Smallest Multiple in Permuted digits | Codevita Solution |Python

Problem:

 Given two integers N,d,find the smallest number that number that is a multiple of d that could be formed by permuting the digits of N .You must use all the digits of N.and if the smallest multiple of d has leading zeros,they can be dropped .If no such number exists,output -1.

Input:

A line containing two space separated integers.representing N and d.

Output:

A single line giving the permutation of N that is the smallest multiple of d,without any leading zeros,if any.If not such permutation exists, the output should be -1.

Constraints:

1<=N<=10^12

1<=d<=1000000

Example 1:

Input:

210 2

Output:

12

Example 2:

Input:

531 2

Output:

1

Understanding of Question:

    We need to find out the minimum number that are divisible by given divisor from the given number.

Pseudocode:

•     Get the input of number and the divisor.
•     Extract the digits and store it in a array.
•     Sort the digits
•     least multiple=-1
•     Find the permutations using import itertools
•     Convert the array into integer.
•     Traverse the array and check if it is divisible by d.
•    Print least multiple.

Program:   

	from itertools import permutations
	def allPer(string):
	permlist=permutations(string)
	for i in permlist:
	k=int("".join(i))
	if k%d==0:
	t.append(k)
	return t
	if __name__ == "__main__":
	num,d=map(int,input().split())
	num=str(num)
	t=[]
	num=list(num)
	num.sort()
	least_multiple=-1
	allPer(num)
	t=[int(i) for i in t ]
	for i in t:
	if i%d==0:
	least_multiple=i
	break
	print(least_multiple)

view raw SmallestMultipleInPermutedDigits.py hosted with ❤ by GitHub