Finding ‘k’ such that its modulus with each array element is same

Given an array of n integers .We need to find all ‘k’ such that 

arr[0] % k = arr[1] % k = ....... = arr[n-1] % k 

Examples:

Input  : arr[] = {6, 38, 34}
Output : 1 2 4
        6%1 = 38%1 = 34%1 = 0
        6%2 = 38%2 = 34%2 = 0
        6%4 = 38%4 = 34%2 = 2

Input  : arr[] = {3, 2}
Output : 1

Suppose the array contains only two elements a and b (b>a). So we can write b = a + d where d is a positive integer and ‘k’ be a number such that b%k = a%k. 

(a + d)%k = a%k
a%k + d%k = a%k 
d%k = 0

Now what we get from the above calculation is that ‘k’ should be a divisor of difference between the two numbers. 

Now what we have to do when we have an array of integers 

1. Find out the difference ‘d’ between maximum and minimum element of the array
2. Find out all the divisors of ‘d’
3. Step 3: For each divisor check if arr[i]%divisor(d) is same or not .if it is same print it.

Implementation:

C++

// C++ implementation of finding all k
// such that arr[i]%k is same for each i
#include<bits/stdc++.h>
using namespace std;

// Prints all k such that arr[i]%k is same for all i
void printEqualModNumbers (int arr[], int n)
{
    // sort the numbers
    sort(arr, arr + n);

    // max difference will be the difference between
    // first and last element of sorted array
    int d = arr[n-1] - arr[0];

    // Case when all the array elements are same
    if(d==0){
        cout<<"Infinite solution";
        return;
    }

    // Find all divisors of d and store in
    // a vector v[]
    vector <int> v;
    for (int i=1; i*i<=d; i++)
    {
        if (d%i == 0)
        {
            v.push_back(i);
            if (i != d/i)
                v.push_back(d/i);
        }
    }

    // check for each v[i] if its modulus with
    // each array element is same or not
    for (int i=0; i<v.size(); i++)
    {
        int temp = arr[0]%v[i];

        // checking for each array element if
        // its modulus with k is equal to k or not
        int j;
        for (j=1; j<n; j++)
            if (arr[j] % v[i] != temp)
                break;

        // if check is true print v[i]
        if (j == n)
            cout << v[i] <<" ";
    }
}

// Driver function
int main()
{
    int arr[] = {38, 6, 34};
    int n = sizeof(arr)/sizeof(arr[0]);
    printEqualModNumbers(arr, n);
    return 0;
}

Java

//  Java implementation of finding all k
// such that arr[i]%k is same for each i

import java.util.Arrays;
import java.util.Vector;

class Test
{
    // Prints all k such that arr[i]%k is same for all i
    static void printEqualModNumbers (int arr[], int n)
    {
        // sort the numbers
        Arrays.sort(arr);

        // max difference will be the difference between
        // first and last element of sorted array
        int d = arr[n-1] - arr[0];
        // Case when all the array elements are same
        if(d==0){
            System.out.println("Infinite solution");
            return;
        }
        // Find all divisors of d and store in
        // a vector v[]
        Vector<Integer> v = new Vector<>();
        for (int i=1; i*i<=d; i++)
        {
            if (d%i == 0)
            {
                v.add(i);
                if (i != d/i)
                    v.add(d/i);
            }
        }

        // check for each v[i] if its modulus with
        // each array element is same or not
        for (int i=0; i<v.size(); i++)
        {
            int temp = arr[0]%v.get(i);

            // checking for each array element if
            // its modulus with k is equal to k or not
            int j;
            for (j=1; j<n; j++)
                if (arr[j] % v.get(i) != temp)
                    break;

            // if check is true print v[i]
            if (j == n)
                System.out.print(v.get(i) + " ");
        }
    }

    // Driver method
    public static void main(String args[])
    {
        int arr[] = {38, 6, 34};

        printEqualModNumbers(arr, arr.length);
    }
}

Python3

# Python3 implementation of finding all k
# such that arr[i]%k is same for each i

# Prints all k such that arr[i]%k is
# same for all i
def printEqualModNumbers(arr, n):

    # sort the numbers
    arr.sort();

    # max difference will be the difference
    # between first and last element of
    # sorted array
    d = arr[n - 1] - arr[0];
    // Case when all the array elements are same
    if(d==0):
        print("Infinite solution")
        return

    # Find all divisors of d and store
    # in a vector v[]
    v = [];
    i = 1;
    while (i * i <= d):
        if (d % i == 0):
                v.append(i);
                if (i != d / i):
                    v.append(d / i);
        i += 1;

    # check for each v[i] if its modulus with
    # each array element is same or not
    for i in range(len(v)):
        temp = arr[0] % v[i];

