# Minimum time required to cover a Binary Array

Given an integer N which represents the size of a boolean array which is initially filled with 0, and also given an array arr[] of size K consisting of (1-based) indices at which ‘1’ are present in the boolean array. Now, at one unit of time the adjacent cells of the arr[i] in the boolean array becomes 1 that is 010 becomes 111. Find the minimum time required to convert the whole array into 1s.
Examples:

Input : N = 5, arr[] = {1, 4}
Output :
Explanation:
Initially the boolean array is of size 5 filled with 5 zeros. arr[] represents places at which 1 is present in the boolean array. Therefore, the boolean array becomes 10010.
Now, at time (t) = 1, the 0s at 3rd and 5th position becomes 1 => 10111 and at the same time, 0 at 2nd position becomes 1 => 11111. All the 1s initially increment their adjacent 0s at the same moment of time. So at t=1, the string gets converted to all 1s.
Input : N=7, arr[] = {1, 7}
Output :
Explanation:
At time (t) = 1, 1000001 becomes 1100011
At time (t) = 2, 1100011 becomes 1110111
At time (t) = 3, 1110111 becomes 1111111
Hence, minimum time is 3 to change the binary array into 1.

Approach:
To solve the problem mentioned above we have to observe that we need to find the longest segment of zeroes until 1 appears. For example, for a binary number 00010000010000, the longest segment of 0s is from 4th to 10th position. Now, observe that there are 5 0s between the indices which is an odd number. Hence we can conclude that to cover 5 zeros we need 5/2 + 1 that is 3 units of time because all the other segments will get filled in less than or equal to 3 units of time.

• If the longest zero segments is odd then we can conclude that, x/2 + 1 units of time is required where x is the number of 0s in the longest segment.
• If the longest zero segments is even then we can conclude that, x/2 units of time is required where x is the number of 0s in the longest segment.

We can calculate the maximum length contiguous segment until 1 occurs in the boolean array and return the answer depending upon whether the length is odd or even.
Below is the implementation of the above approach:

## C++

 `// CPP implementation to find the ` `// Minimum time required to cover a Binary Array ` `#include ` `using` `namespace` `std; ` ` `  `// function to calculate the time ` `int` `solve(vector<``int``> arr, ``int` `n) ` `{ ` ` `  `    ``int` `k = arr.size(); ` ` `  `    ``// Map to mark or store the binary values ` `    ``bool` `mp[n + 2]; ` ` `  `    ``// Firstly fill the boolean ` `    ``// array with all zeroes ` `    ``for` `(``int` `i = 0; i <= n; i++) { ` `        ``mp[i] = 0; ` `    ``} ` ` `  `    ``// Mark the 1s ` `    ``for` `(``int` `i = 0; i < k; i++) { ` `        ``mp[arr[i]] = 1; ` `    ``} ` ` `  `    ``// Number of 0s until first '1' occurs ` `    ``int` `leftSegment = arr - 1; ` ` `  `    ``// Maximum Number of 0s in between 2 '1's. ` `    ``for` `(``int` `i = 1; i < k; i++) { ` `        ``leftSegment = max(leftSegment, arr[i] - arr[i - 1] - 1); ` `    ``} ` ` `  `    ``// Number of 0s from right until first '1' occurs ` `    ``int` `rightSegment = n - arr[k - 1]; ` ` `  `    ``// Return maximum from left and right segment ` `    ``int` `maxSegment = max(leftSegment, rightSegment); ` ` `  `    ``int` `tim; ` ` `  `    ``// check if count is odd ` `    ``if` `(maxSegment & 1) ` `        ``tim = (maxSegment / 2) + 1; ` ` `  `    ``// check ifcount is even ` `    ``else` `        ``tim = maxSegment / 2; ` ` `  `    ``// return the time ` `    ``return` `tim; ` `} ` ` `  `// driver code ` `int` `main() ` `{ ` `    ``// initialise N ` `    ``int` `N = 5; ` ` `  `    ``// array initialisation ` `    ``vector<``int``> arr = { 1, 4 }; ` ` `  `    ``cout << solve(arr, N); ` `} `

