排序给定阵列需要重新排列的子阵列数量
原文:https://www . geeksforgeeks . org/子数组数量-需要重新排列才能对给定数组进行排序/
给定一个由第一个 N 自然数组成的数组arr【】，任务是找到需要重新排列的最小数量的子数组，从而对最终的数组进行排序。
示例:
输入: arr[] = {2，1，4，3，5} 输出: 1 解释: 操作 1: 选择子阵列{arr[0]，arr[3]}，即{2，1，4，3 }。将此子阵列的元素重新排列为{1，2，3，4}。数组修改为{1，2，3，4，5}。
输入: arr[] = {5，2，3，4，1} 输出: 3
方法:给定的问题可以通过观察以下场景来解决:
- 如果给定数组 arr[] 已经排序，则打印 0 。
- 如果第一个和最后一个元素分别是 1 和 N ，那么只需要对 1 个子阵列arr【1，N–2】或arr【2，N–1】进行排序。因此，打印 1 。
- 如果第一个和最后一个元素分别是 N 和 1 ，则需要对 3 个子阵列即arr【0，N–2】、arr【1，N–1】、arr【0，1】进行排序。因此，打印 3 。
- 否则，对两个子阵列进行排序，即 arr[1，N–1]，和 arr[0，N–2]【T3]。
因此，打印需要重新排列的最小子阵列数。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the number
// of subarrays required to be
// rearranged to sort the given array
void countSubarray(int arr[], int n)
{
// Base Case
int ans = 2;
// Check if the given array is
// already sorted
if (is_sorted(arr, arr + n)) {
ans = 0;
}
// Check if the first element of
// array is 1 or last element is
// equal to size of array
else if (arr[0] == 1
|| arr[n - 1] == n) {
ans = 1;
}
else if (arr[0] == n
&& arr[n - 1] == 1) {
ans = 3;
}
// Print the required answer
cout << ans;
}
// Driver Code
int main()
{
int arr[] = { 5, 2, 3, 4, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
countSubarray(arr, N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
class GFG{
// Function that returns 0 if a pair
// is found unsorted
static int arraySortedOrNot(int arr[], int n)
{
// Array has one or no element or the
// rest are already checked and approved.
if (n == 1 || n == 0)
return 1;
// Unsorted pair found (Equal values allowed)
if (arr[n - 1] < arr[n - 2])
return 0;
// Last pair was sorted
// Keep on checking
return arraySortedOrNot(arr, n - 1);
}
// Function to count the number
// of subarrays required to be
// rearranged to sort the given array
static void countSubarray(int arr[], int n)
{
// Base Case
int ans = 2;
// Check if the given array is
// already sorted
if (arraySortedOrNot(arr, arr.length) != 0)
{
ans = 0;
}
// Check if the first element of
// array is 1 or last element is
// equal to size of array
else if (arr[0] == 1 ||
arr[n - 1] == n)
{
ans = 1;
}
else if (arr[0] == n &&
arr[n - 1] == 1)
{
ans = 3;
}
// Print the required answer
System.out.print(ans);
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 5, 2, 3, 4, 1 };
int N = arr.length;
countSubarray(arr, N);
}
}
// This code is contributed by susmitakundugoaldanga
Python 3
# Python3 program for the above approach
# Function to count the number
# of subarrays required to be
# rearranged to sort the given array
def countSubarray(arr, n):
# Base Case
ans = 2
# Check if the given array is
# already sorted
if (sorted(arr) == arr):
ans = 0
# Check if the first element of
# array is 1 or last element is
# equal to size of array
elif (arr[0] == 1 or arr[n - 1] == n):
ans = 1
elif (arr[0] == n and arr[n - 1] == 1):
ans = 3
# Print the required answer
print(ans)
# Driver Code
arr = [ 5, 2, 3, 4, 1 ]
N = len(arr)
countSubarray(arr, N)
# This code is contributed by amreshkumar3
C
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function that returns 0 if a pair
// is found unsorted
static int arraySortedOrNot(int []arr, int n)
{
// Array has one or no element or the
// rest are already checked and approved.
if (n == 1 || n == 0)
return 1;
// Unsorted pair found (Equal values allowed)
if (arr[n - 1] < arr[n - 2])
return 0;
// Last pair was sorted
// Keep on checking
return arraySortedOrNot(arr, n - 1);
}
// Function to count the number
// of subarrays required to be
// rearranged to sort the given array
static void countSubarray(int []arr, int n)
{
// Base Case
int ans = 2;
// Check if the given array is
// already sorted
if (arraySortedOrNot(arr, arr.Length) != 0)
{
ans = 0;
}
// Check if the first element of
// array is 1 or last element is
// equal to size of array
else if (arr[0] == 1 ||
arr[n - 1] == n)
{
ans = 1;
}
else if (arr[0] == n &&
arr[n - 1] == 1)
{
ans = 3;
}
// Print the required answer
Console.Write(ans);
}
// Driver Code
public static void Main()
{
int []arr = { 5, 2, 3, 4, 1 };
int N = arr.Length;
countSubarray(arr, N);
}
}
// This code is contributed by bgangwar59
java 描述语言
<script>
// Javascript program for the above approach
// Function that returns 0 if a pair
// is found unsorted
function arraySortedOrNot(arr, n)
{
// Array has one or no element or the
// rest are already checked and approved.
if (n == 1 || n == 0)
return 1;
// Unsorted pair found (Equal values allowed)
if (arr[n - 1] < arr[n - 2])
return 0;
// Last pair was sorted
// Keep on checking
return arraySortedOrNot(arr, n - 1);
}
// Function to count the number
// of subarrays required to be
// rearranged to sort the given array
function countSubarray(arr, n)
{
// Base Case
var ans = 2;
// Check if the given array is
// already sorted
if (arraySortedOrNot(arr, arr.length) != 0)
{
ans = 0;
}
// Check if the first element of
// array is 1 or last element is
// equal to size of array
else if (arr[0] == 1 ||
arr[n - 1] == n)
{
ans = 1;
}
else if (arr[0] == n &&
arr[n - 1] == 1)
{
ans = 3;
}
// Print the required answer
document.write(ans);
}
// Driver Code
var arr = [ 5, 2, 3, 4, 1 ];
var N = arr.length;
countSubarray(arr, N);
// This code is contributed by SURENDRA_GANGWAR
</script>
Output:
3
时间复杂度:O(N) T5辅助空间:** O(1)
版权属于：月萌API www.moonapi.com，转载请注明出处