        # checking for each array element if
        # its modulus with k is equal to k or not
        j = 1;
        while (j < n):
            if (arr[j] % v[i] != temp):
                break;
            j += 1;

        # if check is true print v[i]
        if (j == n):
            print(v[i], end = " ");

# Driver Code
arr = [38, 6, 34];
printEqualModNumbers(arr, len(arr));

# This code is contributed by mits

C#

// C# implementation of finding all k
// such that arr[i]%k is same for each i
using System;
using System.Collections;
class Test
{
    // Prints all k such that arr[i]%k is same for all i
    static void printEqualModNumbers (int []arr, int n)
    {
        // sort the numbers
        Array.Sort(arr);

        // max difference will be the difference between
        // first and last element of sorted array
        int d = arr[n-1] - arr[0];
        // Case when all the array elements are same
        if(d==0){
            Console.write("Infinite solution");
            return;
        }
        // Find all divisors of d and store in
        // a vector v[]
        ArrayList v = new ArrayList();
        for (int i=1; i*i<=d; i++)
        {
            if (d%i == 0)
            {
                v.Add(i);
                if (i != d/i)
                    v.Add(d/i);
            }
        }

        // check for each v[i] if its modulus with
        // each array element is same or not
        for (int i=0; i<v.Count; i++)
        {
            int temp = arr[0]%(int)v[i];

            // checking for each array element if
            // its modulus with k is equal to k or not
            int j;
            for (j=1; j<n; j++)
                if (arr[j] % (int)v[i] != temp)
                    break;

            // if check is true print v[i]
            if (j == n)
                Console.Write(v[i] + " ");
        }
    }

    // Driver method
    public static void Main()
    {
        int []arr = {38, 6, 34};

        printEqualModNumbers(arr, arr.Length);
    }
}
// This code is contributed by mits

PHP

<?php
// PHP implementation of finding all k
// such that arr[i]%k is same for each i

    // Prints all k such that arr[i]%k is same for all i
    function printEqualModNumbers ($arr, $n)
    {
        // sort the numbers
        sort($arr);

        // max difference will be the difference between
        // first and last element of sorted array
        $d = $arr[$n-1] - $arr[0];
        // Case when all the array elements are same
        if(d==0){
            print("Infinite solution");
            return;
        }
        // Find all divisors of d and store in
        // a vector v[]
        $v = array();
        for ($i=1; $i*$i<=$d; $i++)
        {
            if ($d%$i == 0)
            {
                array_push($v,$i);
                if ($i != $d/$i)
                    array_push($v,$d/$i);
            }
        }

        // check for each v[i] if its modulus with
        // each array element is same or not
        for ($i=0; $i<count($v); $i++)
        {
            $temp = $arr[0]%$v[$i];

            // checking for each array element if
            // its modulus with k is equal to k or not
            $j=1;
            for (; $j<$n; $j++)
                if ($arr[$j] % $v[$i] != $temp)
                    break;

            // if check is true print v[i]
            if ($j == $n)
                print($v[$i]." ");
        }
    }

    // Driver method

        $arr = array(38, 6, 34);

        printEqualModNumbers($arr, count($arr));

// This code is contributed by mits
?>

Javascript

<script>

// JavaScript implementation of finding all k
// such that arr[i]%k is same for each i

    // Prints all k such that arr[i]%k is same for all i
    function printEqualModNumbers (arr, n)
    {
        // sort the numbers
        arr.sort((a, b) => a - b);

        // max difference will be the difference between
        // first and last element of sorted array
        d = arr[n-1] - arr[0];
        // Case when all the array elements are same
        if(d==0){
            document.write("Infinite solution");
            return;
        }
        // Find all divisors of d and store in
        // a vector v[]
        v = new Array();
        for (i=1; i*i<=d; i++)
        {
            if (d%i == 0)
            {
                v.push(i);
                if (i != d/i)
                    v.push(d/i);
            }
        }

        // check for each v[i] if its modulus with
        // each array element is same or not
        for (i=0; i< v.length; i++)
        {
            temp = arr[0]%v[i];

            // checking for each array element if
            // its modulus with k is equal to k or not
            j=1;
            for (; j<n; j++)
                if (arr[j] % v[i] != temp)
                    break;

            // if check is true print v[i]
            if (j == n)
                document.write(v[i] + " ");
        }
    }

    // Driver method

        let arr = new Array(38, 6, 34);

        printEqualModNumbers(arr, arr.length);

// This code is contributed by _saurabh_jaiswal

</script>

Output

1 2 4 

Time Complexity: O(nlog(n))
Since the given array has to be sorted for the given problem, we use the sorting algorithm which takes O(nlog(n)) time.

Space Complexity: O(n)
We use a vector to store all the divisors of the difference of the first and the last element of the sorted array. This has a space complexity of O(n).