# 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 Practice

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` `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

## 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 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

## 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

## PHP

 ``

## Javascript

 ``

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).

If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next