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Maximize K to make given array Palindrome when each element is replaced by its remainder with K

  • Last Updated : 12 Jul, 2021

Given an array A[] containing N positive integers, the task is to find the largest possible number K, such that after replacing all elements by the elements modulo K(A[i]=A[i]%K, for all 0<=i<N), the array becomes a palindrome. If K is infinitely large, print -1.

Examples:

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

 

Observation: The following observations help in solving the problem:



  1. If the array is already a palindrome, K can be infinitely large.
  2. Two numbers, say A and B can be made equal by taking their modulus with their difference(|A-B|) as well as the factors of their difference.

Approach: The problem can be solved by making K equal to the GCD of the absolute differences of A[i] and A[N-i-1]. Follow the steps below to solve the problem:

  1. Check whether A is already a palindrome. If it is, return -1.
  2. Store the absolute difference of the first and last elements of the array in a variable, say K, which will store the largest number required to change A into a palindrome.
  3. Traverse from 1 to N/2-1, and for each current index i, do the following:
    1. Update K with the GCD of K and the absolute difference of A[i] and A[N-i-1].
  4. Return K.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
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




<script>
 
// 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
 
</script>
Output
-1

Time Complexity: O(NLogM), where M is the largest element in the array
Auxiliary Space: O(1)

 

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