# Minimum time required to fill given N slots

Last Updated : 10 Oct, 2023

Given an integer N which denotes the number of slots, and an array arr[] consisting of K integers in the range [1, N] . Each element of the array are in the range [1, N] which represents the indices of the filled slots. At each unit of time, the index with filled slot fills the adjacent empty slots. The task is to find the minimum time taken to fill all the N slots.

Examples:

Input: N = 6, K = 2, arr[] = {2, 6}
Output: 2
Explanation:
Initially there are 6 slots and the indices of the filled slots are slots[] = {0, 2, 0, 0, 0, 6}, where 0 represents unfilled.
After 1 unit of time, slots[] = {1, 2, 3, 0, 5, 6}
After 2 units of time, slots[] = {1, 2, 3, 4, 5, 6}
Therefore, the minimum time required is 2.

Input: N = 5, K = 5, arr[] = {1, 2, 3, 4, 5}
Output: 0

## Minimum time required to fill given N slots using Level Order Traversal:

To solve the given problem, the idea is to perform Level Order Traversal on the given N slots using a Queue. Follow the steps below to solve the problem:

• Initialize a variable, say time as 0, and an auxiliary array visited[] to mark the filled indices in each iteration.
• Now, push the indices of filled slots given in array arr[] in a queue and mark them as visited.
• Now, iterate until the queue is not empty and perform the following steps:
• Remove the front index i from the queue and if the adjacent slots (i – 1) and (i + 1) are in the range [1, N] and are unvisited, then mark them as visited and push them into the queue.
• If any of the non visited index becomes visited in the current process Increment the time by 1 .
• After completing the above steps, print the value of (time – 1) as the minimum time required to fill all the slots.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include `   `using` `namespace` `std;`   `// Function to return the minimum` `// time to fill all the slots` `void` `minTime(vector<``int``> arr, ``int` `N, ``int` `K)` `{` `    `  `    ``// Stores visited slots` `    ``queue<``int``> q;` `    `  `    ``// Checks if a slot is visited or not` `    ``vector<``bool``> vis(N + 1, ``false``);`   `    ``int` `time` `= 0;`   `    ``// Insert all filled slots` `    ``for` `(``int` `i = 0; i < K; i++) {`   `        ``q.push(arr[i]);` `        ``vis[arr[i]] = ``true``;` `    ``}`   `    ``// Iterate until queue is` `    ``// not empty` `    ``while` `(q.size() > 0) {`   `        ``// Iterate through all slots` `        ``// in the queue` `          ``bool` `op = ``false``; ` `        ``for` `(``int` `i = 0; i < q.size(); i++) {`   `            ``// Front index` `            ``int` `curr = q.front();` `            ``q.pop();`   `            ``// If previous slot is` `            ``// present and not visited` `            ``if` `(curr - 1 >= 1 && ` `                ``!vis[curr - 1]) {` `                  ``op = ``true``;` `                ``vis[curr - 1] = ``true``;` `                ``q.push(curr - 1);` `            ``}`   `            ``// If next slot is present` `            ``// and not visited` `            ``if` `(curr + 1 <= N && ` `                ``!vis[curr + 1]) {` `                ``op = ``true``;` `                ``vis[curr + 1] = ``true``;` `                ``q.push(curr + 1);` `            ``}` `        ``}`   `        ``// Increment the time` `        ``// at each level` `        ``time``+=op;` `    ``}`   `    ``// Print the answer` `    ``cout << (``time``);` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 6;` `    ``vector<``int``> arr = { 2,6 };` `    ``int` `K = arr.size();`   `    ``// Function Call` `    ``minTime(arr, N, K);` `}`

