Minimum time required to fill given N slots
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 <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 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<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> |
2
Time Complexity: O(N)
Auxiliary Space: O(N)
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