给定一个表示时隙数的整数N ，以及一个数组arr [] ，该数组arr []由[1，N]范围内的K个整数组成。数组的每个元素都在[1，N]范围内，该范围代表已填充插槽的索引。在每个时间单位，带有填充槽的索引将填充相邻的空槽。任务是找到填充所有N个插槽所需的最短时间。
例子：
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
方法：为了解决给定的问题，想法是执行级别顺序遍历 在给定的N个插槽中使用队列。请按照以下步骤解决问题：
- 初始化一个变量，例如time为0 ，并初始化一个辅助数组Visited []，以在每次迭代中标记填充的索引。
- 现在，将队列arr []中给定的已填充插槽的索引推送到队列中 并将其标记为已访问。
- 现在，迭代直到队列不为空，然后执行以下步骤：
- 从队列中删除前端索引i ，如果相邻的插槽(i – 1)和(i + 1)在[1，N]范围内并且未被访问，则将它们标记为已访问并将它们推入队列。
- 将时间增加1 。
- 完成上述步骤后，将(time – 1)的值打印为填充所有插槽所需的最短时间。
下面是上述方法的实现：
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 arr, int N, int K)
{
// Stores visited slots
queue q;
// Checks if a slot is visited or not
vector 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 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 susmitakundugoaldangaC#
// 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 arr, int N, int K)
{
// Stores visited slots
Queue q = new Queue();
// 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 arr = new List() { 2, 6 };
int K = arr.Count;
// Function Call
minTime(arr, N, K);
}
}
// THIS CODE IS CONTRIBUTED BY SURENDRA_GANGWAR. 输出：
2时间复杂度： O(N)
辅助空间： O(N)