Python3程序计算以非递增顺序对给定数组进行排序所需的旋转
给定一个由N个整数组成的数组arr[] ,任务是通过最小逆时针旋转次数以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,打印旋转计数。
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 ,则数组按非递减顺序排序。所需的步骤正是(N – 1) 。
- 如果count的值为0 ,则数组已经按非递增顺序排序。
- 如果count的值为1并且arr[0] ≤ arr[N – 1] ,则所需的旋转次数等于(index + 1) ,方法是将所有数字移动到该索引。此外,检查arr[0] ≤ arr[N – 1]以确保序列是否不递增。
- 否则,无法按非递增顺序对数组进行排序。
下面是上述方法的实现:
Python3
# 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
# Print 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.
输出:
2
时间复杂度: O(N)
辅助空间: O(1)
有关更多详细信息,请参阅有关以非递增顺序对给定数组进行排序所需的计数旋转的完整文章!