给定非负整数K 、 M和由N 个元素组成的数组arr[ ] ,任务是在K 次右旋转后找到数组的第M个元素。
例子:
Input: arr[] = {3, 4, 5, 23}, K = 2, M = 1
Output: 5
Explanation:
The array after first right rotation a1[ ] = {23, 3, 4, 5}
The array after second right rotation a2[ ] = {5, 23, 3, 4}
1st element after 2 right rotations is 5.
Input: arr[] = {1, 2, 3, 4, 5}, K = 3, M = 2
Output: 4
Explanation:
The array after 3 right rotations has 4 at its second position.
天真的方法:
解决问题最简单的方法是进行K次右旋操作,然后找到最终数组的第M个元素。
时间复杂度: O(N * K)
辅助空间: O(N)
有效的方法:
为了优化问题,需要进行以下观察:
- 如果数组旋转N次,它会再次返回初始数组。
For example, a[ ] = {1, 2, 3, 4, 5}, K=5
Modified array after 5 right rotation a5[ ] = {1, 2, 3, 4, 5}.
- 因此,经过第K次旋转后的数组中的元素与原始数组中索引为K%N处的元素相同。
- 如果K >= M ,则 K 次右旋后数组的第 M 个元素为
{ (N-K) + (M-1) } th element in the original array.
- 如果K < M ,则 K 次右旋后数组的第 M 个元素为:
(M – K – 1) th element in the original array.
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to return Mth element of
// array after k right rotations
int getFirstElement(int a[], int N,
int K, int M)
{
// The array comes to original state
// after N rotations
K %= N;
int index;
// If K is greater or equal to M
if (K >= M)
// Mth element after k right
// rotations is (N-K)+(M-1) th
// element of the array
index = (N - K) + (M - 1);
// Otherwise
else
// (M - K - 1) th element
// of the array
index = (M - K - 1);
int result = a[index];
// Return the result
return result;
}
// Driver Code
int main()
{
int a[] = { 1, 2, 3, 4, 5 };
int N = sizeof(a) / sizeof(a[0]);
int K = 3, M = 2;
cout << getFirstElement(a, N, K, M);
return 0;
}
Java
// Java program to implement
// the above approach
class GFG{
// Function to return Mth element of
// array after k right rotations
static int getFirstElement(int a[], int N,
int K, int M)
{
// The array comes to original state
// after N rotations
K %= N;
int index;
// If K is greater or equal to M
if (K >= M)
// Mth element after k right
// rotations is (N-K)+(M-1) th
// element of the array
index = (N - K) + (M - 1);
// Otherwise
else
// (M - K - 1) th element
// of the array
index = (M - K - 1);
int result = a[index];
// Return the result
return result;
}
// Driver Code
public static void main(String[] args)
{
int a[] = { 1, 2, 3, 4, 5 };
int N = 5;
int K = 3, M = 2;
System.out.println(getFirstElement(a, N, K, M));
}
}
// This code is contributed by Ritik Bansal
Python3
# Python3 program to implement
# the above approach
# Function to return Mth element of
# array after k right rotations
def getFirstElement(a, N, K, M):
# The array comes to original state
# after N rotations
K %= N
# If K is greater or equal to M
if (K >= M):
# Mth element after k right
# rotations is (N-K)+(M-1) th
# element of the array
index = (N - K) + (M - 1)
# Otherwise
else:
# (M - K - 1) th element
# of the array
index = (M - K - 1)
result = a[index]
# Return the result
return result
# Driver Code
if __name__ == "__main__":
a = [ 1, 2, 3, 4, 5 ]
N = len(a)
K , M = 3, 2
print( getFirstElement(a, N, K, M))
# This code is contributed by chitranayal
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to return Mth element of
// array after k right rotations
static int getFirstElement(int []a, int N,
int K, int M)
{
// The array comes to original state
// after N rotations
K %= N;
int index;
// If K is greater or equal to M
if (K >= M)
// Mth element after k right
// rotations is (N-K)+(M-1) th
// element of the array
index = (N - K) + (M - 1);
// Otherwise
else
// (M - K - 1) th element
// of the array
index = (M - K - 1);
int result = a[index];
// Return the result
return result;
}
// Driver Code
public static void Main()
{
int []a = { 1, 2, 3, 4, 5 };
int N = 5;
int K = 3, M = 2;
Console.Write(getFirstElement(a, N, K, M));
}
}
// This code is contributed by Code_Mech
Javascript
4
时间复杂度: O(1)
辅助空间: O(1)
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