## Java

 `// Java program for the above approach`   `import` `java.io.*;` `import` `java.util.*;` `class` `GFG {`   `    ``// Function to return the minimum` `    ``// time to fill all the slots` `    ``public` `static` `void` `minTime(``int` `arr[], ` `                               ``int` `N, ``int` `K)` `    ``{` `        `  `        ``// Stores visited slots` `        ``Queue q = ``new` `LinkedList<>();`   `        ``// Checks if a slot is visited or not` `        ``boolean` `vis[] = ``new` `boolean``[N + ``1``];` `        ``int` `time = ``0``;`   `        ``// Insert all filled slots` `        ``for` `(``int` `i = ``0``; i < K; i++) {`   `            ``q.add(arr[i]);` `            ``vis[arr[i]] = ``true``;` `        ``}`   `        ``// Iterate until queue is` `        ``// not empty` `        ``while` `(q.size() > ``0``) {`   `            ``// Iterate through all slots` `            ``// in the queue` `            ``for` `(``int` `i = ``0``; i < q.size(); i++) {`   `                ``// Front index` `                ``int` `curr = q.poll();`   `                ``// If previous slot is` `                ``// present and not visited` `                ``if` `(curr - ``1` `>= ``1` `&& ` `                    ``!vis[curr - ``1``]) {` `                    ``vis[curr - ``1``] = ``true``;` `                    ``q.add(curr - ``1``);` `                ``}`   `                ``// If next slot is present` `                ``// and not visited` `                ``if` `(curr + ``1` `<= N && ` `                    ``!vis[curr + ``1``]) {`   `                    ``vis[curr + ``1``] = ``true``;` `                    ``q.add(curr + ``1``);` `                ``}` `            ``}`   `            ``// Increment the time` `            ``// at each level` `            ``time++;` `        ``}`   `        ``// Print the answer` `        ``System.out.println(time - ``1``);` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `N = ``6``;` `        ``int` `arr[] = { ``2``, ``6` `};` `        ``int` `K = arr.length;`   `        ``// Function Call` `        ``minTime(arr, N, K);` `    ``}` `}`

## Python3

 `# Python3 program for the above approach`   `# Function to return the minimum` `# time to fill all the slots` `def` `minTime(arr, N, K):` `    `  `    ``# Stores visited slots` `    ``q ``=` `[]` `    `  `    ``# Checks if a slot is visited or not` `    ``vis ``=` `[``False``] ``*` `(N ``+` `1``)`   `    ``time ``=` `0`   `    ``# Insert all filled slots` `    ``for` `i ``in` `range``(K):` `        ``q.append(arr[i])` `        ``vis[arr[i]] ``=` `True` `        `  `    ``# Iterate until queue is` `    ``# not empty` `    ``while` `(``len``(q) > ``0``):` `        `  `        ``# Iterate through all slots` `        ``# in the queue` `        ``for` `i ``in` `range``(``len``(q)):` `            `  `            ``# Front index` `            ``curr ``=` `q[``0``]` `            ``q.pop(``0``)`   `            ``# If previous slot is` `            ``# present and not visited` `            ``if` `(curr ``-` `1` `>``=` `1` `and` `vis[curr ``-` `1``] ``=``=` `0``):` `                ``vis[curr ``-` `1``] ``=` `True` `                ``q.append(curr ``-` `1``)` `            `  `            ``# If next slot is present` `            ``# and not visited` `            ``if` `(curr ``+` `1` `<``=` `N ``and` `vis[curr ``+` `1``] ``=``=` `0``):` `                ``vis[curr ``+` `1``] ``=` `True` `                ``q.append(curr ``+` `1``)` `            `  `        ``# Increment the time` `        ``# at each level` `        ``time ``+``=` `1` `    `  `    ``# Print the answer` `    ``print``(time ``-` `1``)`   `# Driver Code` `N ``=` `6` `arr ``=` `[ ``2``, ``6` `]` `K ``=` `len``(arr)`   `# Function Call` `minTime(arr, N, K)`   `# This code is contributed by susmitakundugoaldanga`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG` `{` `  `  `// Function to return the minimum` `// time to fill all the slots` `static` `void` `minTime(List<``int``> arr, ``int` `N, ``int` `K)` `{` `    `  `    ``// Stores visited slots` `    ``Queue<``int``> q = ``new` `Queue<``int``>();` `    `  `    ``// Checks if a slot is visited or not` `    ``int` `[]vis = ``new` `int``[N + 1];` `    ``Array.Clear(vis, 0, vis.Length);` `    ``int` `time = 0;`   `    ``// Insert all filled slots` `    ``for` `(``int` `i = 0; i < K; i++) ` `    ``{` `        ``q.Enqueue(arr[i]);` `        ``vis[arr[i]] = 1;` `    ``}`   `    ``// Iterate until queue is` `    ``// not empty` `    ``while` `(q.Count > 0) ` `    ``{`   `        ``// Iterate through all slots` `        ``// in the queue` `        ``for` `(``int` `i = 0; i < q.Count; i++) ` `        ``{`   `            ``// Front index` `            ``int` `curr = q.Peek();` `            ``q.Dequeue();`   `            ``// If previous slot is` `            ``// present and not visited` `            ``if` `(curr - 1 >= 1 && ` `                ``vis[curr - 1]==0)` `            ``{` `                ``vis[curr - 1] = 1;` `                ``q.Enqueue(curr - 1);` `            ``}`   `            ``// If next slot is present` `            ``// and not visited` `            ``if` `(curr + 1 <= N && ` `                ``vis[curr + 1] == 0) ` `            ``{`   `                ``vis[curr + 1] = 1;` `                ``q.Enqueue(curr + 1);` `            ``}` `        ``}`   `        ``// Increment the time` `        ``// at each level` `        ``time++;` `    ``}`   `    ``// Print the answer` `    ``Console.WriteLine(time-1);` `}`   `// Driver Code` `public` `static` `void` `Main()` `{` `    ``int` `N = 6;` `    ``List<``int``> arr = ``new` `List<``int``>() { 2, 6 };` `    ``int` `K = arr.Count;`   `    ``// Function Call` `    ``minTime(arr, N, K);` `}` `}`   `// THIS CODE IS CONTRIBUTED BY SURENDRA_GANGWAR.`

