Finding ‘k’ such that its modulus with each array element is same
Last Updated :
12 Dec, 2024
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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
- Find out the absolute 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.
Implementation:
// 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 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 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# 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
<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>
<?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
?>
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).