## Java

 `// Java implementation to find the  ` `// Minimum time required to cover a Binary Array  ` `class` `GFG { ` ` `  `    ``// function to calculate the time  ` `    ``static` `int` `solve(``int` `[]arr, ``int` `n)  ` `    ``{  ` `     `  `        ``int` `k = arr.length;  ` `     `  `        ``// Map to mark or store the binary values  ` `        ``int` `mp[] = ``new` `int``[n + ``2``];  ` `     `  `        ``// Firstly fill the boolean  ` `        ``// array with all zeroes  ` `        ``for` `(``int` `i = ``0``; i <= n; i++) {  ` `            ``mp[i] = ``0``;  ` `        ``}  ` `     `  `        ``// Mark the 1s  ` `        ``for` `(``int` `i = ``0``; i < k; i++) {  ` `            ``mp[arr[i]] = ``1``;  ` `        ``}  ` `     `  `        ``// Number of 0s until first '1' occurs  ` `        ``int` `leftSegment = arr[``0``] - ``1``;  ` `     `  `        ``// Maximum Number of 0s in between 2 '1's.  ` `        ``for` `(``int` `i = ``1``; i < k; i++) {  ` `            ``leftSegment = Math.max(leftSegment, arr[i] - arr[i - ``1``] - ``1``);  ` `        ``}  ` `     `  `        ``// Number of 0s from right until first '1' occurs  ` `        ``int` `rightSegment = n - arr[k - ``1``];  ` `     `  `        ``// Return maximum from left and right segment  ` `        ``int` `maxSegment = Math.max(leftSegment, rightSegment);  ` `     `  `        ``int` `tim;  ` `     `  `        ``// check if count is odd  ` `        ``if` `((maxSegment & ``1``) == ``1``)  ` `            ``tim = (maxSegment / ``2``) + ``1``;  ` `     `  `        ``// check ifcount is even  ` `        ``else` `            ``tim = maxSegment / ``2``;  ` `     `  `        ``// return the time  ` `        ``return` `tim;  ` `    ``}  ` `     `  `    ``// driver code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``// initialise N  ` `        ``int` `N = ``5``;  ` `     `  `        ``// array initialisation  ` `        ``int` `arr[] = { ``1``, ``4` `};  ` `     `  `        ``System.out.println(solve(arr, N));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation to find the  ` `# Minimum time required to cover a Binary Array  ` ` `  `# function to calculate the time  ` `def` `solve(arr, n) : ` ` `  `    ``k ``=` `len``(arr)  ` ` `  `    ``# Map to mark or store the binary values ` `    ``# Firstly fill the boolean  ` `    ``# array with all zeroes  ` `    ``mp ``=` `[``False` `for` `i ``in` `range``(n ``+` `2``)]  ` ` `  `    ``# Mark the 1s  ` `    ``for` `i ``in` `range``(k) : ` `        ``mp[arr[i]] ``=` `True` ` `  `    ``# Number of 0s until first '1' occurs  ` `    ``leftSegment ``=` `arr[``0``] ``-` `1` ` `  `    ``# Maximum Number of 0s in between 2 '1's.  ` `    ``for` `i ``in` `range``(``1``,k) : ` `        ``leftSegment ``=` `max``(leftSegment, arr[i] ``-` `arr[i ``-` `1``] ``-` `1``) ` ` `  `    ``# Number of 0s from right until first '1' occurs  ` `    ``rightSegment ``=` `n ``-` `arr[k ``-` `1``] ` ` `  `    ``# Return maximum from left and right segment  ` `    ``maxSegment ``=` `max``(leftSegment, rightSegment);  ` ` `  `    ``tim ``=` `0` ` `  `    ``# check if count is odd  ` `    ``if` `(maxSegment & ``1``) : ` `        ``tim ``=` `(maxSegment ``/``/` `2``) ``+` `1` ` `  `    ``# check ifcount is even  ` `    ``else` `: ` `        ``tim ``=` `maxSegment ``/``/` `2` ` `  `    ``# return the time  ` `    ``return` `tim ` ` `  `# Driver code  ` `# initialise N  ` `N ``=` `5` ` `  `# array initialisation  ` `arr ``=` `[ ``1``, ``4` `]  ` ` `  `print``(solve(arr, N)) ` ` `  `# This code is contributed by Sanjit_Prasad `

## C#

 `// C# implementation to find the  ` `// Minimum time required to cover ` `// a Binary Array  ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function to calculate the time  ` `static` `int` `solve(``int``[] arr, ``int` `n)  ` `{  ` `    ``int` `k = arr.Length;  ` ` `  `    ``// Map to mark or store the binary values  ` `    ``int``[] mp = ``new` `int``[n + 2];  ` ` `  `    ``// Firstly fill the boolean  ` `    ``// array with all zeroes  ` `    ``for``(``int` `i = 0; i <= n; i++)  ` `    ``{  ` `        ``mp[i] = 0;  ` `    ``}  ` ` `  `    ``// Mark the 1s  ` `    ``for``(``int` `i = 0; i < k; i++)  ` `    ``{  ` `        ``mp[arr[i]] = 1;  ` `    ``}  ` ` `  `    ``// Number of 0s until first '1' occurs  ` `    ``int` `leftSegment = arr - 1;  ` ` `  `    ``// Maximum Number of 0s in between 2 '1's.  ` `    ``for``(``int` `i = 1; i < k; i++) ` `    ``{  ` `        ``leftSegment = Math.Max(leftSegment,  ` `                               ``arr[i] -  ` `                               ``arr[i - 1] - 1);  ` `    ``}  ` ` `  `    ``// Number of 0s from right until first '1' occurs  ` `    ``int` `rightSegment = n - arr[k - 1];  ` ` `  `    ``// Return maximum from left and right segment  ` `    ``int` `maxSegment = Math.Max(leftSegment, ` `                              ``rightSegment);  ` ` `  `    ``int` `tim;  ` ` `  `    ``// Check if count is odd  ` `    ``if` `((maxSegment & 1) == 1)  ` `        ``tim = (maxSegment / 2) + 1;  ` ` `  `    ``// Check ifcount is even  ` `    ``else` `        ``tim = maxSegment / 2;  ` ` `  `    ``// Return the time  ` `    ``return` `tim;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main () ` `{  ` `     `  `    ``// Initialise N  ` `    ``int` `N = 5;  ` ` `  `    ``// Array initialisation  ` `    ``int``[] arr = { 1, 4 };  ` ` `  `    ``Console.Write(solve(arr, N));  ` `}  ` `} ` ` `  `// This code is contributed by chitranayal `

Output:

```1
```

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