通过旋转最大化给定数组中相应相同元素的计数

给定两个包含N 个整数的数组arr1[]和arr2[]并且数组arr1[]具有不同的元素。任务是通过对数组arr1[]执行循环左移或右移来找到给定数组中对应相同元素的最大计数。
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：这个问题可以使用贪心方法来解决。以下是步骤：

1. 将数组arr2[]的所有元素的位置存储在一个数组中(比如store[] )。
2. 对于数组arr1[]中的每个元素，执行以下操作：
   • 找出当前元素在arr2[]中的位置与arr1[] 中的位置之间的差异(比如diff )。
   • 如果diff小于 0，则将 diff 更新为(N – diff) 。
   • 将当前差异diff的频率存储在地图中。
3. 经过以上步骤，map中存储的最大频率就是在arr1[]上旋转后相等元素的最大个数。

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 sapnasingh4991

Python3

# 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 Singh

C#

// 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

Javascript

 
// JavaScript program of the above approach
 
// Function that prlets maximum
// equal elements
function maximumEqual(a, b, n)
{
   
    // Vector to store the index
    // of elements of array b
    let store = Array.from({length: 1e5}, (_, i) => 0); 
   
    // Storing the positions of
    // array B
    for (let i = 0; i < n; i++)
    {
        store[b[i]] = i + 1;
    }
   
    // frequency array to keep count
    // of elements with similar
    // difference in distances
    let ans = Array.from({length: 1e5}, (_, i) => 0); 
   
    // Iterate through all element in arr1[]
    for (let i = 0; i < n; i++)
    {
   
        // Calculate number of
        // shift required to
        // make current element
        // equal
        let 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]++;
    }
   
    let finalans = 0;
   
    // Compute the maximum frequency
    // stored
    for (let i = 0; i < 1e5; i++)
        finalans = Math.max(finalans,
                            ans[i]);
   
    // Printing the maximum number
    // of equal elements
    document.write(finalans + "\n");
}
 
// Driver Code
 
    // Given two arrays
    let A = [ 6, 7, 3, 9, 5 ];
    let B = [ 7, 3, 9, 5, 6 ];
   
    let size = A.length;
   
    // Function Call
    maximumEqual(A, B, size);
           

输出：

5

时间复杂度： O(N)
辅助空间： O(N)

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