## Javascript

 ``

Time Complexity: O(N)
Auxiliary Space: O(N)

## Minimum time required to fill given N slots using Sorting:

Another approach to solve this problem can be done by sorting the array of filled slots and then iterating through the array to find the maximum distance between adjacent filled slots. The minimum time required to fill all the slots would be equal to the maximum distance between adjacent filled slots.

• Declare a function minTime() which takes three arguments: vector of integers arr, integer N, and integer K.
• Sort the vector arr[] using the sort function from STL.
• Declare an integer variable maxDist and initialize it to zero.
• Traverse the sorted array arr and find the maximum distance between adjacent slots, store this distance in the maxDist variable.
• Check the distance from the first and last slot to the closest filled slot and store the maximum of these two distances in the maxDist variable.
• Subtract 1 from maxDist to account for the time it takes to fill each slot.
• Print the value of maxDist.

Below is the implementation of this approach:

## C++

 `// C++ program for the above approach` `#include `   `using` `namespace` `std;`   `// Function to return the minimum` `// time to fill all the slots` `void` `minTime(vector<``int``> arr, ``int` `N, ``int` `K)` `{` `    ``// Sort the array` `    ``sort(arr.begin(), arr.end());` `    ``int` `maxDist = 0;`   `    ``// Find maximum distance` `    ``// between adjacent slots` `    ``for` `(``int` `i = 1; i < K; i++) {` `        ``maxDist = max(maxDist, arr[i] - arr[i - 1] - 1);` `    ``}` `    ``// an area is being filled with both sides it will take` `    ``// half time to fill that area` `    ``maxDist = (maxDist + 1) / 2;` `  `  `    ``// Check the distance from the` `    ``// first and last slot to the` `    ``// closest filled slot`   `    ``maxDist = max(maxDist, arr[0] - 1);` `    ``maxDist = max(maxDist, N - arr[K - 1]);`   `    ``// Print the answer` `    ``cout << maxDist;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 6;` `    ``vector<``int``> arr = { 2, 6 };` `    ``int` `K = arr.size();` `    ``// Function Call` `    ``minTime(arr, N, K);` `}`

## Java

 `import` `java.util.*;`   `public` `class` `Main {`   `    ``// Function to return the minimum` `    ``// time to fill all the slots` `    ``static` `void` `minTime(List arr, ``int` `N, ``int` `K) {` `        ``// Sort the array` `        ``Collections.sort(arr);` `        ``int` `maxDist = ``0``;`   `        ``// Find maximum distance` `        ``// between adjacent slots` `        ``for` `(``int` `i = ``1``; i < K; i++) {` `            ``maxDist = Math.max(maxDist, arr.get(i) - arr.get(i - ``1``) - ``1``);` `        ``}`   `        ``// Check the distance from the` `        ``// first and last slot to the` `        ``// closest filled slot` `        ``maxDist = Math.max(maxDist, arr.get(``0``) - ``1``);` `        ``maxDist = Math.max(maxDist, N - arr.get(K - ``1``));`   `        ``// Subtract 1 from maxDist to` `        ``// account for the time it takes` `        ``// to fill each slot` `        ``maxDist--;`   `        ``// Print the answer` `        ``System.out.println(maxDist);` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args) {` `        ``int` `N = ``6``;` `        ``List arr = ``new` `ArrayList();` `        ``arr.add(``2``);` `        ``arr.add(``6``);` `        ``int` `K = arr.size();` `        ``// Function Call` `        ``minTime(arr, N, K);` `    ``}` `}`

