Question

A binary array contains only 0 and 1 (1-based indexing). You can perform the following operation at most one times. Change an array element from 1 to 0 . You are required to maximize the number of subarrays that contain at least one 0 and find the number of such subarrays that contain at least one 0 . Input format. The first line contains T denoting the number of test cases. The first line of each test case contains N denoting the size of the binary array. The next line of each test case contains N space-separated integers denoting array elements. Output format Print the maximum number of subarrays which will contain at least one 0 in a new line.

Answer 1

To maximize the number of subarrays that contain at least one 0 in a binary array, we can use a sliding window approach. By iterating through the array and maintaining left and right pointers, we can count the subarrays that contain at least one 0.

Step-by-step explanation:

To maximize the number of subarrays that contain at least one 0, we can use a sliding window approach. We can iterate through the binary array and maintain a left pointer and a right pointer. We move the right pointer until we find a 0 or until we reach the end of the array. Once we find a 0, we know that all subarrays from the left pointer to the right pointer will contain at least one 0. We can count the number of these subarrays and update the maximum count if necessary.

Here is the step-by-step approach:

1. Initialize the left pointer and right pointer to 0.
2. Initialize a variable maxCount to 0.
3. Iterate through the binary array:

• Check if the current element is 0.
  • If yes, increment the count of subarrays by (right - left + 1).
  • Update the maximum count if necessary.
  • Increment the right pointer.
• If the current element is 1:
  • Move the left pointer to the right by one.
  • Increment the right pointer.

Print the maximum count of subarrays containing at least one 0.

For example, for the input [1, 0, 1, 0, 1], the algorithm works as follows:

1. Initialize left = 0, right = 0, maxCount = 0.
2. First element is 1. Move left = 1, right = 1.
3. Next element is 0. Increment count by (right - left + 1) = 1. Update maxCount to 1. Move right = 2.
4. Next element is 1. Move left = 2, right = 3.
5. Next element is 0. Increment count by (right - left + 1) = 2. Update maxCount to 2. Move right = 4.
6. Last element is 1. Move left = 3, right = 5.
7. Print maxCount, which is 2.