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 slotsvoid 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 Codeint 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 approachimport 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 slotsdef 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 CodeN = 6arr = [ 2, 6 ]K = len(arr)# Function CallminTime(arr, N, K)# This code is contributed by susmitakundugoaldanga |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{ // Function to return the minimum// time to fill all the slotsstatic 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 Codepublic 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 slotsfunction 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 Codevar N = 6;var arr = [2, 6];var K = arr.length;// Function CallminTime(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 slotsvoid 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 Codeint 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 slotsdef 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 Codeif __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 slotsfunction 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 slotsconst N = 6;// Array representing filled slotsconst arr = [2, 6];// Number of filled slots (K)const K = arr.length;// Call the minTime function with the given inputsminTime(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).
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