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# Minimum time required to fill given N slots

• Last Updated : 30 Nov, 2022

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: 1

Approach: 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.
• 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` `        ``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]) {` `                ``vis[curr - 1] = ``true``;` `                ``q.push(curr - 1);` `            ``}`   `            ``// If next slot is present` `            ``// and not visited` `            ``if` `(curr + 1 <= N && ` `                ``!vis[curr + 1]) {`   `                ``vis[curr + 1] = ``true``;` `                ``q.push(curr + 1);` `            ``}` `        ``}`   `        ``// Increment the time` `        ``// at each level` `        ``time``++;` `    ``}`   `    ``// Print the answer` `    ``cout << (``time` `- 1);` `}`   `// 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

 ``

Output:

`2`

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

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