当每个元素被其余数替换为 K 时，最大化 K 以生成给定的数组 Palindrome

给定一个数组 A[]包含N个正整数，任务是找到可能的最大数K ，使得在将所有元素替换为模K(A[i]=A[i]%K, for all 0<=i< N)，数组变成回文。如果K无限大，则打印 -1。

例子：

Input: A={1, 2, 3, 4}, N=4
Output:
1
Explanation:
For K=1, A becomes {1%1, 2%1, 3%1, 4%1}={0, 0, 0, 0} which is a palindromic array. 

Input: A={1, 2, 3, 2, 1}, N=5
Output:
-1

观察：以下观察有助于解决问题：

1. 如果数组已经是回文数， K可以无限大。
2. 两个数，比如A和B ，可以通过取它们的模数和它们的差(|AB|)以及它们的差的因素来相等。

方法：问题可以通过使K等于GCD来解决 A[i]和A[Ni-1] 的绝对差值。请按照以下步骤解决问题：

1. 检查A是否已经是回文。如果是，则返回-1 。
2. 将数组的第一个和最后一个元素的绝对差存储在一个变量中，比如K ，它将存储将A变为回文所需的最大数。
3. 从1 遍历到 N/2-1 ，对于每个当前索引 i，执行以下操作：
   1. 用GCD更新K K和A[i]和A[Ni-1] 的绝对差。
4. 返回K 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// utility function to calculate the GCD of two numbers
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    else
        return gcd(b, a % b);
}
// Function to calculate the largest K, replacing all
// elements of an array A by their modulus with K, makes A a
// palindromic array
int largestK(int A[], int N)
{
    // check if A is palindrome
    int l = 0, r = N - 1, flag = 0;
    while (l < r) {
        // A is not palindromic
        if (A[l] != A[r]) {
            flag = 1;
            break;
        }
        l++;
        r--;
    }
    // K can be infitely large in this case
    if (flag == 0)
        return -1;
 
    // variable to store the largest K that makes A
    // palindromic
    int K = abs(A[0] - A[N - 1]);
    for (int i = 1; i < N / 2; i++)
        K = gcd(K, abs(A[i] - A[N - i - 1]));
    // return the required answer
    return K;
}
// Driver code
int main()
{
    // Input
    int A[] = { 1, 2, 3, 2, 1 };
    int N = sizeof(A) / sizeof(A[0]);
 
    // Function call
    cout << largestK(A, N) << endl;
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Utility function to calculate the GCD
// of two numbers
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
    else
        return gcd(b, a % b);
}
 
// Function to calculate the largest K,
// replacing all elements of an array A
// by their modulus with K, makes A a
// palindromic array
static int largestK(int A[], int N)
{
     
    // Check if A is palindrome
    int l = 0, r = N - 1, flag = 0;
    while (l < r)
    {
         
        // A is not palindromic
        if (A[l] != A[r])
        {
            flag = 1;
            break;
        }
        l++;
        r--;
    }
     
    // K can be infitely large in this case
    if (flag == 0)
        return -1;
 
    // Variable to store the largest K
    // that makes A palindromic
    int K = Math.abs(A[0] - A[N - 1]);
    for(int i = 1; i < N / 2; i++)
        K = gcd(K, Math.abs(A[i] - A[N - i - 1]));
         
    // Return the required answer
    return K;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Input
    int A[] = { 1, 2, 3, 2, 1 };
    int N = A.length;
     
    // Function call
    System.out.println(largestK(A, N));
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach
 
# utility function to calculate the GCD of two numbers
def gcd(a, b):
    if (b == 0):
        return a
    else:
        return gcd(b, a % b)
       
# Function to calculate the largest K, replacing all
# elements of an array A by their modulus with K, makes A a
# palindromic array
def largestK(A, N):
   
    # check if A is palindrome
    l,r,flag = 0, N - 1, 0
    while (l < r):
        # A is not palindromic
        if (A[l] != A[r]):
            flag = 1
            break
        l += 1
        r -= 1
    # K can be infitely large in this case
    if (flag == 0):
        return -1
 
    # variable to store the largest K that makes A
    # palindromic
    K = abs(A[0] - A[N - 1])
    for i in range(1,N//2):
        K = gcd(K, abs(A[i] - A[N - i - 1]))
     
    # return the required answer
    return K
   
# Driver code
if __name__ == '__main__':
    # Input
    A= [1, 2, 3, 2, 1 ]
    N = len(A)
 
    # Function call
    print (largestK(A, N))
 
# This code is contributed by mohit kumar 29.

C#

// c# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// utility function to calculate the GCD of two numbers
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
    else
        return gcd(b, a % b);
}
   
// Function to calculate the largest K, replacing all
// elements of an array A by their modulus with K, makes A a
// palindromic array
static int largestK(int []A, int N)
{
   
    // check if A is palindrome
    int l = 0, r = N - 1, flag = 0;
    while (l < r) {
        // A is not palindromic
        if (A[l] != A[r]) {
            flag = 1;
            break;
        }
        l++;
        r--;
    }
   
    // K can be infitely large in this case
    if (flag == 0)
        return -1;
 
    // variable to store the largest K that makes A
    // palindromic
    int K = Math.Abs(A[0] - A[N - 1]);
    for (int i = 1; i < N / 2; i++)
        K = gcd(K, Math.Abs(A[i] - A[N - i - 1]));
   
    // return the required answer
    return K;
}
 
// Driver code
public static void Main()
{
   
    // Input
    int []A = { 1, 2, 3, 2, 1 };
    int N = A.Length;
 
    // Function call
    Console.Write(largestK(A, N));
}
}
 
// This code is contributed by ipg2016107.

Javascript

 
// Javascript program for the above approach
 
// utility function to calculate the
// GCD of two numbers
function gcd(a, b)
{
    if (b == 0)
        return a;
    else
        return gcd(b, a % b);
}
 
// Function to calculate the largest
// K, replacing all elements of an
// array A by their modulus with K,
// makes A a palindromic array
function largestK(A, N)
{
     
    // Check if A is palindrome
    let l = 0, r = N - 1, flag = 0;
     
    while (l < r)
    {
         
        // A is not palindromic
        if (A[l] != A[r])
        {
            flag = 1;
            break;
        }
        l++;
        r--;
    }
     
    // K can be infitely large in this case
    if (flag == 0)
        return -1;
 
    // Variable to store the largest K
    // that makes A palindromic
    let K = Math.abs(A[0] - A[N - 1]);
    for(let i = 1; i < N / 2; i++)
        K = gcd(K, Math.abs(A[i] - A[N - i - 1]));
         
    // Return the required answer
    return K;
}
 
// Driver code
 
// Input
let A = [ 1, 2, 3, 2, 1 ];
let N = A.length;
 
// Function call
document.write(largestK(A, N) + "<br>");
 
// This code is contributed by gfgking
 

输出

-1

时间复杂度： O(NLogM)，其中 M 是数组中的最大元素
辅助空间： O(1)