给定一个数组arr []表示范围为[1，N]的前N个自然数的排列，每个第i个索引的任务是检查是否存在一个i长度子数组，使其包含所有[1，i]范围内的数字。
注意： 1 –使用基于索引的索引。
例子：
Input: arr[] = {4, 5, 1, 3, 2}
Output: True False True False True
Explanation:
For i = 1, the subarray {arr[2]} contains all the numbers from [1, i] and of size i(=1). Therefore, the required output is True.
For i = 2, no subarray of size 2 exists which contains all the numbers in the range[1, i]. Therefore, the required output is False.
For i = 3, the subarray {arr[2], arr[3], arr[4]} contains all the numbers from [1, i] and of size i(=3). Therefore, the required output is True.
For i = 4, no subarray of size 4 exists which contains all the numbers in the range[1, i]. Therefore, the required output is False.
For i = 5, the subarray {arr[0], arr[1], arr[2], arr[3], arr[4]} contains all the numbers from [1, i] and of size i(=5). Therefore, the required output is True.
Input: arr = {1, 4, 3, 2}
Output: True False False True
天真的方法：想法是遍历数组，对于每个索引，检查是否存在大小为i的子数组，该子数组包含[1，i]范围内的所有数字。如果发现为true，则打印True 。否则，打印False。
时间复杂度： O(N 2 )
辅助空间： O(N)
高效的方法：使用散列可以有效地存储给定数组的每个元素的位置，可以解决该问题。请按照以下步骤解决问题：
- 创建一个地图，例如Map ，以存储给定数组中每个元素的位置。
- 遍历数组并将数组中每个元素的位置存储到Map中。
- 创建一个集合，比如说st，以存储范围[1，N]中每个元素的索引。
- 初始化两个变量，例如Min和Max ，以存储st中存在的最小和最大元素。
- 迭代[1，N]范围，然后将Map [i]的值插入st并检查Max – Min + 1 = i 。如果发现为true，则打印True 。
- 否则，输出False 。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to check if a subarray of size i exists
// that contain all the numbers in the range [1, i]
void checksubarrayExist1_N(int arr[], int N)
{
// Store the position
// of each element of arr[]
unordered_map pos;
// Traverse the array
for (int i = 0; i < N; i++) {
// Insert the position
// of arr[i]
pos[arr[i]] = i;
}
// Store position of each element
// from the range [1, N]
set st;
// Iterate over the range [1, N]
for (int i = 1; i <= N; i++) {
// Insert the index of i into st
st.insert(pos[i]);
// Find the smallest element of st
int Min = *(st.begin());
// Find the largest element of st
int Max = *(st.rbegin());
// If distance between the largest
// and smallest element of arr[]
// till i-th index is equal to i
if (Max - Min + 1 == i) {
cout << "True ";
}
else {
cout << "False ";
}
}
}
// Driver Code
int main()
{
int arr[] = { 1, 4, 3, 2 };
int N = sizeof(arr) / sizeof(arr[0]);
checksubarrayExist1_N(arr, N);
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG {
// Function to check if a subarray of size i exists
// that contain all the numbers in the range [1, i]
static void checksubarrayExist1_N(int arr[], int N)
{
// Store the position
// of each element of arr[]
Map pos=new HashMap<>();
// Traverse the array
for (int i = 0; i < N; i++) {
// Insert the position
// of arr[i]
pos.put(arr[i],i);
}
// Store position of each element
// from the range [1, N]
Set st=new HashSet<>();
// Iterate over the range [1, N]
for (int i = 1; i <= N; i++) {
// Insert the index of i into st
st.add(pos.get(i));
// Find the smallest element of st
int Min = Collections.min(st);
// Find the largest element of st
int Max = Collections.max(st);
// If distance between the largest
// and smallest element of arr[]
// till i-th index is equal to i
if (Max - Min + 1 == i) {
System.out.print("True ");
}
else {
System.out.print("False ");
}
}
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 4, 3, 2 };
int N = arr.length;
checksubarrayExist1_N(arr, N);
}
}
// This code is contributed by offbeat Python3
# Python3 program to implement
# the above approach
# Function to check if a subarray of size i exists
# that contain all the numbers in the range [1, i]
def checksubarrayExist1_N(arr, N):
# Store the position
# of each element of arr[]
pos = {}
# Traverse the array
for i in range(N):
# Insert the position
# of arr[i]
pos[arr[i]] = i
# Store position of each element
# from the range [1, N]
st = {}
# Iterate over the range [1, N]
for i in range(1, N + 1):
# Insert the index of i into st
st[pos[i]] = 1
# Find the smallest element of st
Min = sorted(list(st.keys()))[0]
# Find the largest element of st
Max = sorted(list(st.keys()))[-1]
# If distance between the largest
# and smallest element of arr[]
# till i-th index is equal to i
if (Max - Min + 1 == i):
print("True", end = " ")
else:
print("False", end = " ")
# Driver Code
if __name__ == '__main__':
arr = [1, 4, 3, 2]
N = len(arr)
checksubarrayExist1_N(arr, N)
# This code is contributed by mohit kumar 29.C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
using System.Linq;
class GFG{
// Function to check if a subarray of size i exists
// that contain all the numbers in the range [1, i]
static void checksubarrayExist1_N(int[] arr, int N)
{
// Store the position
// of each element of arr[]
Dictionary pos = new Dictionary();
// Traverse the array
for(int i = 0; i < N; i++)
{
// Insert the position
// of arr[i]
pos[arr[i]] = i;
}
// Store position of each element
// from the range [1, N]
HashSet st = new HashSet();
// Iterate over the range [1, N]
for(int i = 1; i <= N; i++)
{
// Insert the index of i into st
st.Add(pos[i]);
// Find the smallest element of st
int Min = st.Min();
// Find the largest element of st
int Max = st.Max();
// If distance between the largest
// and smallest element of arr[]
// till i-th index is equal to i
if (Max - Min + 1 == i)
{
Console.Write("True ");
}
else
{
Console.Write("False ");
}
}
}
// Driver code
public static void Main(string[] args)
{
int[] arr = { 1, 4, 3, 2 };
int N = arr.Length;
checksubarrayExist1_N(arr, N);
}
}
// This code is contributed by ukasp 输出：
True False False True时间复杂度： O(N * log(N))
辅助空间： O(N)