Minimum time required to fill given N slots

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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 <bits/stdc++.h>
 
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<Integer> 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

<script>
 
// Javascript program for the above approach
 
// Function to return the minimum
// time to fill all the slots
function minTime(arr, N, K)
{
     
    // Stores visited slots
    var q = [];
     
    // Checks if a slot is visited or not
    var vis = Array(N + 1).fill(false);
 
    var time = 0;
 
    // Insert all filled slots
    for (var i = 0; i < K; i++) {
 
        q.push(arr[i]);
        vis[arr[i]] = true;
    }
 
    // Iterate until queue is
    // not empty
    while (q.length > 0) {
 
        // Iterate through all slots
        // in the queue
        for (var i = 0; i < q.length; i++) {
 
            // Front index
            var curr = q[0];
            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
    document.write(time - 1);
}
 
// Driver Code
var N = 6;
var arr = [2, 6];
var K = arr.length;
 
// Function Call
minTime(arr, N, K);
 
// This code is contributed by noob2000.
</script>

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 <bits/stdc++.h>
 
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<Integer> 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<Integer> arr = new ArrayList<Integer>();
        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:-

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).

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