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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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

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

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]; 
  
    // 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]; 
       
        // 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];  
      
    # 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];  
      
        // 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];  
      
        // 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 
?> 

Output:

1 2 4

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My Personal Notes

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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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