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++ 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 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 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# 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. |
<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++ 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);
} |
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);
}
} |
# 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)
|
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);
}
} |
// 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).