Armstrong Number of Order N

 Coding Problem Keys 

 Armstrong Number of Order N 

 Problem Statement 

Consider a non-empty string instr. Identify and print a string outstr using the below mentioned logic.

• Find the sum of ASCII value of all the characters present in the instr.
• If sum is an Armstrong number of order n, remove duplicate characters present in the string instr and assign the updated instr to outstr. Perform case-sensitive comparison of characters while removing duplicate characters
• A number with digit is said to be an Armstrong number of order n if the sum of its digits each raised to the power of n is equal to the number i.e. abcd = a⁴ + b⁴ + c⁴ + d⁴.
• If the sum is not an Armstrong number of order n, print -1.

 Input Format 

First line contains instr.

Read input from the standard input stream.

 Output Format 

Print outstr or -1 accordingly to the standard output stream.

 Example Input/Output 1 

 Input 

BANaA

 Output 

BANa

 Explanation 

For the given input, instr is 'BANaA'.

The sum of ASCII values of all the characters present in instr is 66 + 65 + 78 + 97 + 65 sum i.e. 371 which has 3 digits.

The sum i.e. 371, is an Armstrong number of order n i.e. 3. So remove the duplicate character in instr i.e. 'A', and assign the updated instr, i.e. 'BANa' to outstr.

 Example Input/Output 2 

 Input 

Van

 Output 

-1

 Explanation 

For the given input, instr is 'Van'.

The sum of ASCII value of all the characters present in instr is 86 + 97 + 110 = 293 which has 3 digits.

The sum i.e. 293, is not an Armstrong number of order n i.e. 3.

Hence the output -1.

 Solution 

 Programming Language: Python 3 Language 

import math
l = input()
s = 0
c = 0
for i in l:
    s+=ord(i)
for j in str(s):
    c+=pow(int(j),len(str(s)))
res = []
flag = False
if c==s:
    res = list(dict.fromkeys(l))
    flag = True
if flag:
    for i in res:
        print(i,end="")
else:
    print(-1)

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 Alter 

l=input()
s=0;c=0;r=0;a=""
for x in l:
    s=s+ord(x)
n=s
while n>0:
    n=n//10
    c=c+1
n=s
while n>0:
    d=n%10
    n=n//10
    r=r+d**c
if s==r:
    for x in w:
        if x not in a:
            a=a+x
    print(a)
else:
    print(-1)

# Published By PKJCODERS

 (Note: Incase If the code doesn't Pass the output kindly comment us with your feedback to help us improvise.)