给定非负整数K , M和具有N个元素的数组arr [] ,在K个左旋转之后,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 left rotation a1[ ] = {4, 5, 23, 3}
The array after second left rotation a2[ ] = {5, 23, 3, 4}
st element after 2 left rotations is 5.
Input: arr[] = {1, 2, 3, 4, 5}, K = 3, M = 2
Output: 5
Explanation:
The array after 3 left rotation has 5 at its second position.
天真的方法:想法是执行K次左旋转操作,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)
高效方法:要优化问题,请注意以下几点:
- 如果将数组旋转N次,它将再次返回初始数组。
For example, a[ ] = {1, 2, 3, 4, 5}, K=5 then the array after 5 left rotation a5[ ] = {1, 2, 3, 4, 5}.
因此,第K次旋转后数组中的元素与原始数组中索引为K%N的元素相同。
- 左旋转K后,数组的第M个元素为
{ (K + M – 1) % N }th
原始数组中的元素。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to return Mth element of
// array after k left rotations
int getFirstElement(int a[], int N,
int K, int M)
{
// The array comes to original state
// after N rotations
K %= N;
// Mth element after k left rotations
// is (K+M-1)%N th element of the
// original array
int index = (K + M - 1) % N;
int result = a[index];
// Return the result
return result;
}
// Driver Code
int main()
{
// Array initialization
int a[] = { 3, 4, 5, 23 };
// Size of the array
int N = sizeof(a) / sizeof(a[0]);
// Given K rotation and Mth element
// to be found after K rotation
int K = 2, M = 1;
// Function call
cout << getFirstElement(a, N, K, M);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to return Mth element of
// array after k left rotations
public static int getFirstElement(int[] a, int N,
int K, int M)
{
// The array comes to original state
// after N rotations
K %= N;
// Mth element after k left rotations
// is (K+M-1)%N th element of the
// original array
int index = (K + M - 1) % N;
int result = a[index];
// Return the result
return result;
}
// Driver code
public static void main(String[] args)
{
// Array initialization
int a[] = { 3, 4, 5, 23 };
// Size of the array
int N = a.length;
// Given K rotation and Mth element
// to be found after K rotation
int K = 2, M = 1;
// Function call
System.out.println(getFirstElement(a, N, K, M));
}
}
// This code is contributed by divyeshrabadiya07
Python3
# Python3 program for the above approach
# Function to return Mth element of
# array after k left rotations
def getFirstElement(a, N, K, M):
# The array comes to original state
# after N rotations
K %= N
# Mth element after k left rotations
# is (K+M-1)%N th element of the
# original array
index = (K + M - 1) % N
result = a[index]
# Return the result
return result
# Driver Code
if __name__ == '__main__':
# Array initialization
a = [ 3, 4, 5, 23 ]
# Size of the array
N = len(a)
# Given K rotation and Mth element
# to be found after K rotation
K = 2
M = 1
# Function call
print(getFirstElement(a, N, K, M))
# This code is contributed by mohit kumar 29
C#
// C# program for the above approach
using System;
class GFG{
// Function to return Mth element of
// array after k left rotations
public static int getFirstElement(int[] a, int N,
int K, int M)
{
// The array comes to original state
// after N rotations
K %= N;
// Mth element after k left rotations
// is (K+M-1)%N th element of the
// original array
int index = (K + M - 1) % N;
int result = a[index];
// Return the result
return result;
}
// Driver code
public static void Main(string[] args)
{
// Array initialization
int []a = { 3, 4, 5, 23 };
// Size of the array
int N = a.Length;
// Given K rotation and Mth element
// to be found after K rotation
int K = 2, M = 1;
// Function call
Console.Write(getFirstElement(a, N, K, M));
}
}
// This code is contributed by rutvik_56
5
时间复杂度: O(1)
辅助空间: O(1)