## Python

 `# Function to return the minimum` `# time to fill all the slots` `def` `minTime(arr, N, K):` `    ``# Sort the array` `    ``arr.sort()` `    ``maxDist ``=` `0`   `    ``# Find maximum distance` `    ``# between adjacent slots` `    ``for` `i ``in` `range``(``1``, K):` `        ``maxDist ``=` `max``(maxDist, arr[i] ``-` `arr[i ``-` `1``] ``-` `1``)`   `    ``# Check the distance from the` `    ``# first and last slot to the` `    ``# closest filled slot` `    ``maxDist ``=` `max``(maxDist, arr[``0``] ``-` `1``)` `    ``maxDist ``=` `max``(maxDist, N ``-` `arr[K ``-` `1``])`   `    ``# Subtract 1 from maxDist to` `    ``# account for the time it takes` `    ``# to fill each slot` `    ``maxDist ``-``=` `1`   `    ``# Print the answer` `    ``print``(maxDist)`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `    ``N ``=` `6` `    ``arr ``=` `[``2``, ``6``]` `    ``K ``=` `len``(arr)` `    ``# Function Call` `    ``minTime(arr, N, K)`

## C#

 `using` `System;` `using` `System.Collections.Generic;` `using` `System.Linq;`   `public` `class` `Program` `{` `    ``// Function to return the minimum` `    ``// time to fill all the slots` `    ``public` `static` `void` `MinTime(List<``int``> arr, ``int` `N, ``int` `K)` `    ``{` `        ``// Sort the array` `        ``arr.Sort();` `        ``int` `maxDist = 0;`   `        ``// Find maximum distance` `        ``// between adjacent slots` `        ``for` `(``int` `i = 1; i < K; i++)` `        ``{` `            ``maxDist = Math.Max(maxDist, arr[i] - arr[i - 1] - 1);` `        ``}`   `        ``// Check the distance from the` `        ``// first and last slot to the` `        ``// closest filled slot` `        ``maxDist = Math.Max(maxDist, arr[0] - 1);` `        ``maxDist = Math.Max(maxDist, N - arr[K - 1]);`   `        ``// Subtract 1 from maxDist to` `        ``// account for the time it takes` `        ``// to fill each slot` `        ``maxDist--;`   `        ``// Print the answer` `        ``Console.WriteLine(maxDist);` `    ``}`   `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `N = 6;` `        ``List<``int``> arr = ``new` `List<``int``> { 2, 6 };` `        ``int` `K = arr.Count;` `        ``// Function Call` `        ``MinTime(arr, N, K);` `    ``}` `}`

## Javascript

 `// Function to calculate the minimum time to fill all slots` `function` `minTime(arr, N, K) {` `    ``// Sort the array of slots in ascending order` `    ``arr.sort((a, b) => a - b);` `    ``let maxDist = 0;`   `    ``// Calculate the maximum distance between adjacent slots` `    ``for` `(let i = 1; i < K; i++) {` `        ``maxDist = Math.max(maxDist, arr[i] - arr[i - 1] - 1);` `    ``}`   `    ``// Check the distance from the first and last slot to the closest filled slot` `    ``maxDist = Math.max(maxDist, arr[0] - 1);` `    ``maxDist = Math.max(maxDist, N - arr[K - 1]);`   `    ``// Subtract 1 from maxDist to account for the time it takes to fill each slot` `    ``maxDist--;`   `    ``// Print the calculated minimum time` `    ``console.log(maxDist);` `}` `//Driver code`   `// Total number of slots` `const N = 6;`   `// Array representing filled slots` `const arr = [2, 6];`   `// Number of filled slots (K)` `const K = arr.length;`   `// Call the minTime function with the given inputs` `minTime(arr, N, K);`

Output:-

```2

```

Time Complexity: O(KlogK), then it loops through the vector once to calculate the maximum distance between adjacent slots and twice to calculate the distance from the first and last slot to the closest filled slot, which takes O(K) time. Therefore, the overall time complexity is O(KlogK).
Auxiliary Space: O(1).