arr[] of zero’s and one’s only. The task is to find the minimum number of one’s to be changed to zero such if there exist any index i in the array such that arr[i] = 0 then arr[i-1] and arr[i+1] both should not be equals to 1 at the same time.

That is, for any index i the below condition should fail:

if (arr[i]== 0):
    (arr[i-1] == 1) && (arr[i+1] == 1)

Note: 1-based indexing is considered for the array.



Examples:

Input : arr[] = { 1, 1, 0, 1, 1, 0, 1, 0, 1, 0 }
Output : 2
Explanation: Indexs 2 and 7 OR 4 and 7 can be changed to zero.

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

Approach: The idea is that, whenever we found condition like arr[i-1] = 1 \;and\; arr[i] = 0 \;and \;arr[i+1] = 1 we simply changed the value of (i+1)th index to zero(0). So that index between (i-1)-th and (i+1)-th index is safe.

Below is the implementation of the above approach:

C++

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// C++ program to find minimum number
// of 1's  to be replaced to 0's
#include <bits/stdc++.h>
using namespace std;
  
// Function to find minimum number
// of 1's  to be replaced to 0's
int minChanges(int A[], int n)
{
    int cnt = 0;
  
    for (int i = 0; i < n - 2; ++i) {
  
        if ((i - 1 >= 0) && A[i - 1] == 1
            && A[i + 1] == 1 && A[i] == 0) {
            A[i + 1] = 0;
            cnt++;
        }
  
    }
  
    // return final answer
    return cnt;
}
  
// Driver program
int main()
{
    int A[] = { 1, 1, 0, 1, 1, 0, 1, 0, 1, 0 };
    int n = sizeof(A) / sizeof(A[0]);
  
    cout << minChanges(A, n);
  
    return 0;
}

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Java

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// Java program to find minimum number
// of 1's to be replaced to 0's
import java.lang.*;
import java.util.*;
  
class GFG
{
// Function to find minimum number
// of 1's to be replaced to 0's
static int minChanges(int[] A, int n)
{
    int cnt = 0;
  
    for (int i = 0; i < n - 2; ++i)
    {
  
        if ((i - 1 >= 0) && A[i - 1] == 1 && 
                            A[i + 1] == 1 && 
                            A[i] == 0
        {
            A[i + 1] = 0;
            cnt++;
        }
  
    }
  
    // return final answer
    return cnt;
}
  
// Driver Code
public static void main(String args[])
{
    int[] A = { 1, 1, 0, 1, 1, 0, 1, 0, 1, 0 };
    int n = A.length;
  
    System.out.print(minChanges(A, n));
}
}
  
// This code is contributed
// by Akanksha Rai

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Python3

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# Python 3 program to find minimum 
# number of 1's to be replaced to 0's
      
# Function to find minimum number
# of 1's to be replaced to 0's
def minChanges(A, n):
    cnt = 0
    for i in range(n - 2):
        if ((i - 1 >= 0) and A[i - 1] == 1 and 
           A[i + 1] == 1 and A[i] == 0):
            A[i + 1] = 0
            cnt = cnt + 1
      
    # return final answer
    return cnt
      
# Driver Code
A = [1, 1, 0, 1, 1, 0, 1, 0, 1, 0]
n = len(A)
print(minChanges(A, n))
          
# This code is contributed 
# by Shashank_Sharma

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C#

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// C# program to find minimum number
// of 1's to be replaced to 0's
using System;
  
class GFG
{
// Function to find minimum number
// of 1's to be replaced to 0's
static int minChanges(int[] A, int n)
{
    int cnt = 0;
  
    for (int i = 0; i < n - 2; ++i)
    {
  
        if ((i - 1 >= 0) && A[i - 1] == 1 && 
                            A[i + 1] == 1 && A[i] == 0) 
        {
            A[i + 1] = 0;
            cnt++;
        }
  
    }
  
    // return final answer
    return cnt;
}
  
// Driver Code
public static void Main()
{
    int[] A = { 1, 1, 0, 1, 1, 0, 1, 0, 1, 0 };
    int n = A.Length;
  
    Console.Write(minChanges(A, n));
}
}
  
// This code is contributed
// by Akanksha Rai

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PHP

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<?php
// PHP program to find minimum number
// of 1's to be replaced to 0's
  
// Function to find minimum number
// of 1's to be replaced to 0's
function minChanges($A, $n)
{
    $cnt = 0;
  
    for ($i = 0; $i < $n - 2; ++$i
    {
        if (($i - 1 >= 0) && $A[$i - 1] == 1 &&
          $A[$i + 1] == 1 && $A[$i] == 0)
        {
            $A[$i + 1] = 0;
            $cnt++;
        }
  
    }
  
    // return final answer
    return $cnt;
}
  
// Driver Code
$A = array(1, 1, 0, 1, 1, 
           0, 1, 0, 1, 0);
$n = sizeof($A);
  
echo minChanges($A, $n);
  
// This code is contributed 
// by Ankita_Saini
?>

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Output:

2

Time Complexity: O(N)



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