Rearrangement of a number which is also divisible by it
Given a number n, we need to rearrange all its digits such that the new arrangement is divisible by n. Also, the new number should not be equal to x. If no such rearrangement is possible, print -1.
Examples:
Input : n = 1035 Output : 3105 The result 3105 is divisible by given n and has the same set of digits. Input : n = 1782 Output : m = 7128
Simple Approach : Find all the permutation of given n and then check whether it is divisible by n or not also check that new permutation should not be equal to n.
Efficient Approach : Let’s suppose that y is our result then y = m * n, also we know that y must be a rearrangement of digits of n so we can say now restrict m (the multiplier) as per given conditions.
1) y has the same number of digits as n has. So, m must be less than 10.
2) y must not be equal to n. So, m will be greater than 1.
So we get the multiplier m in the range [2,9]. So we will find all the possible y and then check that should y has the same digits as n or not.
C++
// CPP program for finding rearrangement of n// that is divisible by n#include<bits/stdc++.h>using namespace std; // perform hashing for given nvoid storeDigitCounts(int n, vector<int> &hash){ // perform hashing while (n) { hash[n%10]++; n /= 10; }} // check whether any arrangement existsint rearrange (int n){ // Create a hash for given number n // The hash is of size 10 and stores // count of each digit in n. vector<int> hash_n(10, 0); storeDigitCounts(n, hash_n); // check for all possible multipliers for (int mult=2; mult<10; mult++) { int curr = n*mult; vector<int> hash_curr(10, 0); storeDigitCounts(curr, hash_curr); // check hash table for both. // Please refer below link for help // of equal() if (equal(hash_n.begin(), hash_n.end(), hash_curr.begin())) return curr; } return -1;} // driver programint main(){ int n = 10035; cout << rearrange(n); return 0;} |
Python3
# Python3 program for finding rearrangement # of n that is divisible by n # Perform hashing for given n def storeDigitCounts(n, Hash): # perform hashing while n > 0: Hash[n % 10] += 1 n //= 10 # check whether any arrangement exists def rearrange(n): # Create a hash for given number n # The hash is of size 10 and stores # count of each digit in n. hash_n = [0] * 10 storeDigitCounts(n, hash_n) # check for all possible multipliers for mult in range(2, 10): curr = n * mult hash_curr = [0] * 10 storeDigitCounts(curr, hash_curr) # check hash table for both. if hash_n == hash_curr: return curr return -1 # Driver Codeif __name__ == "__main__": n = 10035 print(rearrange(n)) # This code is contributed by Rituraj Jain |
Output:
30105
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