给定一个由N个整数组成的数组arr [] ，任务是按照逆时针旋转次数最小的非递增顺序对数组进行排序。如果无法对数组进行排序，则打印“ -1” 。否则，打印转数。
例子：
Input: arr[] = {2, 1, 5, 4, 3}
Output: 2
Explanation: Two anti-clockwise rotations are required to sort the array in decreasing order, i.e. {5, 4, 3, 2, 1}
Input: arr[] = {2, 3, 1}
Output: -1
方法：想法是遍历给定的数组arr []并计算满足arr [i + 1]> arr [i]的索引数。请按照以下步骤解决问题：
- 将arr [i +1]> arr [i]的计数存储在变量中，并在arr [i + 1]> arr [i]时存储索引。
- 如果count的值为N – 1 ，则数组以非降序排序。所需的步骤正好是(N – 1) 。
- 如果count的值为0 ，则该数组为array的数组已经以非递增的顺序排序。
- 如果count的值为1且arr [0] ≤arr [N – 1] ，则通过执行所有数字到该索引的移位，所需的roattaions数量等于(index +1) 。另外，检查arr [0]≤arr [N – 1]以确保序列不递增。
- 否则，将无法以非递增顺序对数组进行排序。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to count minimum anti-
// clockwise rotations required to
// sort the array in non-increasing order
void minMovesToSort(int arr[], int N)
{
// Stores count of arr[i + 1] > arr[i]
int count = 0;
// Store last index of arr[i+1] > arr[i]
int index;
// Traverse the given array
for (int i = 0; i < N - 1; i++) {
// If the adjacent elements are
// in increasing order
if (arr[i] < arr[i + 1]) {
// Increment count
count++;
// Update index
index = i;
}
}
// Print the result according
// to the following conditions
if (count == 0) {
cout << "0";
}
else if (count == N - 1) {
cout << N - 1;
}
else if (count == 1
&& arr[0] <= arr[N - 1]) {
cout << index + 1;
}
// Otherwise, it is not
// possible to sort the array
else {
cout << "-1";
}
}
// Driver Code
int main()
{
// Given array
int arr[] = { 2, 1, 5, 4, 2 };
int N = sizeof(arr)
/ sizeof(arr[0]);
// Function Call
minMovesToSort(arr, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to count minimum anti-
// clockwise rotations required to
// sort the array in non-increasing order
static void minMovesToSort(int arr[], int N)
{
// Stores count of arr[i + 1] > arr[i]
int count = 0;
// Store last index of arr[i+1] > arr[i]
int index = 0;
// Traverse the given array
for(int i = 0; i < N - 1; i++)
{
// If the adjacent elements are
// in increasing order
if (arr[i] < arr[i + 1])
{
// Increment count
count++;
// Update index
index = i;
}
}
// Print the result according
// to the following conditions
if (count == 0)
{
System.out.print("0");
}
else if (count == N - 1)
{
System.out.print(N - 1);
}
else if (count == 1 &&
arr[0] <= arr[N - 1])
{
System.out.print(index + 1);
}
// Otherwise, it is not
// possible to sort the array
else
{
System.out.print("-1");
}
}
// Driver Code
public static void main(String[] args)
{
// Given array
int[] arr = { 2, 1, 5, 4, 2 };
int N = arr.length;
// Function Call
minMovesToSort(arr, N);
}
}
// This code is contributed by susmitakundugoaldangaPython3
# Python program for the above approach
# Function to count minimum anti-
# clockwise rotations required to
# sort the array in non-increasing order
def minMovesToSort(arr, N) :
# Stores count of arr[i + 1] > arr[i]
count = 0
# Store last index of arr[i+1] > arr[i]
index = 0
# Traverse the given array
for i in range(N-1):
# If the adjacent elements are
# in increasing order
if (arr[i] < arr[i + 1]) :
# Increment count
count += 1
# Update index
index = i
# Prthe result according
# to the following conditions
if (count == 0) :
print("0")
elif (count == N - 1) :
print( N - 1)
elif (count == 1
and arr[0] <= arr[N - 1]) :
print(index + 1)
# Otherwise, it is not
# possible to sort the array
else :
print("-1")
# Driver Code
# Given array
arr = [ 2, 1, 5, 4, 2 ]
N = len(arr)
# Function Call
minMovesToSort(arr, N)
# This code i contributed by sanjoy_62.C#
// C# program for the above approach
using System;
class GFG{
// Function to count minimum anti-
// clockwise rotations required to
// sort the array in non-increasing order
static void minMovesToSort(int[] arr, int N)
{
// Stores count of arr[i + 1] > arr[i]
int count = 0;
// Store last index of arr[i+1] > arr[i]
int index = 0;
// Traverse the given array
for(int i = 0; i < N - 1; i++)
{
// If the adjacent elements are
// in increasing order
if (arr[i] < arr[i + 1])
{
// Increment count
count++;
// Update index
index = i;
}
}
// Print the result according
// to the following conditions
if (count == 0)
{
Console.Write("0");
}
else if (count == N - 1)
{
Console.Write(N - 1);
}
else if (count == 1 &&
arr[0] <= arr[N - 1])
{
Console.Write(index + 1);
}
// Otherwise, it is not
// possible to sort the array
else
{
Console.Write("-1");
}
}
// Driver Code
public static void Main()
{
// Given array
int[] arr = { 2, 1, 5, 4, 2 };
int N = arr.Length;
// Function Call
minMovesToSort(arr, N);
}
}
// This code is contributed by code_hunt输出：
2时间复杂度： O(N)
辅助空间： O(1)