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Maximum number of 0s that can be flipped such that Array has no adjacent 1s
• Last Updated : 27 Apr, 2021

Given a binary array arr, the task is to find the maximum number of 0s that can be flipped such that the array has no adjacent 1s, i.e. the array does not contain any two 1s at consecutive indices.
Examples:

Input: arr[] = {1, 0, 0, 0, 1}
Output:
Explanation:
The 0 at index 2 can be replaced by 1.
Input: arr[] = {1, 0, 0, 1}
Output:
Explanation:
No 0 (zeroes) can be replaced by 1 such that no two consecutive indices have 1.

Approach:

• Iterate over the array and for every index which have 0, check if its adjacent two indices have 0 or not. For the last and first index of the array, check for the adjacent left and right index respectively.
• For every such index satisfying the above condition, increase the count.
• Print the final count at the end as the required answer

Below code is the implementation of the above approach:

## C++

 // C++ program for the above approach#includeusing namespace std; // Maximum number of 0s that// can be replaced by 1int canReplace(int array[], int n){    int i = 0, count = 0;     while (i < n)    {                 // Check for three consecutive 0s        if (array[i] == 0 &&            (i == 0 || array[i - 1] == 0) &&            (i == n - 1|| array[i + 1] == 0))        {             // Flip the bit            array[i] = 1;             // Increase the count            count++;        }        i++;    }    return count;} // Driver's Codeint main(){    int array[5] = { 1, 0, 0, 0, 1 };             cout << canReplace(array, 5);} // This code is contributed by spp____

## Java

 // Java program for the above approach public class geeks {     // Maximum number of 0s that    // can be replaced by 1    public static int canReplace(        int[] array)    {        int i = 0, count = 0;         while (i < array.length) {             // Check for three consecutive 0s            if (array[i] == 0                && (i == 0                    || array[i - 1] == 0)                && (i == array.length - 1                    || array[i + 1] == 0)) {                 // Flip the bit                array[i] = 1;                 // Increase the count                count++;            }            i++;        }         return count;    }     // Driver code    public static void main(String[] args)    {        int[] array = { 1, 0, 0, 0, 1 };        System.out.println(canReplace(array));    }}

## Python3

 # Python3 program for the above approach # Maximum number of 0s that# can be replaced by 1def canReplace(arr, n):     i = 0    count = 0     while (i < n):         # Check for three consecutive 0s        if (arr[i] == 0 and                (i == 0 or arr[i - 1] == 0) and                (i == n - 1 or arr[i + 1] == 0)):             # Flip the bit            arr[i] = 1             # Increase the count            count += 1         i += 1    return count # Driver codeif __name__ == '__main__':     arr = [ 1, 0, 0, 0, 1]         print(canReplace(arr, 5)) # This code is contributed by himanshu77

## C#

 // C# program for the above approachusing System; class GFG{ // Maximum number of 0s that// can be replaced by 1public static int canReplace(int[] array){    int i = 0, count = 0;    while (i < array.Length)    {         // Check for three consecutive 0s        if (array[i] == 0 &&           (i == 0 || array[i - 1] == 0) &&           (i == array.Length - 1 || array[i + 1] == 0))        {                         // Flip the bit            array[i] = 1;             // Increase the count            count++;        }        i++;    }     return count;} // Driver codepublic static void Main(String []args){    int[] array = { 1, 0, 0, 0, 1 };         Console.WriteLine(canReplace(array));}} // This code is contributed by Rajput-Ji

## Javascript


Output:
1

Time Complexity: O(N)

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