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Minimum possible value of D which when added to or subtracted from K repeatedly obtains every array element

  • Last Updated : 06 Jul, 2021

Given an array arr[] of size N and an integer K, the task is to find the maximum possible value of D, such that every array element can be obtained, starting from the initial value of K, by either changing K to K – D or K + D at each step.

Examples:

Input: arr[ ] = {1, 7, 11}, K = 3
Output: 2
Explanation:
Considering the value of D to be 2, every array element can be obtained by the following operations:

  • arr[0](= 1):  Decrementing 2 from K(=3) obtains arr[0].
  • arr[1](= 7): Incrementing K(=3) by 2 times D obtains arr[1].
  • arr[2](= 11): Incrementing K(=3) by 4 times D obtains arr[2].

Therefore, D (=2) satisfies the conditions. Also, it is the maximum possible value of D.

Input: arr[ ] = {33, 105, 57}, K = 81
Output: 24



Approach: The problem can be solved by finding the Greatest Common Divisor of the absolute difference between each array element and K. Follow the steps below to solve the problem

  • Traverse the array arr[], and change the value of the current element arr[i] to abs(arr[i] – K).
  • Initialize a variable, say D as arr[0], to store the result.
  • Iterate in the range [1, N – 1] using a variable, say i, and in each iteration, update the value of D to gcd(D, arr[i]).
  • After completing the above steps, print the value of D as the answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Recursive function tox
// previous gcd of a and b
int gcd(int a, int b)
{
    if (b == 0)
        return a;
 
    return gcd(b, a % b);
}
 
// Function to find the maximum value
// of D such that every element in
// the array can be obtained by
// performing K + D or K - D
int findMaxD(int arr[], int N, int K)
{
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
 
        // Update arr[i]
        arr[i] = abs(arr[i] - K);
    }
 
    // Stores GCD of the array
    int D = arr[0];
 
    // Iterate over the range [1, N]
    for (int i = 1; i < N; i++) {
 
        // Update the value of D
        D = gcd(D, arr[i]);
    }
 
    // Print the value of D
    return D;
}
 
// Driver Code
int main()
{
 
    int arr[] = { 1, 7, 11 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 3;
 
    cout << findMaxD(arr, N, K);
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Recursive function tox
// previous gcd of a and b
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
 
    return gcd(b, a % b);
}
 
// Function to find the maximum value
// of D such that every element in
// the array can be obtained by
// performing K + D or K - D
static int findMaxD(int arr[], int N, int K)
{
     
    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
         
        // Update arr[i]
        arr[i] = Math.abs(arr[i] - K);
    }
 
    // Stores GCD of the array
    int D = arr[0];
 
    // Iterate over the range [1, N]
    for(int i = 1; i < N; i++)
    {
         
        // Update the value of D
        D = gcd(D, arr[i]);
    }
 
    // Print the value of D
    return D;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 7, 11 };
    int N = arr.length;
    int K = 3;
 
    System.out.print(findMaxD(arr, N, K));
}
}
 
// This code is contributed by rishavmahato348

Python3




# // python program for the above approach
 
# // Recursive function tox
# // previous gcd of a and b
def gcd(a, b):
    if (b == 0):
        return a
    return gcd(b, a % b)
 
# // Function to find the maximum value
# // of D such that every element in
# // the array can be obtained by
# // performing K + D or K - D
def findMaxD(arr, N, K):
   
    # // Traverse the array arr[]
    for i in range(0, N):
       
        # // Update arr[i]
        arr[i] = abs(arr[i] - K)
 
    # // Stores GCD of the array
    D = arr[0]
 
    # // Iterate over the range[1, N]
    for i in range(1, N):
       
        # // Update the value of D
        D = gcd(D, arr[i])
 
    # // Print the value of D
    return D
 
# // Driver Code
arr = [1, 7, 11]
N = len(arr)
K = 3
print(findMaxD(arr, N, K))
 
# This code is contributed by amreshkumar3

C#




// C# program for the above approach
using System;
 
class GFG {
 
    // Recursive function tox
    // previous gcd of a and b
    static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
 
        return gcd(b, a % b);
    }
 
    // Function to find the maximum value
    // of D such that every element in
    // the array can be obtained by
    // performing K + D or K - D
    static int findMaxD(int[] arr, int N, int K)
    {
 
        // Traverse the array arr[]
        for (int i = 0; i < N; i++) {
 
            // Update arr[i]
            arr[i] = Math.Abs(arr[i] - K);
        }
 
        // Stores GCD of the array
        int D = arr[0];
 
        // Iterate over the range [1, N]
        for (int i = 1; i < N; i++) {
 
            // Update the value of D
            D = gcd(D, arr[i]);
        }
 
        // Print the value of D
        return D;
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 1, 7, 11 };
        int N = arr.Length;
        int K = 3;
 
        Console.Write(findMaxD(arr, N, K));
    }
}
 
// This code is contributed by subhammahato348.

Javascript




<script>
// Javascript program for the above approach
 
// Recursive function tox
// previous gcd of a and b
function gcd(a, b) {
    if (b == 0)
        return a;
 
    return gcd(b, a % b);
}
 
// Function to find the maximum value
// of D such that every element in
// the array can be obtained by
// performing K + D or K - D
function findMaxD(arr, N, K) {
    // Traverse the array arr[]
    for (let i = 0; i < N; i++) {
 
        // Update arr[i]
        arr[i] = Math.abs(arr[i] - K);
    }
 
    // Stores GCD of the array
    let D = arr[0];
 
    // Iterate over the range [1, N]
    for (let i = 1; i < N; i++) {
 
        // Update the value of D
        D = gcd(D, arr[i]);
    }
 
    // Print the value of D
    return D;
}
 
// Driver Code
 
 
let arr = [1, 7, 11];
let N = arr.length;
let K = 3;
 
document.write(findMaxD(arr, N, K));
 
// This code is contributed by _saurabh_jaiswal.
</script>
Output
2

Time Complexity: O(N*log(N))
Auxiliary Space: O(1)

 

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