给定两个N个整数的数组arr1 []和arr2 [] ，数组arr1 []具有不同的元素。任务是通过对数组arr1 []进行循环左移或右移，找到给定数组中相应相同元素的最大数量。
例子：
Input: arr1[] = { 6, 7, 3, 9, 5 }, arr2[] = { 7, 3, 9, 5, 6 }
Output: 5
Explanation:
By performing cyclic left shift on array arr1[] by 1.
Updated array arr1[] = {7, 3, 9, 5, 6}.
This rotation contains a maximum number of equal elements between array arr1[] and arr2[].
Input: arr1[] = {1, 3, 2, 4}, arr2[] = {4, 2, 3, 1}
Output: 2
Explanation:
By performing cyclic left shift on array arr1[] by 1.
Updated array arr1[] = {3, 2, 4, 1}
This rotation contains a maximum number of equal elements between array arr1[] and arr2[].
方法：此问题可以使用贪婪方法解决。步骤如下：
- 将数组arr2 []的所有元素的位置存储在数组中(例如store [] )。
- 对于数组arr1 []中的每个元素，请执行以下操作：
- 求出当前元素在arr2 []中的位置与arr1 []中的位置之间的差异(例如diff )。
- 如果diff小于0，则将diff更新为(N – diff) 。
- 将电流差diff的频率存储在映射中。
- 完成上述步骤后，映射中存储的最大频率是在arr1 []上旋转之后的相等元素的最大数量。
下面是上述方法的实现：
C++
// C++ program of the above approach
#include
using namespace std;
// Function that prints maximum
// equal elements
void maximumEqual(int a[], int b[],
int n)
{
// Vector to store the index
// of elements of array b
vector store(1e5);
// Storing the positions of
// array B
for (int i = 0; i < n; i++) {
store[b[i]] = i + 1;
}
// frequency array to keep count
// of elements with similar
// difference in distances
vector ans(1e5);
// Iterate through all element in arr1[]
for (int i = 0; i < n; i++) {
// Calculate number of
// shift required to
// make current element
// equal
int d = abs(store[a[i]]
- (i + 1));
// If d is less than 0
if (store[a[i]] < i + 1) {
d = n - d;
}
// Store the frequency
// of current diff
ans[d]++;
}
int finalans = 0;
// Compute the maximum frequency
// stored
for (int i = 0; i < 1e5; i++)
finalans = max(finalans,
ans[i]);
// Printing the maximum number
// of equal elements
cout << finalans << "\n";
}
// Driver Code
int main()
{
// Given two arrays
int A[] = { 6, 7, 3, 9, 5 };
int B[] = { 7, 3, 9, 5, 6 };
int size = sizeof(A) / sizeof(A[0]);
// Function Call
maximumEqual(A, B, size);
return 0;
} Java
// Java program of the above approach
import java.util.*;
class GFG{
// Function that prints maximum
// equal elements
static void maximumEqual(int a[],
int b[], int n)
{
// Vector to store the index
// of elements of array b
int store[] = new int[(int) 1e5];
// Storing the positions of
// array B
for (int i = 0; i < n; i++)
{
store[b[i]] = i + 1;
}
// frequency array to keep count
// of elements with similar
// difference in distances
int ans[] = new int[(int) 1e5];
// Iterate through all element in arr1[]
for (int i = 0; i < n; i++)
{
// Calculate number of
// shift required to
// make current element
// equal
int d = Math.abs(store[a[i]] - (i + 1));
// If d is less than 0
if (store[a[i]] < i + 1)
{
d = n - d;
}
// Store the frequency
// of current diff
ans[d]++;
}
int finalans = 0;
// Compute the maximum frequency
// stored
for (int i = 0; i < 1e5; i++)
finalans = Math.max(finalans,
ans[i]);
// Printing the maximum number
// of equal elements
System.out.print(finalans + "\n");
}
// Driver Code
public static void main(String[] args)
{
// Given two arrays
int A[] = { 6, 7, 3, 9, 5 };
int B[] = { 7, 3, 9, 5, 6 };
int size = A.length;
// Function Call
maximumEqual(A, B, size);
}
}
// This code is contributed by sapnasingh4991Python3
# Python3 program for the above approach
# Function that prints maximum
# equal elements
def maximumEqual(a, b, n):
# List to store the index
# of elements of array b
store = [0] * 10 ** 5
# Storing the positions of
# array B
for i in range(n):
store[b[i]] = i + 1
# Frequency array to keep count
# of elements with similar
# difference in distances
ans = [0] * 10 ** 5
# Iterate through all element
# in arr1[]
for i in range(n):
# Calculate number of shift
# required to make current
# element equal
d = abs(store[a[i]] - (i + 1))
# If d is less than 0
if (store[a[i]] < i + 1):
d = n - d
# Store the frequency
# of current diff
ans[d] += 1
finalans = 0
# Compute the maximum frequency
# stored
for i in range(10 ** 5):
finalans = max(finalans, ans[i])
# Printing the maximum number
# of equal elements
print(finalans)
# Driver Code
if __name__ == '__main__':
# Given two arrays
A = [ 6, 7, 3, 9, 5 ]
B = [ 7, 3, 9, 5, 6 ]
size = len(A)
# Function Call
maximumEqual(A, B, size)
# This code is contributed by Shivam SinghC#
// C# program of the above approach
using System;
class GFG{
// Function that prints maximum
// equal elements
static void maximumEqual(int[] a,
int[] b, int n)
{
// Vector to store the index
// of elements of array b
int[] store = new int[(int) 1e5];
// Storing the positions of
// array B
for(int i = 0; i < n; i++)
{
store[b[i]] = i + 1;
}
// Frequency array to keep count
// of elements with similar
// difference in distances
int[] ans = new int[(int) 1e5];
// Iterate through all element in arr1[]
for(int i = 0; i < n; i++)
{
// Calculate number of
// shift required to
// make current element
// equal
int d = Math.Abs(store[a[i]] - (i + 1));
// If d is less than 0
if (store[a[i]] < i + 1)
{
d = n - d;
}
// Store the frequency
// of current diff
ans[d]++;
}
int finalans = 0;
// Compute the maximum frequency
// stored
for(int i = 0; i < 1e5; i++)
finalans = Math.Max(finalans, ans[i]);
// Printing the maximum number
// of equal elements
Console.Write(finalans + "\n");
}
// Driver Code
public static void Main()
{
// Given two arrays
int[]A = { 6, 7, 3, 9, 5 };
int[]B = { 7, 3, 9, 5, 6 };
int size = A.Length;
// Function Call
maximumEqual(A, B, size);
}
}
// This code is contributed by chitranayal输出：
5
时间复杂度： O(N)
辅助空间： O(N)