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
- 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]; // 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 ?> |
Output:
1 2 4
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