Given an integer n, return whether it is equal to the sum of its own digits raised to the power of the number of digits.
Example 1
Input
n = 153
Output
True
Explanation
153 = 1 ** 3 + 5 ** 3 + 3 ** 3
Check Narcissistic Number in Python
The key part is to obtain the length of the number (e.g. the number of digits): We can either do this by converting it to a string, or iteratively divide by 10:
Converting a number to string, then compute the sum of powers. Once we convert the number into a string, we can easily iterate the digits:
1 2 3 4 5 6 7 | class Solution: def isNarcissisticNumber(self, n): l = len(str(n)) a = 0 for i in str(n): a += int(i) ** l return a == n |
class Solution:
def isNarcissisticNumber(self, n):
l = len(str(n))
a = 0
for i in str(n):
a += int(i) ** l
return a == nAnd below is the log(N) method to compute the length of a number then compute the sum of powers:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 | class Solution: def isNarcissisticNumber(self, n): def getLen(n): if n == 0: return 1 l = 0 while n > 0: l += 1 n //= 10 return l l = getLen(n) a = 0 for i in str(n): a += int(i) ** l return a == n |
class Solution:
def isNarcissisticNumber(self, n):
def getLen(n):
if n == 0:
return 1
l = 0
while n > 0:
l += 1
n //= 10
return l
l = getLen(n)
a = 0
for i in str(n):
a += int(i) ** l
return a == nCheck Narcissistic Number in C++
we can also get the length by using the 
function. And compute the sum of powers by using the pow function.
1 2 3 4 5 6 7 8 9 10 11 | bool isNarcissisticNumber(int n) { using ll = long long; ll sum = 0; int len = ceil(log10(n+1)); int m = n; while (m) { sum += pow(m%10, len); m /= 10; } return sum == n; } |
bool isNarcissisticNumber(int n) {
using ll = long long;
ll sum = 0;
int len = ceil(log10(n+1));
int m = n;
while (m) {
sum += pow(m%10, len);
m /= 10;
}
return sum == n;
}–EOF (The Ultimate Computing & Technology Blog) —
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